在python中压缩多个if语句

Question

I'm trying to write a function to check if an object is found in multiple lists and to remove the object from any list its found in. I want to know if there is a way to make it cleaner or smarter using some form of generic variable where you predefine the format or something along those lines. my code in its ugly form:

def create_newlist(choice):

    if choice in list_a:
        list_a.remove(choice)
    if choice in list_b:
        list_b.remove(choice)
    if choice in list_c:
        list_c.remove(choice)
    if choice in list_d:
        list_d.remove(choice)
    if choice in list_e:
        list_e.remove(choice)

What I'm hoping for is something like:

if choice in list_x:
   list_x.remove(choice)

I would like it to work for each list, would I need to loop through? any suggestions would be great! I have the workaround but I would love to learn a more elegant way of coding this!

Answer 1

How about creating a list of lists and looping over that?

Something like:

lists = [list_a, list_b, list_c, list_d, list_e]
for lst in lists: 
    if choice in lst: 
        lst.remove(choice)

Answer 2

In a one-liner

If you use some_list.remove(item) , only the first found match of item is removed from the list. Therefore, it depends if the lists possibly include duplicated items, which (all) need to be removed:

1. If all items in all lists are unique

list1 = ["a", "b" , "c", "d"]
list2 = ["k", "l", "m", "n"]
list3 = ["c", "b", "a", "e"]

[l.remove("a") for l in [list1, list2, list3] if "a" in l]

print(list3)
> ["c", "b", "e"]

However

2. If one or more lists possibly includes duplicated items

In that case itertools ' filterfalse() will come in handy:

from itertools import filterfalse

def remove_val(val, lists):
    return [list(filterfalse(lambda w: w == val, l)) for l in lists]

l1 = ["a", "b" , "c", "d"]
l2 = ["k", "l", "m", "n"]
l3 = ["a", "c", "b", "a", "a", "e"]

newlists = remove_val("a", [l1, l2, l3])

Then the test:

print(newlists)
> [['b', 'c', 'd'], ['k', 'l', 'm', 'n'], ['c', 'b', 'e']]

Answer 3

Make your list_x a list of all your lists

Then do it this way

for each in list_x:
    if choice in each:
        # if is actually not required
        each.remove(choice)

Answer 4

An alternative to the former suggestions, which is a bit clearer, is to define a helper function. So, instead of:

def create_newlist(choice):

    if choice in list_a:
        list_a.remove(choice)
    if choice in list_b:
        list_b.remove(choice)
    if choice in list_c:
        list_c.remove(choice)
    if choice in list_d:
        list_d.remove(choice)
    if choice in list_e:
        list_e.remove(choice)

You'd have:

def create_newlist(_list, choice):
    if choice in _list:
        _list.remove(choice)

lists = [list_a, list_b, list_c, list_d, list_e]

for _lst in lists:
    create_newlist(_lst, choice)