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
.
Use numpy. isin() to return a boolean array of the same shape as Y that is True where an element of Y is in X and False otherwise. Use numpy. in1d() to test whether each element of a 1-D array X is also present in a second array Y.
Use the inbuilt ES6 function some() to iterate through each and every element of first array and to test the array. Use the inbuilt function includes() with second array to check if element exist in the first array or not. If element exist then return true else return false.
To check if two NumPy arrays A and B are equal: Use a comparison operator (==) to form a comparison array. Check if all the elements in the comparison array are True.
Using Numpy array, we can easily find whether specific values are present or not. For this purpose, we use the “in” operator. “in” operator is used to check whether certain element and values are present in a given sequence and hence return Boolean values 'True” and “False“.
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
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.
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