Python - List returning [[…], 6]












12














If I run the following code



data = [[1,2],[3,4],[5,6]]



for x in data:

    print(x[0])



for x[0] in data:

    print(x)


I get the following output



1

3

5



[[1, 2], 6]

[[3, 4], 6]

[[...], 6]


I end up with a list containing [[...], 6], but what is this [...] list?



It doesn't behave normally, because calling y = [[...], 6] and then the following statements show [...] to be 0



>>> print(y)

[[Ellipsis], 6]



>>> print(y[0])

[0]


However when I run the code at the top, and type the following statements the results don't make sense:



>>> print(x)

[[...], 6]



>>> print(x[0])

[[...], 6]



>>> print(x[0][0])

[[...], 6]



>>> print(x[0][0][0])

[[...], 6]


and yet somehow both of these result in 6



>>> print(x[1])

6



>>> print(x[0][1])

6


To review the question: How is this possible, and what does [...] represent, and how can the for loop at the top create such a list?






python python-3.x







share|improve this question




















  • 3




    Did you do any research? stackoverflow.com/questions/17160162/…
    – jonrsharpe
    yesterday










  • Strangely I couldn't find that question from my searches, it seems to answer the question perfectly. The only question that remains unanswered is how does the for loop create such a list? @jonrsharpe
    – Ruler Of The World
    yesterday










  • Because each step of the loop assigns the next element from x to x[0]. Those two loops are not the same.
    – jonrsharpe
    yesterday












  • A very worthy Hot Network Question. Here's another useful question that compliments the one linked above by Jon: stackoverflow.com/questions/30462352/…
    – vaultah
    yesterday
















12














If I run the following code



data = [[1,2],[3,4],[5,6]]



for x in data:

    print(x[0])



for x[0] in data:

    print(x)


I get the following output



1

3

5



[[1, 2], 6]

[[3, 4], 6]

[[...], 6]


I end up with a list containing [[...], 6], but what is this [...] list?



It doesn't behave normally, because calling y = [[...], 6] and then the following statements show [...] to be 0



>>> print(y)

[[Ellipsis], 6]



>>> print(y[0])

[0]


However when I run the code at the top, and type the following statements the results don't make sense:



>>> print(x)

[[...], 6]



>>> print(x[0])

[[...], 6]



>>> print(x[0][0])

[[...], 6]



>>> print(x[0][0][0])

[[...], 6]


and yet somehow both of these result in 6



>>> print(x[1])

6



>>> print(x[0][1])

6


To review the question: How is this possible, and what does [...] represent, and how can the for loop at the top create such a list?






python python-3.x







share|improve this question




















  • 3




    Did you do any research? stackoverflow.com/questions/17160162/…
    – jonrsharpe
    yesterday










  • Strangely I couldn't find that question from my searches, it seems to answer the question perfectly. The only question that remains unanswered is how does the for loop create such a list? @jonrsharpe
    – Ruler Of The World
    yesterday










  • Because each step of the loop assigns the next element from x to x[0]. Those two loops are not the same.
    – jonrsharpe
    yesterday












  • A very worthy Hot Network Question. Here's another useful question that compliments the one linked above by Jon: stackoverflow.com/questions/30462352/…
    – vaultah
    yesterday














12












12








12


3





If I run the following code



data = [[1,2],[3,4],[5,6]]



for x in data:

    print(x[0])



for x[0] in data:

    print(x)


I get the following output



1

3

5



[[1, 2], 6]

[[3, 4], 6]

[[...], 6]


I end up with a list containing [[...], 6], but what is this [...] list?



It doesn't behave normally, because calling y = [[...], 6] and then the following statements show [...] to be 0



>>> print(y)

[[Ellipsis], 6]



>>> print(y[0])

[0]


However when I run the code at the top, and type the following statements the results don't make sense:



>>> print(x)

[[...], 6]



>>> print(x[0])

[[...], 6]



>>> print(x[0][0])

[[...], 6]



>>> print(x[0][0][0])

[[...], 6]


and yet somehow both of these result in 6



>>> print(x[1])

6



>>> print(x[0][1])

6


To review the question: How is this possible, and what does [...] represent, and how can the for loop at the top create such a list?






python python-3.x







share|improve this question















If I run the following code



data = [[1,2],[3,4],[5,6]]



for x in data:

    print(x[0])



for x[0] in data:

    print(x)


I get the following output



1

3

5



[[1, 2], 6]

[[3, 4], 6]

[[...], 6]


I end up with a list containing [[...], 6], but what is this [...] list?



