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Why can I use a list index as an indexing variable in a for loop?



The 2019 Stack Overflow Developer Survey Results Are InHow do I check if a list is empty?Finding the index of an item given a list containing it in PythonAccessing the index in 'for' loops?Understanding Python super() with __init__() methodsHow do I sort a dictionary by value?How to make a flat list out of list of lists?How do I pass a variable by reference?Loop through an array in JavaScriptHow do I list all files of a directory?Why is printing “B” dramatically slower than printing “#”?



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12















I have the following code:



a = [0,1,2,3]

for a[-1] in a:
print(a[-1])


The output is:



0
1
2
2


I'm confused about why a list index can be used as an indexing variable in a for a loop?










share|improve this question









New contributor




Kundan Verma is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.















  • 7





    Somehow this question looks like a bad and newbie question, but I don't get the logic either lol

    – gameon67
    1 hour ago






  • 4





    I don't know why you would ever want to do this, but now I know you can

    – Nathan
    1 hour ago






  • 8





    This would be a great question for an awful coding interview

    – Nathan
    1 hour ago






  • 1





    OP, since it seems like your post may be gaining some traffic, I edited your post in an attempt to make your question clearer. If I'm misinterpreting your question, you can rollback the edit.

    – Christian Dean
    40 mins ago






  • 1





    This question first got two down votes :) I guess you shouldn't panic by down votes!

    – Amir A. Shabani
    19 mins ago

















12















I have the following code:



a = [0,1,2,3]

for a[-1] in a:
print(a[-1])


The output is:



0
1
2
2


I'm confused about why a list index can be used as an indexing variable in a for a loop?










share|improve this question









New contributor




Kundan Verma is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.















  • 7





    Somehow this question looks like a bad and newbie question, but I don't get the logic either lol

    – gameon67
    1 hour ago






  • 4





    I don't know why you would ever want to do this, but now I know you can

    – Nathan
    1 hour ago






  • 8





    This would be a great question for an awful coding interview

    – Nathan
    1 hour ago






  • 1





    OP, since it seems like your post may be gaining some traffic, I edited your post in an attempt to make your question clearer. If I'm misinterpreting your question, you can rollback the edit.

    – Christian Dean
    40 mins ago






  • 1





    This question first got two down votes :) I guess you shouldn't panic by down votes!

    – Amir A. Shabani
    19 mins ago













12












12








12


7






I have the following code:



a = [0,1,2,3]

for a[-1] in a:
print(a[-1])


The output is:



0
1
2
2


I'm confused about why a list index can be used as an indexing variable in a for a loop?










share|improve this question









New contributor




Kundan Verma is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












I have the following code:



a = [0,1,2,3]

for a[-1] in a:
print(a[-1])


The output is:



0
1
2
2


I'm confused about why a list index can be used as an indexing variable in a for a loop?







python for-loop






share|improve this question









New contributor




Kundan Verma is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question









New contributor




Kundan Verma is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question








edited 32 mins ago









Amir A. Shabani

457616




457616






New contributor




Kundan Verma is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 1 hour ago









Kundan VermaKundan Verma

712




712




New contributor




Kundan Verma is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Kundan Verma is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Kundan Verma is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







  • 7





    Somehow this question looks like a bad and newbie question, but I don't get the logic either lol

    – gameon67
    1 hour ago






  • 4





    I don't know why you would ever want to do this, but now I know you can

    – Nathan
    1 hour ago






  • 8





    This would be a great question for an awful coding interview

    – Nathan
    1 hour ago






  • 1





    OP, since it seems like your post may be gaining some traffic, I edited your post in an attempt to make your question clearer. If I'm misinterpreting your question, you can rollback the edit.

    – Christian Dean
    40 mins ago






  • 1





    This question first got two down votes :) I guess you shouldn't panic by down votes!

    – Amir A. Shabani
    19 mins ago












  • 7





    Somehow this question looks like a bad and newbie question, but I don't get the logic either lol

    – gameon67
    1 hour ago






  • 4





    I don't know why you would ever want to do this, but now I know you can

    – Nathan
    1 hour ago






  • 8





    This would be a great question for an awful coding interview

    – Nathan
    1 hour ago






  • 1





    OP, since it seems like your post may be gaining some traffic, I edited your post in an attempt to make your question clearer. If I'm misinterpreting your question, you can rollback the edit.

    – Christian Dean
    40 mins ago






  • 1





    This question first got two down votes :) I guess you shouldn't panic by down votes!

    – Amir A. Shabani
    19 mins ago







7




7





Somehow this question looks like a bad and newbie question, but I don't get the logic either lol

– gameon67
1 hour ago





Somehow this question looks like a bad and newbie question, but I don't get the logic either lol

– gameon67
1 hour ago




4




4





I don't know why you would ever want to do this, but now I know you can

– Nathan
1 hour ago





I don't know why you would ever want to do this, but now I know you can

– Nathan
1 hour ago




8




8





This would be a great question for an awful coding interview

– Nathan
1 hour ago





This would be a great question for an awful coding interview

– Nathan
1 hour ago




1




1





OP, since it seems like your post may be gaining some traffic, I edited your post in an attempt to make your question clearer. If I'm misinterpreting your question, you can rollback the edit.

– Christian Dean
40 mins ago





OP, since it seems like your post may be gaining some traffic, I edited your post in an attempt to make your question clearer. If I'm misinterpreting your question, you can rollback the edit.

– Christian Dean
40 mins ago




1




1





This question first got two down votes :) I guess you shouldn't panic by down votes!

– Amir A. Shabani
19 mins ago





This question first got two down votes :) I guess you shouldn't panic by down votes!

