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list1 = [1,2,3,4,5,6]
list2 = [3, 5, 7, 9]
list(set(list1).intersection(list2))
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list1 = ['little','blue','widget']
list2 = ['there','is','a','little','blue','cup','on','the','table']
list3 = set(list1)&set(list2)
list4 = sorted(list3, key = lambda k : list1.index(k))
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>>> a = [1, 2, 3, 4]
>>> b = [2, 3, 4, 5]
>>> c = [3, 4, 5, 6]
>>> set(a) & set(b) & set(c)
{3, 4}
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# using numpy
import numpy as np
a = np.array([1,2,3,2,3,4,3,4,5,6])
b = np.array([7,2,10,2,7,4,9,4,9,8])
# using intersect1d
print(np.intersect1d(a,b))
# output
# [2 4]
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// Declare two array
let array1 = ['a', 'b', 'x', 'z'];
let array2 = ['k', 'x', 'c']
// Iterate through each element in the
// first array and if some of them
// include the elements in the second
// array then return true.
function findCommonElements3(arr1, arr2) {
return arr1.some(item => arr2.includes(item))
}
console.log(findCommonElements3(array1, array2))