how to remove character in one string if present in another Python -

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• remove elements occur in 1 list another 5 answers

i'm trying create function remove characters 1 string if present in another.

for example:

l1 = ['a', 'b', 'c', 'd', 'e', 'f']

l2 = ['b', 'c', 'e']

i want return ['a', 'd', 'f']

i have within function. thank help!

here's simple list comp approach:

def f(l1, l2):     return [x x in l1 if x not in l2]  l1 = ['a', 'b', 'c', 'd', 'e', 'f'] l2 = ['b', 'c', 'e'] print(f(l1, l2)) >>> ['a', 'd', 'f'] 

here few more (using filter can say):

f = lambda l1, l2: list(filter(lambda elem: elem not in l2, l1)) 

if want modify original list:

def f(l1, l2):     elem in l2:         l1.remove(elem)     return l1  l1 = ['a', 'b', 'c', 'd', 'e', 'f'] l2 = ['b', 'c', 'e'] print(l1) # prints ['a', 'b', 'c', 'd', 'e', 'f'] print(f(l1, l2)) # modifies l1 , returns it, printing ['a', 'd', 'f'] print(l1) # prints ['a', 'd', 'f'] (notice list has been modified) 

if need strings (and not lists posted in question), here's lambda:

s1 = 'abcdef' s2 = 'bce' # both of below work strings , lists alike (and return string) fn = lambda s1, s2: "".join(char char in s1 if char not in s2) # or, using filter: fn = lambda s1, s2: "".join(filter(lambda char: char not in s2, s1))  print(fn(s1, s2) >>> 'adf'