– Amir A. Shabani
19 mins ago












5 Answers
5






active

oldest

votes


















10














To bring a language-lawyer aspect to the question, let's look at documentation:




for_stmt ::= "for" target_list "in" expression_list ":" suite


The expression list is evaluated once; it should yield an iterable object. An iterator is created for the result of the expression_list. The suite is then executed once for each item provided by the iterator, in the order returned by the iterator. Each item in turn is assigned to the target list using the standard rules for assignments (see Assignment statements), and then the suite is executed.




(emphasis not originally in docs)
The suite refers to the statements under the for-block, print(a[-1]) in our particular case.



Simply, on each iteration, the loop variable (target_list) gets assigned to the next item in the iterable (expression_list).



Let's extend the print statement:



a = [0, 1, 2, 3]
for a[-1] in a:
print(a, a[-1])


This gives the following output:



[0, 1, 2, 0] 0 # a[-1] assigned 0
[0, 1, 2, 1] 1 # a[-1] assigned 1
[0, 1, 2, 2] 2 # a[-1] assigned 2
[0, 1, 2, 2] 2 # a[-1] assigned itself (2)


As a[-1] is a valid form left-hand side, assignments to a[-1] will mutate a, modifying the list during iteration. In this particular example, a[-1] retains -2 from the previous evaluation. If we trace the assignment on each iteration, we have



a[-1] = 0 # a[0] # a[3] is 0 now
a[-1] = 1 # a[1] # a[3] is 1 now
a[-1] = 2 # a[2] # a[3] is 2 now
a[-1] = 2 # a[3] # a[3] is itself (2)





share|improve this answer




















  • 2





    Might be useful to bold Each item [...] is assigned to the target list using the standard rules for assignments.

    – Mateen Ulhaq
    57 mins ago







  • 2





    Essentially, the important thing to note here is that Python considers a[-1] to be a valid form of the left-hand side of an assignment statement (e.g. a[-1] = 1 is valid grammar). Thus a[-1] is a valid "variable" name, because as the documentation stated, it evaluates the binding variable(s) in a for loop declaration as it would the left-hand side of an assignment.

    – Christian Dean
    45 mins ago


















2














it is an interesting question, you can understand it by that:



for v in a:
a[-1] = v
print(a[-1])

print(a)


actually a becomes: [0, 1, 2, 2] after loop



output:



0
1
2
2
[0, 1, 2, 2]


Hope that helps you, and comment if you have further questions. : )






share|improve this answer






























    2














    (this is more of a long comment than an answer - there are a couple of good ones already, especially @TrebledJ's. But I had to think of it explicitly in terms of overwriting variables that already have values before it clicked for me.)



    If you had



    x = 0
    l = [1, 2, 3]
    for x in l:
    print(x)


    you wouldn't be surprised that x is overriden each time through the loop. Even though x existed before, its value isn't used (i.e. for 0 in l:, which would throw an error). Rather, we assign the values from l to x.



    When we do



    a = [0, 1, 2, 3]

    for a[-1] in a:
    print(a[-1])


    even though a[-1] already exists and has a value, we don't put that value in but rather assign to a[-1] each time through the loop.






    share|improve this answer

























    • Somehow I thought the variable used in for loop is immutable. +1 for pointing it out.

      – Amir A. Shabani
      48 mins ago


















    1














    The left expression of a for loop statement gets assigned with each item in the iterable on the right in each iteration, so



    for n in a:
    print(n)


    is just a fancy way of doing:



    for i in range(len(a)):
    n = a[i]
    print(n)


    Likewise,



    for a[-1] in a:
    print(a[-1])


    is just a fancy way of doing:



    for i in range(len(a)):
    a[-1] = a[i]
    print(a[-1])


    where in each iteration, the last item of a gets assigned with the next item in a, so when the iteration finally comes to the last item, its value got last assigned with the second-last item, 2.






    share|improve this answer






























      1














      a[-1] refers to the last element of a, in this case a[3]. The for loop is a bit unusual in that it is using this element as the loop variable.
      It's not evaluating that element upon loop entry, but rather it is assigning to it on each iteration through the loop.



      So first a[-1] gets set to 0, then 1, then 2. Finally, on the last iteration, the for loop retrieves a[3] which at that point is 2, so the list ends up as [0, 1, 2, 2].



      A more typical for loop uses a simple local variable name as the loop variable, e.g. for x .... In that case, x is set to the next value upon each iteration. This case is no different, except that a[-1] is set to the next value upon each iteration. You don't see this very often, but it's consistent.






      share|improve this answer

























      • How does it actually gets set to 0, then 1, then 2 and finally 2? That's the confusing part!

        – Amir A. Shabani
        1 hour ago






      • 1





        Agree, I don't really understand by reading this answer

        – gameon67
        1 hour ago











      • I expected it would be like saying for 3 in a: print(a[-1]) (because a[-1] was 3 at the start of the loop) and give an error, but clearly that's not the case. You're obviously correct about what's happening, but I'm surprised this evaluates this way.

        – Nathan
        1 hour ago











      • It's set by the for loop. Each time through, it gets set to the next element of the list. What's the first element? Right, it's 0. What's the second? Right again, it's 1. etc.

        – Tom Karzes
        1 hour ago






      • 1





        "The for loop is a bit unusual in that it is using this element as the loop variable." - This may be where the confusion is coming from. Python isn't using the element a[-1] evaluates to, rather, it's using the index a[-1] itself. If you rephrase that statement, your answer becomes much more clear.

        – Christian Dean
        37 mins ago












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      5 Answers
      5






      active

      oldest

      votes








      5 Answers
      5






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      10














      To bring a language-lawyer aspect to the question, let's look at documentation:




      for_stmt ::= "for" target_list "in" expression_list ":" suite


      The expression list is evaluated once; it should yield an iterable object. An iterator is created for the result of the expression_list. The suite is then executed once for each item provided by the iterator, in the order returned by the iterator. Each item in turn is assigned to the target list using the standard rules for assignments (see Assignment statements), and then the suite is executed.




