问题
I'm sure there's a way of doing this, but I haven't been able to find it. Say I have:
foo = [
[1, 2],
[3, 4],
[5, 6]
]
def add(num1, num2):
return num1 + num2
Then how can I use map(add, foo)
such that it passes num1=1
, num2=2
for the first iteration, i.e., it does add(1, 2)
, then add(3, 4)
for the second, etc.?
- Trying
map(add, foo)
obviously doesadd([1, 2], #nothing)
for the first iteration - Trying
map(add, *foo)
doesadd(1, 3, 5)
for the first iteration
I want something like map(add, foo)
to do add(1, 2)
on the first iteration.
Expected output: [3, 7, 11]
回答1:
It sounds like you need starmap:
>>> import itertools
>>> list(itertools.starmap(add, foo))
[3, 7, 11]
This unpacks each argument [a, b]
from the list foo
for you, passing them to the function add
. As with all the tools in the itertools
module, it returns an iterator which you can consume with the list
built-in function.
From the documents:
Used instead of
map()
when argument parameters are already grouped in tuples from a single iterable (the data has been “pre-zipped”). The difference betweenmap()
andstarmap()
parallels the distinction betweenfunction(a,b)
andfunction(*c)
.
回答2:
try this:
foo = [
[1, 2],
[3, 4],
[5, 6]]
def add(num1, num2):
return num1 + num2
print(map(lambda x: add(x[0], x[1]), foo))
回答3:
There was another answer with a perfectly valid method (even if not as readable as ajcr's answer), but for some reason it was deleted. I'm going to reproduce it, as it may be useful for certain situations
>>> map(add, *zip(*foo))
[3, 7, 11]
来源:https://stackoverflow.com/questions/34110317/unpack-nested-list-for-arguments-to-map