How can I remove all occurences of two values in a list?

Question

I want to remove all occurences of two values in a list using one line. I have tried the filter + lambda method and list comprehension, but it's not working the way I would expect.

Something like:

s2 = [0, 0, 3, 11, 14, 17, 18, 18]
# I want to remove all 0 and 18 from s2 list
# i.e. I want output: s2_filter = [3, 11, 14, 17]

s2_filter = filter (lambda a: a != 0 or 18, s2)
>>> s2 = [0, 0, 3, 11, 14, 17, 18, 18] # not what I'm expecting

s2_filter = [i for i in s2 if i != 0 or 18]
>>> s2 = [0, 0, 3, 11, 14, 17, 18, 18] # not what I'm expecting

I guess there is something wrong with the or but I cannot figure out how I should write this correctly.

Can you help?

Bad example, I'll edit it. Sometimes they occur just once, sometimes more than once. — francoiskroll, Mar 06 '17 at 09:41
Why are you looking for a solution to this private case? Why not looking for a generic solution that handles any set of number instead? — Lior Magen, Mar 06 '17 at 09:54
FWIW, `0 or 18` evaluates to `18`. Check it out in the interactive interpreter. — PM 2Ring, Mar 06 '17 at 09:56

score 0 · Accepted Answer · answered Mar 06 '17 at 09:45

0

You can just filter over if i is not in a non desirable list:

s2 = [0, 0, 3, 11, 14, 17, 18, 18]
s2_filter = [i for i in s2 if i not in [0, 18]]
s2_filter
[3, 11, 14, 17]

answered Mar 06 '17 at 09:45

Netwave

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How can I remove all occurences of two values in a list?

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