      (emphasis not originally in docs)
      The suite refers to the statements under the for-block, print(a[-1]) in our particular case.



      Simply, on each iteration, the loop variable (target_list) gets assigned to the next item in the iterable (expression_list).



      Let's extend the print statement:



      a = [0, 1, 2, 3]
      for a[-1] in a:
      print(a, a[-1])


      This gives the following output:



      [0, 1, 2, 0] 0 # a[-1] assigned 0
      [0, 1, 2, 1] 1 # a[-1] assigned 1
      [0, 1, 2, 2] 2 # a[-1] assigned 2
      [0, 1, 2, 2] 2 # a[-1] assigned itself (2)


      As a[-1] is a valid form left-hand side, assignments to a[-1] will mutate a, modifying the list during iteration. In this particular example, a[-1] retains -2 from the previous evaluation. If we trace the assignment on each iteration, we have



      a[-1] = 0 # a[0] # a[3] is 0 now
      a[-1] = 1 # a[1] # a[3] is 1 now
      a[-1] = 2 # a[2] # a[3] is 2 now
      a[-1] = 2 # a[3] # a[3] is itself (2)





      share|improve this answer




















      • 2





        Might be useful to bold Each item [...] is assigned to the target list using the standard rules for assignments.

        – Mateen Ulhaq
        57 mins ago







      • 2





        Essentially, the important thing to note here is that Python considers a[-1] to be a valid form of the left-hand side of an assignment statement (e.g. a[-1] = 1 is valid grammar). Thus a[-1] is a valid "variable" name, because as the documentation stated, it evaluates the binding variable(s) in a for loop declaration as it would the left-hand side of an assignment.

        – Christian Dean
        45 mins ago















      10














      To bring a language-lawyer aspect to the question, let's look at documentation:




      for_stmt ::= "for" target_list "in" expression_list ":" suite


      The expression list is evaluated once; it should yield an iterable object. An iterator is created for the result of the expression_list. The suite is then executed once for each item provided by the iterator, in the order returned by the iterator. Each item in turn is assigned to the target list using the standard rules for assignments (see Assignment statements), and then the suite is executed.




      (emphasis not originally in docs)
      The suite refers to the statements under the for-block, print(a[-1]) in our particular case.



      Simply, on each iteration, the loop variable (target_list) gets assigned to the next item in the iterable (expression_list).



      Let's extend the print statement:



      a = [0, 1, 2, 3]
      for a[-1] in a:
      print(a, a[-1])


      This gives the following output:



      [0, 1, 2, 0] 0 # a[-1] assigned 0
      [0, 1, 2, 1] 1 # a[-1] assigned 1
      [0, 1, 2, 2] 2 # a[-1] assigned 2
      [0, 1, 2, 2] 2 # a[-1] assigned itself (2)


      As a[-1] is a valid form left-hand side, assignments to a[-1] will mutate a, modifying the list during iteration. In this particular example, a[-1] retains -2 from the previous evaluation. If we trace the assignment on each iteration, we have



      a[-1] = 0 # a[0] # a[3] is 0 now
      a[-1] = 1 # a[1] # a[3] is 1 now
      a[-1] = 2 # a[2] # a[3] is 2 now
      a[-1] = 2 # a[3] # a[3] is itself (2)





      share|improve this answer




















      • 2





        Might be useful to bold Each item [...] is assigned to the target list using the standard rules for assignments.

        – Mateen Ulhaq
        57 mins ago







      • 2





        Essentially, the important thing to note here is that Python considers a[-1] to be a valid form of the left-hand side of an assignment statement (e.g. a[-1] = 1 is valid grammar). Thus a[-1] is a valid "variable" name, because as the documentation stated, it evaluates the binding variable(s) in a for loop declaration as it would the left-hand side of an assignment.

        – Christian Dean
        45 mins ago













      10












      10








      10







      To bring a language-lawyer aspect to the question, let's look at documentation:




      for_stmt ::= "for" target_list "in" expression_list ":" suite


      The expression list is evaluated once; it should yield an iterable object. An iterator is created for the result of the expression_list. The suite is then executed once for each item provided by the iterator, in the order returned by the iterator. Each item in turn is assigned to the target list using the standard rules for assignments (see Assignment statements), and then the suite is executed.




      (emphasis not originally in docs)
      The suite refers to the statements under the for-block, print(a[-1]) in our particular case.



      Simply, on each iteration, the loop variable (target_list) gets assigned to the next item in the iterable (expression_list).



      Let's extend the print statement:



      a = [0, 1, 2, 3]
      for a[-1] in a:
      print(a, a[-1])


      This gives the following output:



      [0, 1, 2, 0] 0 # a[-1] assigned 0
      [0, 1, 2, 1] 1 # a[-1] assigned 1
      [0, 1, 2, 2] 2 # a[-1] assigned 2
      [0, 1, 2, 2] 2 # a[-1] assigned itself (2)


      As a[-1] is a valid form left-hand side, assignments to a[-1] will mutate a, modifying the list during iteration. In this particular example, a[-1] retains -2 from the previous evaluation. If we trace the assignment on each iteration, we have



      a[-1] = 0 # a[0] # a[3] is 0 now
      a[-1] = 1 # a[1] # a[3] is 1 now
      a[-1] = 2 # a[2] # a[3] is 2 now
      a[-1] = 2 # a[3] # a[3] is itself (2)





      share|improve this answer















      To bring a language-lawyer aspect to the question, let's look at documentation:




      for_stmt ::= "for" target_list "in" expression_list ":" suite


      The expression list is evaluated once; it should yield an iterable object. An iterator is created for the result of the expression_list. The suite is then executed once for each item provided by the iterator, in the order returned by the iterator. Each item in turn is assigned to the target list using the standard rules for assignments (see Assignment statements), and then the suite is executed.




