How to remove all occurrences of an element from list in Python?

Question

ls = [[1], [1, 2], [1, 2, 3], [], [2], [2, 3], [], [], [3]]

Is there any python list method that will enable me to remove all the similar items at once. For example, I have a 2d list 'ls' which has three empty list items []. I want to remove all the empty list items at once. I know it can be done with 'remove' and 'loop.' But is there any method to do the operation at once? To be simple: I want all the '[]' to be deleted. And thus the answer would be like this. [[1], [1, 2], [1, 2, 3], [2], [2, 3], [3]]

Answer 1

There is no remove_all method or anything like that, but the best ways to accomplish this are with a list comprehension or filter .

Assuming that ls only contains other lists you can write the following to remove all occurrences of the empty list [] :

List Comprehension

ls = [[1], [1, 2], [1, 2, 3], [], [2], [2, 3], [], [], [3]]
ls = [x for x in ls if x]
# now ls =  [[1], [1, 2], [1, 2, 3], [2], [2, 3], [3]]

Filter

ls = [[1], [1, 2], [1, 2, 3], [], [2], [2, 3], [], [], [3]]
ls = list(filter(None, ls))
# now ls =  [[1], [1, 2], [1, 2, 3], [2], [2, 3], [3]]

Generalizing for more than just the empty list

If you wanted to remove all occurrences of a given element elem (that is not the empty list) you can modify the above code as follows:

ls = [x for x in ls if x != elem]

##### or #####

ls = list(filter(lambda x: x != elem, ls))

Answer 2

A set object is an unordered collection of distinct hashable objects. Common uses include membership testing, removing duplicates from a sequence, and computing mathematical operations such as intersection, union, difference, and symmetric difference. (For other containers see the built-in dict, list, and tuple classes, and the collections module.)

https://docs.python.org/3.8/library/stdtypes.html#set-types-set-frozenset

Answer 3

You can do something like this. But it is specific to this case:

ls = [[1], [1, 2], [1, 2, 3], [], [2], [2, 3], [], [], [3]]
[l for l in ls if len(l)!=0]

Answer 4

You can use filter here to remove all the [] lists and maintaining the order.

list(filter(None,ls))
# [[1], [1, 2], [1, 2, 3], [2], [2, 3], [3]]

List comprehension of the above would be

[lst for lst in ls if lst]
# [[1], [1, 2], [1, 2, 3], [2], [2, 3], [3]]

Answer 5

You can try the followings approaches:

1)

l = [x for x in ls if x != []]

or

l = [x for x in ls if x]

2)

l = filter(None, ls)

Answer 6

you could use collections.Counter with the built-in function filter :

from collections import Counter

count = Counter(map(str, ls))
ls = list(filter(lambda x: count[str(x)] == 1, ls))
ls

output:

[[1], [1, 2], [1, 2, 3], [2], [2, 3], [3]]