It doesn't behave normally, because calling y = [[...], 6] and then the following statements show [...] to be 0



>>> print(y)

[[Ellipsis], 6]



>>> print(y[0])

[0]


However when I run the code at the top, and type the following statements the results don't make sense:



>>> print(x)

[[...], 6]



>>> print(x[0])

[[...], 6]



>>> print(x[0][0])

[[...], 6]



>>> print(x[0][0][0])

[[...], 6]


and yet somehow both of these result in 6



>>> print(x[1])

6



>>> print(x[0][1])

6


To review the question: How is this possible, and what does [...] represent, and how can the for loop at the top create such a list?






python python-3.x




python python-3.x






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited yesterday









Mad Physicist

34.2k156895




34.2k156895










asked yesterday









Ruler Of The WorldRuler Of The World

502318




502318








  • 3




    Did you do any research? stackoverflow.com/questions/17160162/…
    – jonrsharpe
    yesterday










  • Strangely I couldn't find that question from my searches, it seems to answer the question perfectly. The only question that remains unanswered is how does the for loop create such a list? @jonrsharpe
    – Ruler Of The World
    yesterday










  • Because each step of the loop assigns the next element from x to x[0]. Those two loops are not the same.
    – jonrsharpe
    yesterday












  • A very worthy Hot Network Question. Here's another useful question that compliments the one linked above by Jon: stackoverflow.com/questions/30462352/…
    – vaultah
    yesterday














  • 3




    Did you do any research? stackoverflow.com/questions/17160162/…
    – jonrsharpe
    yesterday










  • Strangely I couldn't find that question from my searches, it seems to answer the question perfectly. The only question that remains unanswered is how does the for loop create such a list? @jonrsharpe
    – Ruler Of The World
    yesterday










  • Because each step of the loop assigns the next element from x to x[0]. Those two loops are not the same.
    – jonrsharpe
    yesterday












  • A very worthy Hot Network Question. Here's another useful question that compliments the one linked above by Jon: stackoverflow.com/questions/30462352/…
    – vaultah
    yesterday








3




3




Did you do any research? stackoverflow.com/questions/17160162/…
– jonrsharpe
yesterday




Did you do any research? stackoverflow.com/questions/17160162/…
– jonrsharpe
yesterday












Strangely I couldn't find that question from my searches, it seems to answer the question perfectly. The only question that remains unanswered is how does the for loop create such a list? @jonrsharpe
– Ruler Of The World
yesterday




Strangely I couldn't find that question from my searches, it seems to answer the question perfectly. The only question that remains unanswered is how does the for loop create such a list? @jonrsharpe
– Ruler Of The World
yesterday












Because each step of the loop assigns the next element from x to x[0]. Those two loops are not the same.
– jonrsharpe
yesterday






Because each step of the loop assigns the next element from x to x[0]. Those two loops are not the same.
– jonrsharpe
yesterday














A very worthy Hot Network Question. Here's another useful question that compliments the one linked above by Jon: stackoverflow.com/questions/30462352/…
– vaultah
yesterday




A very worthy Hot Network Question. Here's another useful question that compliments the one linked above by Jon: stackoverflow.com/questions/30462352/…
– vaultah
yesterday












2 Answers
2






active

oldest

votes


















9














Let's give your sublists names:



a = [1, 2]

b = [3, 4]

c = [5, 6]

data = [a, b, c]


Your first loop binds a, b and c successively to x. When the loop terminates, you have effectively set x = c.



The second loop now binds a, b and c successively to x[0]. This is fine for a and b, but for c you are effectively doing c[0] = c, creating a circular reference. Since list is able to catch that, it won't try to print [[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[...






share|improve this answer





























    5














    that's because you're using x[0] as your loop variable (which is bad practice) which exists as a list and not a new name like you're supposed to when iterating with for



    for x[0] in data:
    
    print(x)


    and x is in data so there's a cyclic reference (hence the ellipsis representation to avoid infinite recursion when printing the same data over and over)



    More in detail:



    The ellipsis happens on the last element because of the previous loop that binds x on the last element of data ([5,6]).



    So the second loop assigns [5,6] to x[0] but it's also x. On way to get rid of this is to create a copy of x just before the second loop: x = x[:]






    share|improve this answer























    • Great answer, so the […] represents an infinite loop of the same list?
      – Ruler Of The World
      yesterday










    • ... is the way to represent something that cycles.
      – Jean-François Fabre
      yesterday











    Your Answer






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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    9














    Let's give your sublists names:



    a = [1, 2]
    
b = [3, 4]
    
c = [5, 6]
    
data = [a, b, c]


    Your first loop binds a, b and c successively to x. When the loop terminates, you have effectively set x = c.



    The second loop now binds a, b and c successively to x[0]. This is fine for a and b, but for c you are effectively doing c[0] = c, creating a circular reference. Since list is able to catch that, it won't try to print [[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[...






    share|improve this answer


























      9














      Let's give your sublists names:



      a = [1, 2]
      
b = [3, 4]
      
c = [5, 6]
      
data = [a, b, c]


      Your first loop binds a, b and c successively to x. When the loop terminates, you have effectively set x = c.