      (emphasis not originally in docs)
      The suite refers to the statements under the for-block, print(a[-1]) in our particular case.



      Simply, on each iteration, the loop variable (target_list) gets assigned to the next item in the iterable (expression_list).



      Let's extend the print statement:



      a = [0, 1, 2, 3]
      for a[-1] in a:
      print(a, a[-1])


      This gives the following output:



      [0, 1, 2, 0] 0 # a[-1] assigned 0
      [0, 1, 2, 1] 1 # a[-1] assigned 1
      [0, 1, 2, 2] 2 # a[-1] assigned 2
      [0, 1, 2, 2] 2 # a[-1] assigned itself (2)


      As a[-1] is a valid form left-hand side, assignments to a[-1] will mutate a, modifying the list during iteration. In this particular example, a[-1] retains -2 from the previous evaluation. If we trace the assignment on each iteration, we have



      a[-1] = 0 # a[0] # a[3] is 0 now
      a[-1] = 1 # a[1] # a[3] is 1 now
      a[-1] = 2 # a[2] # a[3] is 2 now
      a[-1] = 2 # a[3] # a[3] is itself (2)






      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited 2 mins ago

























      answered 1 hour ago









      TrebledJTrebledJ

      3,76421329




      3,76421329







      • 2





        Might be useful to bold Each item [...] is assigned to the target list using the standard rules for assignments.

        – Mateen Ulhaq
        57 mins ago







      • 2





        Essentially, the important thing to note here is that Python considers a[-1] to be a valid form of the left-hand side of an assignment statement (e.g. a[-1] = 1 is valid grammar). Thus a[-1] is a valid "variable" name, because as the documentation stated, it evaluates the binding variable(s) in a for loop declaration as it would the left-hand side of an assignment.

        – Christian Dean
        45 mins ago












      • 2





        Might be useful to bold Each item [...] is assigned to the target list using the standard rules for assignments.

        – Mateen Ulhaq
        57 mins ago







      • 2





        Essentially, the important thing to note here is that Python considers a[-1] to be a valid form of the left-hand side of an assignment statement (e.g. a[-1] = 1 is valid grammar). Thus a[-1] is a valid "variable" name, because as the documentation stated, it evaluates the binding variable(s) in a for loop declaration as it would the left-hand side of an assignment.

        – Christian Dean
        45 mins ago







      2




      2





      Might be useful to bold Each item [...] is assigned to the target list using the standard rules for assignments.

      – Mateen Ulhaq
      57 mins ago






      Might be useful to bold Each item [...] is assigned to the target list using the standard rules for assignments.

      – Mateen Ulhaq
      57 mins ago





      2




      2





      Essentially, the important thing to note here is that Python considers a[-1] to be a valid form of the left-hand side of an assignment statement (e.g. a[-1] = 1 is valid grammar). Thus a[-1] is a valid "variable" name, because as the documentation stated, it evaluates the binding variable(s) in a for loop declaration as it would the left-hand side of an assignment.

      – Christian Dean
      45 mins ago





      Essentially, the important thing to note here is that Python considers a[-1] to be a valid form of the left-hand side of an assignment statement (e.g. a[-1] = 1 is valid grammar). Thus a[-1] is a valid "variable" name, because as the documentation stated, it evaluates the binding variable(s) in a for loop declaration as it would the left-hand side of an assignment.

      – Christian Dean
      45 mins ago













      2














      it is an interesting question, you can understand it by that:



      for v in a:
      a[-1] = v
      print(a[-1])

      print(a)


      actually a becomes: [0, 1, 2, 2] after loop



      output:



      0
      1
      2
      2
      [0, 1, 2, 2]


      Hope that helps you, and comment if you have further questions. : )






      share|improve this answer



























        2














        it is an interesting question, you can understand it by that:



        for v in a:
        a[-1] = v
        print(a[-1])

        print(a)


        actually a becomes: [0, 1, 2, 2] after loop



        output:



        0
        1
        2
        2
        [0, 1, 2, 2]


        Hope that helps you, and comment if you have further questions. : )






        share|improve this answer

























          2












          2








          2







          it is an interesting question, you can understand it by that:



          for v in a:
          a[-1] = v
          print(a[-1])

          print(a)


          actually a becomes: [0, 1, 2, 2] after loop



          output:



          0
          1
          2
          2
          [0, 1, 2, 2]


          Hope that helps you, and comment if you have further questions. : )






          share|improve this answer













          it is an interesting question, you can understand it by that:



          for v in a:
          a[-1] = v
          print(a[-1])

          print(a)


          actually a becomes: [0, 1, 2, 2] after loop



          output:



          0
          1
          2
          2
          [0, 1, 2, 2]


          Hope that helps you, and comment if you have further questions. : )







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 1 hour ago









          recnacrecnac

          1,218123




          1,218123





















              2














              (this is more of a long comment than an answer - there are a couple of good ones already, especially @TrebledJ's. But I had to think of it explicitly in terms of overwriting variables that already have values before it clicked for me.)



              If you had



              x = 0
              l = [1, 2, 3]
              for x in l:
              print(x)


              you wouldn't be surprised that x is overriden each time through the loop. Even though x existed before, its value isn't used (i.e. for 0 in l:, which would throw an error). Rather, we assign the values from l to x.



              When we do



              a = [0, 1, 2, 3]

              for a[-1] in a:
              print(a[-1])


              even though a[-1] already exists and has a value, we don't put that value in but rather assign to a[-1] each time through the loop.






              share|improve this answer

























              • Somehow I thought the variable used in for loop is immutable. +1 for pointing it out.

