理解：每次迭代多个值

Question

Is there a way to output two (or more) items per iteration in a list/dictionary/set comprehension? As a simple example, to output all the positive and negative doubles of the integers from 1 to 3 (that is to say, {x | x = ±2n, n ∈ {1...3}} ), is there a syntax similar to the following?

>>> [2*i, -2*i for i in range(1, 4)]
[2, -2, 4, -4, 6, -6]

I know I could output tuples of (+i,-i) and flatten that, but I was wondering if there was any way to completely solve the problem using a single comprehension.

Currently, I am producing two lists and concatenating them (which works, provided the order isn't important):

>>> [2*i for i in range(1, 4)] + [-2*i for i in range(1, 4)]
[2, 4, 6, -2, -4, -6]

Answer 1

Another option is a nested comprehension:

r = [2*i*s for i in range(1, 4) for s in 1, -1]

For a more general case:

r = [item for tpl in (<something that yields tuples>) for item in tpl]

with your original example:

r = [item for tpl in ((2*i, -2*i) for i in range(1, 4)) for item in tpl]

although I'd really suggest itertools.chain.from_iterable as @Lattyware said.

Answer 2

Another form of nested comprehension:

>>> [sub for i in range(1, 4) for sub in (2*i, -2*i)]
[2, -2, 4, -4, 6, -6]

Answer 3

The best answer here is to simply use itertools.chain.from_iterable() to, as you mention, flatten the list:

itertools.chain.from_iterable((2*i, -2*i) for i in range(1, 4))

This is pretty readable, and doesn't require iterating over the source twice (which may be problematic given some iterators can be exhausted, and it means extra computational effort).

Answer 4

Although I would use the itertools method @Lattyware suggested, here is a more general approach using a generator that may also be helpful.

>>> def nums():
        for i in range(1, 4):
            yield 2*i
            yield -2*i


>>> list(nums())
[2, -2, 4, -4, 6, -6]

Answer 5

According to PEP202 there is no way to output more than one object from a list comprehension:

- The form [x, y for ...] is disallowed; one is required to write [(x, y) for ...] .