将变量分配给整个可迭代对象，同时在 for 循环中执行可迭代对象解包

Question

Is there a way, using Extended Iterable Unpacking , in a loop to assign multiple values within an iterable as well as a variable holding all of the iterable.

This is possible not in a loop using:

abc = a, b, c = range(3)

But in a loop I don't know of a syntactically equivalent option. Currently I would use:

it = zip(*[iter(range(9))]*3) #for example

for abc in it:
    a, b, c = abc
    ...

Or this:

for a, b, c in it:
    abc = [a, b, c]
    ...

I want to know if there is a way to have the best of both worlds here. Something along the lines of the following (which is not valid although an example of what I mean) :

for (a, b, c := abc) in it:
    ...

Answer 1

The closest you can get, due to the constraints on where an assignment expression is allowed, is

for a, b, c in (abc := x for x in it):
    ...

x is bound to the "scope" of the generator expressions, but abc is bound in the scope containing the generator expression, ie, the same scope in which a , b , and c are bound.

If I saw anyone doing this in a code review, though, I'd tell them to just write

for abc in it:
    a, b, c = abc
    ...

if they really need all four names defined.

Answer 2

Although this does answer my question there would likely be added overhead, and to me it is just unnecessary extra code.

from itertools import tee

for abc, (a, b, c) in zip(*tee(it)):
    …

Answer 3

You could do this:

for *abc,a,b,c in (x*2 for x in it):
    print(abc,a,b,c)

(0, 1, 2) 0 1 2
(3, 4, 5) 3 4 5
(6, 7, 8) 6 7 8

Or have abc predefined outside the scope of the comprehension:

abc = []
for a,b,c in (abc:=x for x in it):
    print(abc,a,b,c)

(0, 1, 2) 0 1 2
(3, 4, 5) 3 4 5
(6, 7, 8) 6 7 8