                – Amir A. Shabani
                48 mins ago















              2














              (this is more of a long comment than an answer - there are a couple of good ones already, especially @TrebledJ's. But I had to think of it explicitly in terms of overwriting variables that already have values before it clicked for me.)



              If you had



              x = 0
              l = [1, 2, 3]
              for x in l:
              print(x)


              you wouldn't be surprised that x is overriden each time through the loop. Even though x existed before, its value isn't used (i.e. for 0 in l:, which would throw an error). Rather, we assign the values from l to x.



              When we do



              a = [0, 1, 2, 3]

              for a[-1] in a:
              print(a[-1])


              even though a[-1] already exists and has a value, we don't put that value in but rather assign to a[-1] each time through the loop.






              share|improve this answer

























              • Somehow I thought the variable used in for loop is immutable. +1 for pointing it out.

                – Amir A. Shabani
                48 mins ago













              2












              2








              2







              (this is more of a long comment than an answer - there are a couple of good ones already, especially @TrebledJ's. But I had to think of it explicitly in terms of overwriting variables that already have values before it clicked for me.)



              If you had



              x = 0
              l = [1, 2, 3]
              for x in l:
              print(x)


              you wouldn't be surprised that x is overriden each time through the loop. Even though x existed before, its value isn't used (i.e. for 0 in l:, which would throw an error). Rather, we assign the values from l to x.



              When we do



              a = [0, 1, 2, 3]

              for a[-1] in a:
              print(a[-1])


              even though a[-1] already exists and has a value, we don't put that value in but rather assign to a[-1] each time through the loop.






              share|improve this answer















              (this is more of a long comment than an answer - there are a couple of good ones already, especially @TrebledJ's. But I had to think of it explicitly in terms of overwriting variables that already have values before it clicked for me.)



              If you had



              x = 0
              l = [1, 2, 3]
              for x in l:
              print(x)


              you wouldn't be surprised that x is overriden each time through the loop. Even though x existed before, its value isn't used (i.e. for 0 in l:, which would throw an error). Rather, we assign the values from l to x.



              When we do



              a = [0, 1, 2, 3]

              for a[-1] in a:
              print(a[-1])


              even though a[-1] already exists and has a value, we don't put that value in but rather assign to a[-1] each time through the loop.







              share|improve this answer














              share|improve this answer



              share|improve this answer








              edited 45 mins ago









              Amir A. Shabani

              457616




              457616










              answered 52 mins ago









              NathanNathan

              2,02511326




              2,02511326












              • Somehow I thought the variable used in for loop is immutable. +1 for pointing it out.

                – Amir A. Shabani
                48 mins ago

















              • Somehow I thought the variable used in for loop is immutable. +1 for pointing it out.

                – Amir A. Shabani
                48 mins ago
















              Somehow I thought the variable used in for loop is immutable. +1 for pointing it out.

              – Amir A. Shabani
              48 mins ago





              Somehow I thought the variable used in for loop is immutable. +1 for pointing it out.

              – Amir A. Shabani
              48 mins ago











              1














              The left expression of a for loop statement gets assigned with each item in the iterable on the right in each iteration, so



              for n in a:
              print(n)


              is just a fancy way of doing:



              for i in range(len(a)):
              n = a[i]
              print(n)


              Likewise,



              for a[-1] in a:
              print(a[-1])


              is just a fancy way of doing:



              for i in range(len(a)):
              a[-1] = a[i]
              print(a[-1])


              where in each iteration, the last item of a gets assigned with the next item in a, so when the iteration finally comes to the last item, its value got last assigned with the second-last item, 2.






              share|improve this answer



























                1














                The left expression of a for loop statement gets assigned with each item in the iterable on the right in each iteration, so



                for n in a:
                print(n)


                is just a fancy way of doing:



                for i in range(len(a)):
                n = a[i]
                print(n)


                Likewise,



                for a[-1] in a:
                print(a[-1])


                is just a fancy way of doing:



                for i in range(len(a)):
                a[-1] = a[i]
                print(a[-1])


                where in each iteration, the last item of a gets assigned with the next item in a, so when the iteration finally comes to the last item, its value got last assigned with the second-last item, 2.






                share|improve this answer

























                  1












                  1








                  1







                  The left expression of a for loop statement gets assigned with each item in the iterable on the right in each iteration, so



                  for n in a:
                  print(n)


                  is just a fancy way of doing:



                  for i in range(len(a)):
                  n = a[i]
                  print(n)


                  Likewise,



                  for a[-1] in a:
                  print(a[-1])


                  is just a fancy way of doing:



                  for i in range(len(a)):
                  a[-1] = a[i]
                  print(a[-1])


                  where in each iteration, the last item of a gets assigned with the next item in a, so when the iteration finally comes to the last item, its value got last assigned with the second-last item, 2.






                  share|improve this answer













                  The left expression of a for loop statement gets assigned with each item in the iterable on the right in each iteration, so



                  for n in a:
                  print(n)


                  is just a fancy way of doing:



                  for i in range(len(a)):
                  n = a[i]
                  print(n)


                  Likewise,



                  for a[-1] in a:
                  print(a[-1])


                  is just a fancy way of doing:



                  for i in range(len(a)):
                  a[-1] = a[i]
                  print(a[-1])


                  where in each iteration, the last item of a gets assigned with the next item in a, so when the iteration finally comes to the last item, its value got last assigned with the second-last item, 2.







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered 57 mins ago









                  blhsingblhsing

                  43.5k41744




                  43.5k41744





















                      1














                      a[-1] refers to the last element of a, in this case a[3]. The for loop is a bit unusual in that it is using this element as the loop variable.
                      It's not evaluating that element upon loop entry, but rather it is assigning to it on each iteration through the loop.



