array filter in python?

Question

For example, I have two lists

 A           = [6, 7, 8, 9, 10, 11, 12]
subset_of_A  = [6, 9, 12]; # the subset of A


the result should be [7, 8, 10, 11]; the remaining elements

Is there a built-in function in python to do this?

Chinmay Kanchi · Accepted Answer · 2011-04-12T23:04:06.180

168

If the order is not important, you should use set.difference. However, if you want to retain order, a simple list comprehension is all it takes.

result = [a for a in A if a not in subset_of_A]

EDIT: As delnan says, performance will be substantially improved if subset_of_A is an actual set, since checking for membership in a set is O(1) as compared to O(n) for a list.

A = [6, 7, 8, 9, 10, 11, 12]
subset_of_A = set([6, 9, 12]) # the subset of A

result = [a for a in A if a not in subset_of_A]

edited Apr 12 '11 at 23:04

answered Apr 12 '11 at 19:45

Chinmay Kanchi

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And this can be improved vastly by making `subset_of_A` a real `set`, which gives `O(1)` membership test (instead of `O(n)` as with lists). – Apr 12 '11 at 19:47

score 102 · Answer 2 · edited Dec 19 '16 at 19:16

102

Yes, the filter function:

filter(lambda x: x not in subset_of_A, A)

edited Dec 19 '16 at 19:16

Letharion

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answered Apr 12 '11 at 19:42

carlpett

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Note that in Python 2, `filter` returns the list itself, while in Python 3, it returns an iterator. – modulitos Oct 02 '17 at 01:54
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@modulitos `list(filter(...))` – Paweł Kaczorowski Jul 06 '18 at 06:41

score 10 · Answer 3 · answered Apr 12 '11 at 19:43

10

set(A)-set(subset_of_A) gives your the intended result set, but it won't retain the original order. The following is order preserving:

[a for a in A if not a in subset_of_A]

answered Apr 12 '11 at 19:43

Alexander Gessler

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This is wrong, it should be "if a not in", NOT "if not a in" – Life after Guest Mar 15 '22 at 11:30

score 9 · Answer 4 · answered Apr 12 '11 at 19:46

9

No, there is no build in function in python to do this, because simply:

set(A)- set(subset_of_A)

will provide you the answer.

answered Apr 12 '11 at 19:46

eat

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While this works for his example, there may be problems if elements are repeated in the list A. – Alok Mysore Sep 20 '14 at 14:31
@AlokMysore Yet, doesn't `set()` remove duplicates by precisely making the array a set? – vmonteco Sep 30 '21 at 15:52

score 5 · Answer 5 · answered Apr 12 '11 at 19:43

5

tuple(set([6, 7, 8, 9, 10, 11, 12]).difference([6, 9, 12]))

answered Apr 12 '11 at 19:43

NPE

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score 4 · Answer 6 · answered Apr 12 '11 at 19:42

4

How about

set(A).difference(subset_of_A)

answered Apr 12 '11 at 19:42

JoshAdel

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score 3 · Answer 7 · answered Apr 12 '11 at 19:43

This was just asked a couple of days ago (but I cannot find it):

>>> A = [6, 7, 8, 9, 10, 11, 12]
>>> subset_of_A = set([6, 9, 12])
>>> [i for i in A if i not in subset_of_A]
[7, 8, 10, 11]

It might be better to use sets from the beginning, depending on the context. Then you can use set operations like other answers show.

However, converting lists to sets and back only for these operations is slower than list comprehension.

score 2 · Answer 8 · answered Apr 12 '11 at 19:43

2

Use the Set type:

A_set = Set([6,7,8,9,10,11,12])
subset_of_A_set = Set([6,9,12])

result = A_set - subset_of_A_set

answered Apr 12 '11 at 19:43

Platinum Azure

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score 1 · Answer 9 · answered Apr 12 '11 at 19:43

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>>> a = set([6, 7, 8, 9, 10, 11, 12])
>>> sub_a = set([6, 9, 12])
>>> a - sub_a
set([8, 10, 11, 7])

answered Apr 12 '11 at 19:43

Jake

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score 1 · Answer 10 · answered Apr 12 '11 at 19:45

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>>> A           = [6, 7, 8, 9, 10, 11, 12]
>>> subset_of_A  = [6, 9, 12];
>>> set(A) - set(subset_of_A)
set([8, 10, 11, 7])
>>>

answered Apr 12 '11 at 19:45

array filter in python?

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