list append gives None as result [duplicate]

问题

i just wrote a function that should print out all the values 2 dictionaries have in common. so if use the following line in my function:

print list_intersection([1, 3, 5], [5, 3, 1])

The output should be:

[1, 3, 5]

I wrote the following code to solve this problem:

def list_intersection(list_1, list_2):
    empty_list = []
    for number in list_1:
        if number in list_2:
            return empty_list.append(number)

The problem is that i only get None as output, but if i use the following code:

def list_intersection(list_1, list_2):
    empty_list = []
    for number in list_1:
        if number in list_2:
           return number

I get the numbers printed out one by one that are in both lists. I have no idea why my program isn't just putting the numbers both lists have in common into my empty_list and return me my empty_list

回答1:

I suppose the assertion could be made that this isn't exactly a duplicate. For the reason why .append() returns None please see Alex Martelli's erudite answer.

For your code instead do:

def list_intersection(list_1, list_2):
    intersection = []
    for number in list_1:
        if number in list_2:
            intersection.append(number)
    return intersection

This avoids the following pitfalls:

Returning None instead of the list intersection.
Returning None for each element of list_2.

来源：https://stackoverflow.com/questions/26151795/list-append-gives-none-as-result

标签

python

list

python-2.7

append