      The second loop now binds a, b and c successively to x[0]. This is fine for a and b, but for c you are effectively doing c[0] = c, creating a circular reference. Since list is able to catch that, it won't try to print [[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[...






      share|improve this answer
























        9












        9








        9






        Let's give your sublists names:



        a = [1, 2]
        
b = [3, 4]
        
c = [5, 6]
        
data = [a, b, c]


        Your first loop binds a, b and c successively to x. When the loop terminates, you have effectively set x = c.



        The second loop now binds a, b and c successively to x[0]. This is fine for a and b, but for c you are effectively doing c[0] = c, creating a circular reference. Since list is able to catch that, it won't try to print [[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[...






        share|improve this answer












        Let's give your sublists names:



        a = [1, 2]
        
b = [3, 4]
        
c = [5, 6]
        
data = [a, b, c]


        Your first loop binds a, b and c successively to x. When the loop terminates, you have effectively set x = c.



        The second loop now binds a, b and c successively to x[0]. This is fine for a and b, but for c you are effectively doing c[0] = c, creating a circular reference. Since list is able to catch that, it won't try to print [[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[...







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered yesterday









        Mad PhysicistMad Physicist

        34.2k156895




        34.2k156895

























            5














            that's because you're using x[0] as your loop variable (which is bad practice) which exists as a list and not a new name like you're supposed to when iterating with for



            for x[0] in data:
            
    print(x)


            and x is in data so there's a cyclic reference (hence the ellipsis representation to avoid infinite recursion when printing the same data over and over)



            More in detail:



            The ellipsis happens on the last element because of the previous loop that binds x on the last element of data ([5,6]).



            So the second loop assigns [5,6] to x[0] but it's also x. On way to get rid of this is to create a copy of x just before the second loop: x = x[:]






            share|improve this answer























            • Great answer, so the […] represents an infinite loop of the same list?
              – Ruler Of The World
              yesterday










            • ... is the way to represent something that cycles.
              – Jean-François Fabre
              yesterday
















            5














            that's because you're using x[0] as your loop variable (which is bad practice) which exists as a list and not a new name like you're supposed to when iterating with for



            for x[0] in data:
            
    print(x)


            and x is in data so there's a cyclic reference (hence the ellipsis representation to avoid infinite recursion when printing the same data over and over)



            More in detail:



            The ellipsis happens on the last element because of the previous loop that binds x on the last element of data ([5,6]).



            So the second loop assigns [5,6] to x[0] but it's also x. On way to get rid of this is to create a copy of x just before the second loop: x = x[:]






            share|improve this answer























            • Great answer, so the […] represents an infinite loop of the same list?
              – Ruler Of The World
              yesterday










            • ... is the way to represent something that cycles.
              – Jean-François Fabre
              yesterday














            5












            5








            5






            that's because you're using x[0] as your loop variable (which is bad practice) which exists as a list and not a new name like you're supposed to when iterating with for



            for x[0] in data:
            
    print(x)


            and x is in data so there's a cyclic reference (hence the ellipsis representation to avoid infinite recursion when printing the same data over and over)



            More in detail:



            The ellipsis happens on the last element because of the previous loop that binds x on the last element of data ([5,6]).



            So the second loop assigns [5,6] to x[0] but it's also x. On way to get rid of this is to create a copy of x just before the second loop: x = x[:]






            share|improve this answer














            that's because you're using x[0] as your loop variable (which is bad practice) which exists as a list and not a new name like you're supposed to when iterating with for



            for x[0] in data:
            
    print(x)


            and x is in data so there's a cyclic reference (hence the ellipsis representation to avoid infinite recursion when printing the same data over and over)



            More in detail:



            The ellipsis happens on the last element because of the previous loop that binds x on the last element of data ([5,6]).



            So the second loop assigns [5,6] to x[0] but it's also x. On way to get rid of this is to create a copy of x just before the second loop: x = x[:]







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited yesterday

























            answered yesterday









            Jean-François FabreJean-François Fabre

            101k954109




            101k954109












            • Great answer, so the […] represents an infinite loop of the same list?
              – Ruler Of The World
              yesterday










            • ... is the way to represent something that cycles.
              – Jean-François Fabre
              yesterday


















            • Great answer, so the […] represents an infinite loop of the same list?
              – Ruler Of The World
              yesterday










            • ... is the way to represent something that cycles.
              – Jean-François Fabre
              yesterday
















            Great answer, so the […] represents an infinite loop of the same list?
            – Ruler Of The World
            yesterday




            Great answer, so the […] represents an infinite loop of the same list?
            – Ruler Of The World
            yesterday












            ... is the way to represent something that cycles.
            – Jean-François Fabre
            yesterday




            ... is the way to represent something that cycles.
            – Jean-François Fabre
            yesterday


















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