                      So first a[-1] gets set to 0, then 1, then 2. Finally, on the last iteration, the for loop retrieves a[3] which at that point is 2, so the list ends up as [0, 1, 2, 2].



                      A more typical for loop uses a simple local variable name as the loop variable, e.g. for x .... In that case, x is set to the next value upon each iteration. This case is no different, except that a[-1] is set to the next value upon each iteration. You don't see this very often, but it's consistent.






                      share|improve this answer

























                      • How does it actually gets set to 0, then 1, then 2 and finally 2? That's the confusing part!

                        – Amir A. Shabani
                        1 hour ago






                      • 1





                        Agree, I don't really understand by reading this answer

                        – gameon67
                        1 hour ago











                      • I expected it would be like saying for 3 in a: print(a[-1]) (because a[-1] was 3 at the start of the loop) and give an error, but clearly that's not the case. You're obviously correct about what's happening, but I'm surprised this evaluates this way.

                        – Nathan
                        1 hour ago











                      • It's set by the for loop. Each time through, it gets set to the next element of the list. What's the first element? Right, it's 0. What's the second? Right again, it's 1. etc.

                        – Tom Karzes
                        1 hour ago






                      • 1





                        "The for loop is a bit unusual in that it is using this element as the loop variable." - This may be where the confusion is coming from. Python isn't using the element a[-1] evaluates to, rather, it's using the index a[-1] itself. If you rephrase that statement, your answer becomes much more clear.

                        – Christian Dean
                        37 mins ago
















                      1














                      a[-1] refers to the last element of a, in this case a[3]. The for loop is a bit unusual in that it is using this element as the loop variable.
                      It's not evaluating that element upon loop entry, but rather it is assigning to it on each iteration through the loop.



                      So first a[-1] gets set to 0, then 1, then 2. Finally, on the last iteration, the for loop retrieves a[3] which at that point is 2, so the list ends up as [0, 1, 2, 2].



                      A more typical for loop uses a simple local variable name as the loop variable, e.g. for x .... In that case, x is set to the next value upon each iteration. This case is no different, except that a[-1] is set to the next value upon each iteration. You don't see this very often, but it's consistent.






                      share|improve this answer

























                      • How does it actually gets set to 0, then 1, then 2 and finally 2? That's the confusing part!

                        – Amir A. Shabani
                        1 hour ago






                      • 1





                        Agree, I don't really understand by reading this answer

                        – gameon67
                        1 hour ago











                      • I expected it would be like saying for 3 in a: print(a[-1]) (because a[-1] was 3 at the start of the loop) and give an error, but clearly that's not the case. You're obviously correct about what's happening, but I'm surprised this evaluates this way.

                        – Nathan
                        1 hour ago











                      • It's set by the for loop. Each time through, it gets set to the next element of the list. What's the first element? Right, it's 0. What's the second? Right again, it's 1. etc.

                        – Tom Karzes
                        1 hour ago






                      • 1





                        "The for loop is a bit unusual in that it is using this element as the loop variable." - This may be where the confusion is coming from. Python isn't using the element a[-1] evaluates to, rather, it's using the index a[-1] itself. If you rephrase that statement, your answer becomes much more clear.

                        – Christian Dean
                        37 mins ago














                      1












                      1








                      1







                      a[-1] refers to the last element of a, in this case a[3]. The for loop is a bit unusual in that it is using this element as the loop variable.
                      It's not evaluating that element upon loop entry, but rather it is assigning to it on each iteration through the loop.



                      So first a[-1] gets set to 0, then 1, then 2. Finally, on the last iteration, the for loop retrieves a[3] which at that point is 2, so the list ends up as [0, 1, 2, 2].



                      A more typical for loop uses a simple local variable name as the loop variable, e.g. for x .... In that case, x is set to the next value upon each iteration. This case is no different, except that a[-1] is set to the next value upon each iteration. You don't see this very often, but it's consistent.






                      share|improve this answer















                      a[-1] refers to the last element of a, in this case a[3]. The for loop is a bit unusual in that it is using this element as the loop variable.
                      It's not evaluating that element upon loop entry, but rather it is assigning to it on each iteration through the loop.



                      So first a[-1] gets set to 0, then 1, then 2. Finally, on the last iteration, the for loop retrieves a[3] which at that point is 2, so the list ends up as [0, 1, 2, 2].



                      A more typical for loop uses a simple local variable name as the loop variable, e.g. for x .... In that case, x is set to the next value upon each iteration. This case is no different, except that a[-1] is set to the next value upon each iteration. You don't see this very often, but it's consistent.







                      share|improve this answer














                      share|improve this answer



                      share|improve this answer








                      edited 18 mins ago

























                      answered 1 hour ago









                      Tom KarzesTom Karzes

                      11.2k1926




                      11.2k1926












                      • How does it actually gets set to 0, then 1, then 2 and finally 2? That's the confusing part!

                        – Amir A. Shabani
                        1 hour ago






                      • 1





                        Agree, I don't really understand by reading this answer

                        – gameon67
                        1 hour ago











                      • I expected it would be like saying for 3 in a: print(a[-1]) (because a[-1] was 3 at the start of the loop) and give an error, but clearly that's not the case. You're obviously correct about what's happening, but I'm surprised this evaluates this way.

                        – Nathan
                        1 hour ago











                      • It's set by the for loop. Each time through, it gets set to the next element of the list. What's the first element? Right, it's 0. What's the second? Right again, it's 1. etc.

                        – Tom Karzes
                        1 hour ago






                      • 1





                        "The for loop is a bit unusual in that it is using this element as the loop variable." - This may be where the confusion is coming from. Python isn't using the element a[-1] evaluates to, rather, it's using the index a[-1] itself. If you rephrase that statement, your answer becomes much more clear.

                        – Christian Dean
                        37 mins ago


















                      • How does it actually gets set to 0, then 1, then 2 and finally 2? That's the confusing part!

