解压嵌套列表以获取map（）的参数

Question

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, ie, it does add(1, 2) , then add(3, 4) for the second, etc.?

• Trying map(add, foo) obviously does add([1, 2], #nothing) for the first iteration
• Trying map(add, *foo) does add(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]

Answer 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 between map() and starmap() parallels the distinction between function(a,b) and function(*c) .

Answer 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))

Answer 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]