Unpack nested list for arguments to map()

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, i.e., 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]

can you check your question as the answers and what you say you want as expected output are not the same — Padraic Cunningham, Dec 05 '15 at 20:29

Alex Riley · Accepted Answer · 2015-12-05T20:36:11.487

8

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).

edited Dec 05 '15 at 20:36

answered Dec 05 '15 at 20:13

Alex Riley

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after reading the question again it seems we are all wrong as the OP wants to `add(1, 2`), then `add(2, 3)` ... unless that is a mistake by the OP – Padraic Cunningham Dec 05 '15 at 20:28
`[3, 7 ,11]` is indeed the expected result for my example :) – binaryfunt Dec 05 '15 at 20:33
lol well they half clarified, the left the *it does `add(1, 2)`, then `add(2, 3)`* but added expected output – Padraic Cunningham Dec 05 '15 at 20:33
1

@BinaryFunt: I think now you've clarified the expected output it's clear - I edited the typo which caused the doubt originally. – Alex Riley Dec 05 '15 at 20:40
@PadraicCunningham Oh I see my mistake. Woops! ajcr has corrected it now – binaryfunt Dec 05 '15 at 20:40

score 1 · Answer 2 · answered Dec 05 '15 at 20:16

1

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))

answered Dec 05 '15 at 20:16

linluk

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This can be slightly quicker than ajcr's answer – binaryfunt Dec 06 '15 at 17:22

score 1 · Answer 3 · answered Dec 07 '15 at 11:50

1

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]

answered Dec 07 '15 at 11:50

binaryfunt

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I found this super useful to help with passing lists of arguments to tqdm's process_map function. – Lars Larsson Mar 05 '21 at 11:07

Unpack nested list for arguments to map()

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