On Sun, 23 Oct 2016 10:57 am, Chris Angelico wrote:
> This surprised me.
>
> Python 3.4.2 (default, Oct 8 2014, 10:45:20)
> [GCC 4.9.1] on linux
> Type "help", "copyright", "credits" or "license" for more information.
>>>> y=6
>>>> [(x,y) for x in range(y) for y in range(3)]
> [(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2,
> 2), (3, 0), (3, 1), (3, 2), (4, 0), (4, 1), (4, 2), (5, 0), (5, 1),
> (5, 2)]
That surprises me too. I wouldn't have expected that y is both global and
local to the comprehension at the same time.
I think that this is a bug in list comprehensions and should at least give a
warning that y is being used as both local and non-local.
This is what happens if you try to access a local before it exists:
py> [(y, x) for x in (1, y) for y in (10, 20)]
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
NameError: name 'y' is not defined
Swap the order, and it works:
py> [(y, x) for y in (10, 20) for x in (1, y)]
[(10, 1), (10, 10), (20, 1), (20, 20)]
But if y happens to exist as a global, the first version mysteriously works!
py> y = 999
py> [(y, x) for x in (1, y) for y in (10, 20)]
[(10, 1), (20, 1), (10, 999), (20, 999)]
Let's expand the list comp to a function:
y = 999
def list_comp():
result = []
for x in (1, y):
for y in (10, 20):
result.append((y, x))
return result
Calling that function gives a NameError, specifically:
UnboundLocalError: local variable 'y' referenced before assignment
> Normally, a comprehension is described as being equivalent to an
> unrolled loop, inside a nested function.
I don't think that's quite the right description, but something like that.
> That would be like this:
>
> def temp():
> ret = []
> for x in range(y):
> for y in range(3):
> ret.append((x,y))
> return ret
> temp()
Indeed.
--
Steve
“Cheer up,” they said, “things could be worse.” So I cheered up, and sure
enough, things got worse.
--
https://mail.python.org/mailman/listinfo/python-list