                        – Amir A. Shabani
                        1 hour ago






                      • 1





                        Agree, I don't really understand by reading this answer

                        – gameon67
                        1 hour ago











                      • I expected it would be like saying for 3 in a: print(a[-1]) (because a[-1] was 3 at the start of the loop) and give an error, but clearly that's not the case. You're obviously correct about what's happening, but I'm surprised this evaluates this way.

                        – Nathan
                        1 hour ago











                      • It's set by the for loop. Each time through, it gets set to the next element of the list. What's the first element? Right, it's 0. What's the second? Right again, it's 1. etc.

                        – Tom Karzes
                        1 hour ago






                      • 1





                        "The for loop is a bit unusual in that it is using this element as the loop variable." - This may be where the confusion is coming from. Python isn't using the element a[-1] evaluates to, rather, it's using the index a[-1] itself. If you rephrase that statement, your answer becomes much more clear.

                        – Christian Dean
                        37 mins ago

















                      How does it actually gets set to 0, then 1, then 2 and finally 2? That's the confusing part!

                      – Amir A. Shabani
                      1 hour ago





                      How does it actually gets set to 0, then 1, then 2 and finally 2? That's the confusing part!

                      – Amir A. Shabani
                      1 hour ago




                      1




                      1





                      Agree, I don't really understand by reading this answer

                      – gameon67
                      1 hour ago





                      Agree, I don't really understand by reading this answer

                      – gameon67
                      1 hour ago













                      I expected it would be like saying for 3 in a: print(a[-1]) (because a[-1] was 3 at the start of the loop) and give an error, but clearly that's not the case. You're obviously correct about what's happening, but I'm surprised this evaluates this way.

                      – Nathan
                      1 hour ago





                      I expected it would be like saying for 3 in a: print(a[-1]) (because a[-1] was 3 at the start of the loop) and give an error, but clearly that's not the case. You're obviously correct about what's happening, but I'm surprised this evaluates this way.

                      – Nathan
                      1 hour ago













                      It's set by the for loop. Each time through, it gets set to the next element of the list. What's the first element? Right, it's 0. What's the second? Right again, it's 1. etc.

                      – Tom Karzes
                      1 hour ago





                      It's set by the for loop. Each time through, it gets set to the next element of the list. What's the first element? Right, it's 0. What's the second? Right again, it's 1. etc.

                      – Tom Karzes
                      1 hour ago




                      1




                      1





                      "The for loop is a bit unusual in that it is using this element as the loop variable." - This may be where the confusion is coming from. Python isn't using the element a[-1] evaluates to, rather, it's using the index a[-1] itself. If you rephrase that statement, your answer becomes much more clear.

                      – Christian Dean
                      37 mins ago






                      "The for loop is a bit unusual in that it is using this element as the loop variable." - This may be where the confusion is coming from. Python isn't using the element a[-1] evaluates to, rather, it's using the index a[-1] itself. If you rephrase that statement, your answer becomes much more clear.

