list comprehension misbehaving

duncan smith buzzard at invalid.invalid
Thu Mar 28 16:54:23 CET 2013


On 28/03/13 15:25, Wolfgang Maier wrote:
> Dear all, with
> a=list(range(1,11))
>
> why (in Python 2.7 and 3.3) is this explicit for loop working:
> for i in a[:-1]:
>      a.pop() and a
>
> giving:
> [1, 2, 3, 4, 5, 6, 7, 8, 9]
> [1, 2, 3, 4, 5, 6, 7, 8]
> [1, 2, 3, 4, 5, 6, 7]
> [1, 2, 3, 4, 5, 6]
> [1, 2, 3, 4, 5]
> [1, 2, 3, 4]
> [1, 2, 3]
> [1, 2]
> [1]
>
> but the equivalent comprehension failing:
> [a.pop() and a for i in a[:-1]]
>
> giving:
> [[1], [1], [1], [1], [1], [1], [1], [1], [1]]
>
> ???
> Especially, since these two things *do* work as expected:
> [a.pop() and a[:] for i in a[:-1]]
> [a.pop() and print(a) for i in a[:-1]] # Python 3 only
>
> Thanks for your help,
> Wolfgang
>

With the for loop the list is printed each time you pop an element. With 
the list comprehension all but one of the elements are popped before the 
string representation of the resulting list (containing several 
references to a) is printed.

The two list comprehensions that you say "work" behave differently 
because the first contains copies of a (which are unaffected by 
subsequent pops), and the second because (I imagine) it does something 
similar to,

[a.pop() and repr(a) for i in a[:-1]]

on 2.7 (I haven't migrated to 3 yet). i.e. The list contains a 
representation of a after each element is popped.

Duncan



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