[solved](python)Shortcut to express list1-list2 ?

Question

I have 2 lists:
list1 = [1,2,3,4]
list2 = [3,4]
If I want to prevent the numbers in list2 to be shown in list1, is there a shortcut for that?
Or I have to write long code like below:
list1 = [1, 2, 3, 4]
list2 = [3, 4]
temp_list = []
i = 0
while i < len(list1):
if list1[i] not in list2:
temp_list.append(list1[i])
i += 1
list1 = temp_list
print(list1)

Answer 1

[i for i in list1 if i not in list2]
https://code.sololearn.com/cbn7mwGiujgk/?ref=app

Answer 2

[edited]
sorry, the lists sequence in the lambda was wrong. here the correct version:
list1 = [1, 2, 3, 4]
list2 = [3, 4]
print(list(filter(lambda x:x not in list2, list1)))
Lea ,
just for the sake of completeness: using filter with a lambda function can also do this job:
list1 = [1, 2, 3, 4]
list2 = [3, 4]
print(list(filter(lambda x:x not in list1, list2))) # not correct!

Answer 3

print(list(set(list1) - set(list2)))

Answer 4

Lea oh no,
Converting lists into sets has some pitfalls.
Just add a 2nd element 1 to list1.
It will disappear after converting to set.
Simba s solution is cool.Hatsy Rei s solution is safe.

Answer 5

the modules name to import is 'collections' not 'collection'
^

Answer 6

Hatsy Rei Simba Both codes works! Thank you!

Answer 7

Oma Falk I just tried! You are right! Thank you for pointing it out!
https://code.sololearn.com/cGX90sT9QgaI/?ref=app

Answer 8

Prabhas Koya
Thank you for providing more solutions!
By the way,
Is “collection” a moduler i need to add to my library before I run the code?

Answer 9

Simba This is so cool! It means I can apply all the mathematical operations for sets on lists!

Answer 10

I wonder why lists don’t have mathematical opertations like sets.
Sets and lists, it seems both of them are for holding data, but their methods are so different…..

Answer 11

for i in list2:
try:
list1.remove(i)
except:
pass
'''
Looping only through list2, I think this is efficient. Hatsy Rei is looping through extra items in list1 . IF YOU KNOW YOUR LIST2 CONTAINS ELEMENTS WHICH ARE DEFINITELY IN LIST1 , THEN SIMPLY
for i in list2: list1.remove(i)
'''
making more faster
import collections
list1 = collections.deque([1,1,2,3])
list2 = [3,4]
for i in list2:
try:
list1.remove(i)
except:
pass

Answer 12

☺️

Answer 13

Hi, Lothar I ran you code and get the result below:
[]

Answer 14

Yes , you need to import before running your code.

Answer 15

Here it goes
list1 = [1, 1, 2, 3, 4]; list2 = [3, 4]
list0 = list(set(list1 + list2)); print(list0)

Answer 16

https://code.sololearn.com/cczEI6377e22/?ref=app

Answer 17

Lothar yeah that was a typo thanks for correcting

Answer 18

I need your help guys to improve my code