                      – Christian Dean
                      37 mins ago











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There's a third YouTube co-founder"سایت یوتیوب برای چندمین بار در ایران فیلتر شدنسخهٔ اصلیسالار کمانگر جوان آمریکایی ایرانی الاصل مدیر سایت یوتیوب شدنسخهٔ اصلیVideo websites pop up, invite postingsthe originalthe originalYouTube: Overnight success has sparked a backlashthe original"Me at the zoo"YouTube serves up 100 million videos a day onlinethe originalcomScore Releases May 2010 U.S. Online Video Rankingsthe originalYouTube hits 4 billion daily video viewsthe originalYouTube users uploading two days of video every minutethe originalEric Schmidt, Princeton Colloquium on Public & Int'l Affairsthe original«Streaming Dreams»نسخهٔ اصلیAlexa Traffic Rank for YouTube (three month average)the originalHelp! YouTube is killing my business!the originalUtube sues YouTubethe originalGoogle closes $A2b YouTube dealthe originalFlash moves on to smart phonesthe originalYouTube HTML5 Video Playerنسخهٔ اصلیYouTube HTML5 Video Playerthe originalGoogle tries freeing Web video with WebMthe originalVideo length for uploadingthe originalYouTube caps video lengths to reduce infringementthe originalAccount Types: Longer videosthe originalYouTube bumps video limit to 15 minutesthe originalUploading large files and resumable uploadingthe originalVideo Formats: File formatsthe originalGetting Started: File formatsthe originalThe quest for a new video codec in Flash 8the originalAdobe Flash Video File Format Specification Version 10.1the originalYouTube Mobile goes livethe originalYouTube videos go HD with a simple hackthe originalYouTube now supports 4k-resolution videosthe originalYouTube to get high-def 1080p playerthe original«Approximate YouTube Bitrates»نسخهٔ اصلی«Bigger and Better: Encoding for YouTube 720p HD»نسخهٔ اصلی«YouTube's 1080p – Failure Depends on How You Look At It»نسخهٔ اصلیYouTube in 3Dthe originalYouTube in 3D?the originalYouTube 3D Videosthe originalYouTube adds a dimension, 3D goggles not includedthe originalYouTube Adds Stereoscopic 3D Video Support (And 3D Vision Support, Too)the original«Sharing YouTube Videos»نسخهٔ اصلی«Downloading videos from YouTube is not supported, except for one instance when it is permitted.»نسخهٔ اصلی«Terms of Use, 5.B»نسخهٔ اصلی«Some YouTube videos get download option»نسخهٔ اصلی«YouTube looks out for content owners, disables video ripping»«Downloading videos from YouTube is not supported, except for one instance when it is permitted.»نسخهٔ اصلی«YouTube Hopes To Boost Revenue With Video Downloads»نسخهٔ اصلی«YouTube Mobile»نسخهٔ اصلی«YouTube Live on Apple TV Today; Coming to iPhone on June 29»نسخهٔ اصلی«Goodbye Flash: YouTube mobile goes HTML5 on iPhone and Android»نسخهٔ اصلی«YouTube Mobile Goes HTML5, Video Quality Beats Native Apps Hands Down»نسخهٔ اصلی«TiVo Getting YouTube Streaming Today»نسخهٔ اصلی«YouTube video comes to Wii and PlayStation 3 game consoles»نسخهٔ اصلی«Coming Up Next... YouTube on Your TV»نسخهٔ اصلی«Experience YouTube XL on the Big Screen»نسخهٔ اصلی«Xbox Live Getting Live TV, YouTube & Bing Voice Search»نسخهٔ اصلی«YouTube content locations»نسخهٔ اصلی«April fools: YouTube turns the world up-side-down»نسخهٔ اصلی«YouTube goes back to 1911 for April Fools' Day»نسخهٔ اصلی«Simon Cowell's bromance, the self-driving Nascar and Hungry Hippos for iPad... the best April Fools' gags»نسخهٔ اصلی"YouTube Announces It Will Shut Down""YouTube Adds Darude 'Sandstorm' Button To Its Videos For April Fools' Day"«Censorship fears rise as Iran blocks access to top websites»نسخهٔ اصلی«China 'blocks YouTube video site'»نسخهٔ اصلی«YouTube shut down in Morocco»نسخهٔ اصلی«Thailand blocks access to YouTube»نسخهٔ اصلی«Ban on YouTube lifted after deal»نسخهٔ اصلی«Google's Gatekeepers»نسخهٔ اصلی«Turkey goes into battle with Google»نسخهٔ اصلی«Turkey lifts two-year ban on YouTube»نسخهٔ اصلیسانسور در ترکیه به یوتیوب رسیدلغو فیلترینگ یوتیوب در ترکیه«Pakistan blocks YouTube website»نسخهٔ اصلی«Pakistan lifts the ban on YouTube»نسخهٔ اصلی«Pakistan blocks access to YouTube in internet crackdown»نسخهٔ اصلی«Watchdog urges Libya to stop blocking websites»نسخهٔ اصلی«YouTube»نسخهٔ اصلی«Due to abuses of religion, customs Emirates, YouTube is blocked in the UAE»نسخهٔ اصلی«Google Conquered The Web - An Ultimate Winner»نسخهٔ اصلی«100 million videos are viewed daily on YouTube»نسخهٔ اصلی«Harry and Charlie Davies-Carr: Web gets taste for biting baby»نسخهٔ اصلی«Meet YouTube's 224 million girl, Natalie Tran»نسخهٔ اصلی«YouTube to Double Down on Its 'Channel' Experiment»نسخهٔ اصلی«13 Some Media Companies Choose to Profit From Pirated YouTube Clips»نسخهٔ اصلی«Irate HK man unlikely Web hero»نسخهٔ اصلی«Web Guitar Wizard Revealed at Last»نسخهٔ اصلی«Charlie bit my finger – again!»نسخهٔ اصلی«Lowered Expectations: Web Redefines 'Quality'»نسخهٔ اصلی«YouTube's 50 Greatest Viral Videos»نسخهٔ اصلیYouTube Community Guidelinesthe original«Why did my YouTube account get closed down?»نسخهٔ اصلی«Why do I have a sanction on my account?»نسخهٔ اصلی«Is YouTube's three-strike rule fair to users?»نسخهٔ اصلی«Viacom will sue YouTube for $1bn»نسخهٔ اصلی«Mediaset Files EUR500 Million Suit Vs Google's YouTube»نسخهٔ اصلی«Premier League to take action against YouTube»نسخهٔ اصلی«YouTube law fight 'threatens net'»نسخهٔ اصلی«Google must divulge YouTube log»نسخهٔ اصلی«Google Told to Turn Over User Data of YouTube»نسخهٔ اصلی«US judge tosses out Viacom copyright suit against YouTube»نسخهٔ اصلی«Google and Viacom: YouTube copyright lawsuit back on»نسخهٔ اصلی«Woman can sue over YouTube clip de-posting»نسخهٔ اصلی«YouTube loses court battle over music clips»نسخهٔ اصلیYouTube to Test Software To Ease Licensing Fightsthe original«Press Statistics»نسخهٔ اصلی«Testing YouTube's Audio Content ID System»نسخهٔ اصلی«Content ID disputes»نسخهٔ اصلیYouTube Community Guidelinesthe originalYouTube criticized in Germany over anti-Semitic Nazi videosthe originalFury as YouTube carries sick Hillsboro video insultthe originalYouTube attacked by MPs over sex and violence footagethe originalAl-Awlaki's YouTube Videos Targeted by Rep. Weinerthe originalYouTube Withdraws Cleric's Videosthe originalYouTube is letting users decide on terrorism-related videosthe original«Time's Person of the Year: You»نسخهٔ اصلی«Our top 10 funniest YouTube comments – what are yours?»نسخهٔ اصلی«YouTube's worst comments blocked by filter»نسخهٔ اصلی«Site Info YouTube»نسخهٔ اصلیوبگاه YouTubeوبگاه موبایل YouTubeوووووو

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                      Rest API with Magento using PHP with example. Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?How to update product using magento client library for PHP?Oauth Error while extending Magento Rest APINot showing my custom api in wsdl(url) and web service list?Using Magento API(REST) via IXMLHTTPRequest COM ObjectHow to login in Magento website using REST APIREST api call for Guest userMagento API calling using HTML and javascriptUse API rest media management by storeView code (admin)Magento REST API Example ErrorsHow to log all rest api calls in magento2?How to update product using magento client library for PHP?