【Python】列表：sum求和含None的list

Basically my question is say you have an list containing ‘None’ how would you try retrieving the sum of the list. Below is an example I tried which doesn’t work and I get the error: TypeError: unsupported operand type(s) for +: 'int' and 'NoneType'. Thanks

def sumImport(self):
    my_list = [[1,2,3,None],[1,2,3],[1,1],[1,1,2,2]]
    k = sum(chain.from_iterable(my_list))
    return k

[

](https://stackoverflow.com/users/771848/alecxe)

alecxe

435k109 gold badges1000 silver badges1135 bronze badges

asked Sep 1, 2012 at 17:33

[

](https://stackoverflow.com/users/1103714/sam)

You can use filter function

>>> sum(filter(None, [1,2,3,None]))
6

Updated from comments

Typically filter usage is filter(func, iterable), but passing None as first argument is a special case, described in Python docs. Quoting:

If function is None, the identity function is assumed, that is, all elements of iterable that are false are removed.

answered Sep 1, 2012 at 17:34

[

](https://stackoverflow.com/users/523217/alexey-kachayev)

3

Remove None (and zero) elements before summing by using filter:

>>> k = sum(filter(None, chain.from_iterable(my_list)))
>>> k
20

To see why this works, see the documentation for filter:

filter(function, iterable)

Construct a list from those elements of iterable for which function returns true. iterable may be either a sequence, a container which supports iteration, or an iterator. If iterable is a string or a tuple, the result also has that type; otherwise it is always a list. If function is None, the identity function is assumed, that is, all elements of iterable that are false are removed.

Note that filter(function, iterable) is equivalent to [item for item in iterable if function(item)] if function is not None and [item for item in iterable if item] if function is None.

answered Sep 1, 2012 at 17:35

[

](https://stackoverflow.com/users/61974/mark-byers)

Mark ByersMark Byers

756k172 gold badges1533 silver badges1433 bronze badges

1

Another suggestion:

from itertools import chain
k = sum(x for x in chain.from_iterable(my_list) if x)

answered Sep 1, 2012 at 17:41

[

](https://stackoverflow.com/users/20670/tim-pietzcker)

Tim PietzckerTim Pietzcker

310k56 gold badges481 silver badges542 bronze badges

5

Assuming you want to treat None as zero, a simple way is

sum(x if x is not None else 0 for x in chain.from_iterable(my_list))

answered Sep 1, 2012 at 17:34

[

](https://stackoverflow.com/users/1427416/brenbarn)

BrenBarnBrenBarn

224k33 gold badges383 silver badges368 bronze badges

3

Explicitly, this is equivalent to filter:

k = sum([x for x in chain.from_iterable(my_list) if x])

That saves me from remembering another function. :P

answered Sep 1, 2012 at 18:23

[

](https://stackoverflow.com/users/515392/taro-sato)

Taro SatoTaro Sato

1,4141 gold badge15 silver badges19 bronze badges

You always have the option of just writing the loop you want:

k = 0
for sublist in my_list:
    for val in sublist:
        if val is not None:
            k += val

But it certainly doesn’t hurt to know about filter either.

answered Sep 1, 2012 at 17:41

[

](https://stackoverflow.com/users/94977/jason-orendorff)

Jason OrendorffJason Orendorff

39.4k3 gold badges58 silver badges95 bronze badges

Just using sum and map:

sum(map(lambda x: x or 0, [1,2,3,None]))

answered Oct 9, 2021 at 11:12

[

](https://stackoverflow.com/users/9210255/tigertv-ru)

TigerTV.ruTigerTV.ru

1,0142 gold badges14 silver badges32 bronze badges

Not the answer you’re looking for? Browse other questions tagged python sum or ask your own question.

https://stackoverflow.com/questions/12229902/sum-a-list-which-contains-none-using-python