Reputation: 235
What is the best way to check if an numpy array contains any element of another array?
example:
array1 = [10,5,4,13,10,1,1,22,7,3,15,9]
array2 = [3,4,9,10,13,15,16,18,19,20,21,22,23]`
I want to get a True
if array1
contains any value of array2
, otherwise a False
.
Upvotes: 17
Views: 20395
Reputation: 109696
Using Pandas, you can use isin
:
a1 = np.array([10,5,4,13,10,1,1,22,7,3,15,9])
a2 = np.array([3,4,9,10,13,15,16,18,19,20,21,22,23])
>>> pd.Series(a1).isin(a2).any()
True
And using the in1d numpy function(per the comment from @Norman):
>>> np.any(np.in1d(a1, a2))
True
For small arrays such as those in this example, the solution using set is the clear winner. For larger, dissimilar arrays (i.e. no overlap), the Pandas and Numpy solutions are faster. However, np.intersect1d
appears to excel for larger arrays.
Small arrays (12-13 elements)
%timeit set(array1) & set(array2)
The slowest run took 4.22 times longer than the fastest. This could mean that an intermediate result is being cached
1000000 loops, best of 3: 1.69 µs per loop
%timeit any(i in a1 for i in a2)
The slowest run took 12.29 times longer than the fastest. This could mean that an intermediate result is being cached
100000 loops, best of 3: 1.88 µs per loop
%timeit np.intersect1d(a1, a2)
The slowest run took 10.29 times longer than the fastest. This could mean that an intermediate result is being cached
100000 loops, best of 3: 15.6 µs per loop
%timeit np.any(np.in1d(a1, a2))
10000 loops, best of 3: 27.1 µs per loop
%timeit pd.Series(a1).isin(a2).any()
10000 loops, best of 3: 135 µs per loop
Using an array with 100k elements (no overlap):
a3 = np.random.randint(0, 100000, 100000)
a4 = a3 + 100000
%timeit np.intersect1d(a3, a4)
100 loops, best of 3: 13.8 ms per loop
%timeit pd.Series(a3).isin(a4).any()
100 loops, best of 3: 18.3 ms per loop
%timeit np.any(np.in1d(a3, a4))
100 loops, best of 3: 18.4 ms per loop
%timeit set(a3) & set(a4)
10 loops, best of 3: 23.6 ms per loop
%timeit any(i in a3 for i in a4)
1 loops, best of 3: 34.5 s per loop
Upvotes: 23
Reputation: 1819
You can use any
built-in function and list comprehension:
>>> array1 = [10,5,4,13,10,1,1,22,7,3,15,9]
>>> array2 = [3,4,9,10,13,15,16,18,19,20,21,22,23]
>>> any(i in array2 for i in array1)
True
Upvotes: 3
Reputation: 21315
You can try this
>>> array1 = [10,5,4,13,10,1,1,22,7,3,15,9]
>>> array2 = [3,4,9,10,13,15,16,18,19,20,21,22,23]
>>> set(array1) & set(array2)
set([3, 4, 9, 10, 13, 15, 22])
If you get result means there are common elements in both array.
If result is empty means no common elements.
Upvotes: 9