one-liner to check if at least one item in list exists in another list?

Question

Lets say I have a list a=[1,2,3] And I want to know if at least one of the numbers in it exist in another list, like this one: b=[4,5,6,7,8,1] In other words, I want to know if 1,2 or 3 exist(s) in list b. I now I could do something like

def func(a, b):
    for i in a:
       if i in b:
          return True
    return False

But is there possibly a way to put that in one line to make things tidy?

score 83 · Answer 1 · answered May 19 '12 at 19:40

83

Python 2.6 and above:

def func(a, b):
  return not set(a).isdisjoint(b)

For 2.4 or 2.5:

def func(a, b):
  return len(set(a).intersection(b)) != 0

For 2.3 and below:

sudo apt-get update
sudo apt-get upgrade

;)

answered May 19 '12 at 19:40

Thomas

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19

upvote for 2.3 and below. that was funny :D...correct answer though – mkab Mar 18 '16 at 09:45
5

Upvote for the LOLs – Reed Jessen Oct 19 '16 at 23:30
2

probably time to update the 2.3 advice to `2.6 and below` or even `3.4 and below` – Davos Sep 25 '18 at 07:09

score 76 · Answer 2 · answered May 19 '12 at 19:39

76

a simple one-liner would be:

any(i in b for i in a)

answered May 19 '12 at 19:39

mata

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best pythonic solution – clg4 Jan 14 '17 at 20:13

Danica · Accepted Answer · 2012-05-19T20:05:25.880

There are many ways to do this. The most direct translation is:

any_in = lambda a, b: any(i in b for i in a)

You could also use various things involving sets, such as:

any_in = lambda a, b: bool(set(a).intersection(b))

(which depends on the elements of a being hashable, but if that's true, it'll probably be faster to make a set of the larger out of a and b for either of these approaches).

Edit: isdisjoint is better than intersection for Python 2.6 and above, as noted by various people below. Glad to learn about that. :)

score 19 · Answer 4 · answered May 19 '12 at 19:43

19

This is a set problem, not a list problem. With the right data type, the answer is often immediately obvious :-)

def func(a, b):
    return not set(a).isdisjoint(b)

answered May 19 '12 at 19:43

Raymond Hettinger

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@MattO'Brien the OP asks "I want to know if at least one of the numbers in it exist in another list" which logically reduces to a set problem. I'm intrigued by how in your example element a is not in the second list? It seems to be in it, _at least once_. I wonder if you are thinking about the count of elements which would be a `collections.Counter` problem – Davos Sep 25 '18 at 07:15
good point. I have no idea what I was thinking 2 years ago when I wrote that...I've deleted it... – tumultous_rooster Sep 27 '18 at 06:24
what is the right data type ? – FabioSpaghetti Jan 23 '19 at 13:07
@FabioSpaghetti The "set" datatype is more useful for this problem than a "list". – Raymond Hettinger Jan 30 '19 at 17:33

score 7 · Answer 5 · answered May 19 '12 at 19:39

7

By converting your lists to sets you can perform set operations on them. If the intersection is larger than 0, you have at least one element matching:

len(set(a) & set(b)) > 0

answered May 19 '12 at 19:39

Maehler

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Same idea, other variations: `bool( set(a) & set(b) )` or simply `set(a) & set(b)` – johnP Aug 18 '15 at 22:58
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not set(a).isdisjoint(set(b)) – jaboja Jul 02 '18 at 15:41

rabbidrabbit · Answer 6 · 2012-05-19T19:53:01.543

5

This should work.

def func(a, b):
    return any([i in b for i in a])

edited May 19 '12 at 19:53

answered May 19 '12 at 19:40

rabbidrabbit

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what is the output ? – FabioSpaghetti Jan 23 '19 at 14:16
error : 'Cell' object is not iterable – FabioSpaghetti Jan 23 '19 at 14:57

one-liner to check if at least one item in list exists in another list?

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