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






    StackExchange.ifUsing("editor", function () {
    StackExchange.using("externalEditor", function () {
    StackExchange.using("snippets", function () {
    StackExchange.snippets.init();
    });
    });
    }, "code-snippets");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "1"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f54063064%2fpython-list-returning-6%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    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


















            draft saved

            draft discarded




















































            Thanks for contributing an answer to Stack Overflow!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f54063064%2fpython-list-returning-6%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            An IMO inspired problem

            Image
            14 4 This problem from IMO 1988 is said to be one of the most elegant ones in functional equations. Problem : The function $f$ is defined on the set of all positive integers as follows: begin{align} f(1) = 1, f(3) &= 3, f(2n) = f(n), \ f(4n+1) &= 2f(2n+1) - f(n), \ f(4n+3) &= 3 f(2n+1) - 2f(n) end{align} Find the number of $n$ with $f(n) = n, 1 leq n leq 1988$. The main idea towards the solution is realizing that these conditions stand for the fact that $f(n)$ just reverses the digits in the binary representation of the number $n$. So, essentially the solution is finding the number of binary pallindromes $leq 1988_{10}$. My question is the following : How to reformulate the problem so that $f(n)$ reverses the digits of $n$ in its ternary representation. Or even better, can we reformulate it for a
            Read more

            Management

            Image
            Management From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Manager" redirects here. For other uses, see Management (disambiguation) and Manager (disambiguation). An organization chart for the United States Coast Guard shows the hierarchy of managerial roles in that organization. Business administration Management of a business Accounting Management accounting Financial accounting Financial audit Business entities Cooperative Corporation Limited liability company Partnership Sole proprietorship State-owned enterprise Corporate governance Annual general meeting Board of directors Supervisory board Advisory board Audit committee Corporate law Commercial law Constitutional documents Contract Corporate crime Corporate liability Insolvency law International trade law Mergers and acquisitions Economics Commodity Pu
            Read more

            Has there ever been an instance of an active nuclear power plant within or near a war zone?

            Image
            36 3 The question Has there ever been a military combat going on around an active (running or in outage but not decommissioned) commercial nuclear power plant? Or dangerously near it? If yes, how did the plant and its personnel fare in such situation? Why I ask Let's assume for the scope of this question modern commercial nuclear power plants (CANDU, PWR, BWR of generation II plus all of gen III and newer) are usually reasonably safe to operate given that the engineering has been done right and there is a whole cohort of very well trained personnel on-site at all times, well rested and with considerable resources on their hands. Things hard to get with an armed conflict like the recent Syrian war raging around. I can't seem to find any reference or a comprehensive article about this, hence this question.
            Read more