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]]

You can use `set(ls)` to eliminate duplicate entries. But that would convert the list to a set. — amanb, Mar 17 '20 at 06:54
Do you want to remove only the duplicates or the original item as well? In other words, do you want to remove three 1's or just two? — DYZ, Mar 17 '20 at 06:55
@amanb Yeah set do works. list(set(ls)) will also make the set to list again. But I am working with a 2d list. So set won't work for me. — Saykat, Mar 17 '20 at 06:56
@amanb `[[1], [1, 2], [1, 2, 3], [], [2], [2, 3], [], [], [3]]`. I was trying to get the subset of of [1, 2, 3]. — Saykat, Mar 17 '20 at 06:58
@Saykat do you intend to remove the original item as well if it has duplicates? — Raghul Raj, Mar 17 '20 at 06:58
@Saykat, okay. The requirement is still not clear. Please update the question as well. — amanb, Mar 17 '20 at 06:59
@Saykat Are you trying to remove all occurrences of a particular element or everything that is not a particular element (like everything that isn't `[1, 2, 3]`)? — jeremye, Mar 17 '20 at 07:00
The question is very clear at the moment. Please take time to [edit] the question and add [mcve] so that we can be helpful. — Ch3steR, Mar 17 '20 at 07:01
@illiteratecoder edited the question and hopefully, it is now more clear. — Saykat, Mar 17 '20 at 07:04
also similar elements like `[2, 2, 2], [2, 2, 2]]` should be removed ? — kederrac, Mar 17 '20 at 07:14

jeremye · Accepted Answer · 2020-03-17T07:18:57.483

3

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))

edited Mar 17 '20 at 07:18

answered Mar 17 '20 at 06:56

jeremye

1,368
9
19

Yes, you could substitute any other value you want to remove for `1`. But as the question was worded, the desired behavior was to remove all occurrences of one element. – jeremye Mar 17 '20 at 06:59
I have updated the answer to reflect the change in the question – jeremye Mar 17 '20 at 07:06
if you have similar elements like `[[2, 2,2], [2,2,2], [3,3,3], [3,3,3]]` ? – kederrac Mar 17 '20 at 07:12
@kederrac While the title of the question might be misleading, the question is asking about removing all occurrences of a specific element, in this case all occurrences of `[]`. I will update my answer to show how to remove all occurrences of any given element though. – jeremye Mar 17 '20 at 07:15

score 0 · Answer 2 · answered Mar 17 '20 at 06:58

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

score 0 · Answer 3 · answered Mar 17 '20 at 07:03

0

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]

answered Mar 17 '20 at 07:03

Raghul Raj

1,428
9
24

Please look at this [answer](https://stackoverflow.com/questions/53513/how-do-i-check-if-a-list-is-empty). `if len(l)!=0` is not recommended and shouldn't be. – Ch3steR Mar 17 '20 at 07:05

score 0 · Answer 4 · answered Mar 17 '20 at 07:03

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]]

score 0 · Answer 5 · answered Mar 17 '20 at 07:09

0

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)

answered Mar 17 '20 at 07:09

Pygirl

12,969
5
30
43

kederrac · Answer 6 · 2020-03-17T07:35:52.007

0

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]]

edited Mar 17 '20 at 07:35

answered Mar 17 '20 at 07:10

kederrac

16,819
6
32
55

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

6 Answers6

List Comprehension

Filter

Generalizing for more than just the empty list