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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.
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… *dies*. The worst of it is, I can remember mentioning exactly this interpreter behavior to someone perhaps a week ago.
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It doesnt. The loop printsallof the items in the list, without skipping any.
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I get the same result as @lvc (only on IDLE however, when executed as script I get this))
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Something is wrong with your Python/Computer.
it is behavior of putting code via interactive shell. If function returns value without being used, interpreter would print it to shell as debug output
What you see is theinterpreterechoing back the return value ofnext()in addition toibeing printed each iteration:
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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.
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So, why doesnextnot appear to advance the iterator in this case?
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Tested in Python 2.7 and in Python 3+ . Works properly in both
Since theyre printing alternating elements of the original sequence, and its unexpected that the next(a) statement is printing, it appears as if the print i statement is printing all the values.
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If you assign the output ofnext()things work as expected:
What is happening is thatnext(a)returns the next value of a, which is printed to the console because it is not affected.
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It behaves the way you want if called as a function:
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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.
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This being the case, I would expect:
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My first thought was that this might happen because the loop callsiteron what it is passed, and this might give an independent iterator – this isnt the case, as we haveiter(a) is a.
What you can do is affect a variable with this value:
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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. Its interesting…
or printextrainformation to differentiate theprint()output from the interactive interpreter echo:
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(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).)
to skip every second element: the call tonextshould 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 toi.
interesting. I tried for i in a: next(a) ;print i and thought i would jump to 1 and print 1,3,5,7,9. But still it is 0,2,4,6,8. Why?
I was not aware of this behavior from the interpreter. I am glad that I discovered that before losing a lot of time wondering about it while solving some real problem.
@Inbar Rose Only if you run as script.
a = iter(list(range(10))) 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
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So0is the output ofprint(i),1the return value fromnext(), echoed by the interactive interpreter, etc. There are just 5 iterations, each iteration resulting in 2 lines being written to the terminal.
Why is the difference between these functions constant?
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So, advancing the iterator is, as expected, handled by mutating that same object.
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