Python list iterator behavior and next iterator

Question

Consider    gt  gt  gt  lst   iter  1 2 3    gt  gt  gt  next lst  1  gt  gt  gt  next lst  2   So  advancing the iterator is  as expected  handled by mutating that same object    This being the case  I would expect   a   iter list range 10    for i in a     print i     next a    to skip every second element  the call to next should advance the iterator once  then the implicit call made by the loop should advance it a second time - and the result of this second call would be assigned to i    It doesn t  The loop prints all of the items in the list  without skipping any    My first thought was that this might happen because the loop calls iter on what it is passed  and this might give an independent iterator - this isn t the case  as we have iter a  is a    So  why does next not appear to advance the iterator in this case

User · Answer

What is happening is that next a  returns the next value of a  which is printed to the console because it is not affected   What you can do is affect a variable with this value    gt  gt  gt  a   iter list range 10     gt  gt  gt  for i in a         print i         b next a      0 2 4 6 8

User · Answer

It behaves the way you want if called as a function    gt  gt  gt  def test            a   iter list range 10            for i in a              print i              next a        gt  gt  gt  test   0 2 4 6 8

User · Answer

Something is wrong with your Python Computer   a   iter list range 10    for i in a     print i     next a    gt  gt  gt   0 2 4 6 8   Works like expected    Tested in Python 2 7 and in Python 3    Works properly in both

User · Answer

I find the existing answers a little confusing  because they only indirectly indicate the essential mystifying thing in the code example  both  the  print i  and the  next a   are causing their results to be printed   Since they re printing alternating elements of the original sequence  and it s unexpected that the  next a   statement is printing  it appears as if the  print i  statement is printing all the values   In that light  it becomes more clear that assigning the result of  next a   to a variable inhibits the printing of its  result  so that just the alternate values that the  i  loop variable are printed   Similarly  making the  print  statement emit something more distinctive disambiguates it  as well    One of the existing answers refutes the others because that answer is having the example code evaluated as a block  so that the interpreter is not reporting the intermediate values for  next a      The beguiling thing in answering questions  in general  is being explicit about what is obvious once you know the answer   It can be elusive   Likewise critiquing answers once you understand them   It s interesting

User · Answer

What you see is the interpreter echoing back the return value of next   in addition to i being printed each iteration    gt  gt  gt  a   iter list range 10     gt  gt  gt  for i in a         print i         next a       0 1 2 3 4 5 6 7 8 9   So 0 is the output of print i   1 the return value from next    echoed by the interactive interpreter  etc  There are just 5 iterations  each iteration resulting in 2 lines being written to the terminal   If you assign the output of next   things work as expected    gt  gt  gt  a   iter list range 10     gt  gt  gt  for i in a         print i             next a       0 2 4 6 8   or print extra information to differentiate the print   output from the interactive interpreter echo    gt  gt  gt  a   iter list range 10     gt  gt  gt  for i in a         print  Printing      format i          next a       Printing  0 1 Printing  2 3 Printing  4 5 Printing  6 7 Printing  8 9   In other words  next   is working as expected  but because it returns the next value from the iterator  echoed by the interactive interpreter  you are led to believe that the loop has its own iterator copy somehow

User · Answer

For those who still do not understand    gt  gt  gt  a   iter list range 10     gt  gt  gt  for i in a         print i         next a       0   print i  printed this 1   next a  printed this 2   print i  printed this 3   next a  printed this 4   print i  printed this 5   next a  printed this 6   print i  printed this 7   next a  printed this 8   print i  printed this 9   next a  printed this   As others have already said  next increases the iterator by 1 as expected  Assigning its returned value to a variable doesn t magically changes its behaviour

[python] Python list iterator behavior and next(iterator)

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