Remove all occurrences of an integer from list

Question

In python, how can I remove all occurrences of an integer from a list. For example:

list_a = [5, 20, 15, 10, 55, 30, 65]
list_a.METHOD(5)
print(list_a)
[20, 10, 30]

Essentially, the method removes anything that contains a 5. Not specifically multiples of 5, given that 10 and 20 do not contain a 5. Thanks.

Possible duplicate of [Remove all occurrences of a value from a Python list](http://stackoverflow.com/questions/1157106/remove-all-occurrences-of-a-value-from-a-python-list) — utsav_deep, Feb 08 '17 at 18:30

score 1 · Answer 1 · answered Feb 08 '17 at 18:33

1

list_a = [5, 20, 15, 10, 55, 30, 65]
list_a = [x for x in list_a if "5" not in str(x)]
print(list_a)
[20, 10, 30]

answered Feb 08 '17 at 18:33

yedpodtrzitko

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score 0 · Answer 2 · answered Feb 08 '17 at 18:30

0

This is one way:

[x for x in list_a if str(x).find('5') < 0]

That's a list comprehension that converts each number to a string and then searches the string for the character 5. find returns -1 if not found so we keep only things where the 5 wasn't found.

answered Feb 08 '17 at 18:30

Oliver Dain

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if you want to test only if a substring is in a string, use `in`, and not `find`. – Daniel Feb 08 '17 at 18:32

score 0 · Answer 3 · answered Feb 08 '17 at 18:30

0

You can use filter and lambda.

filter(lambda x:"5" not in str(x) , [5, 20, 15, 10, 55, 30, 65])

answered Feb 08 '17 at 18:30

Bipul Jain

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Remove all occurrences of an integer from list

3 Answers3