I have a dataset:

    list1         list2
    
0    [1,3,4]      [4,3,2]
    
1    [1,3,2]      [0,4,6]
    
2   [4,5,8]      NA
    
3    [6,3,7]     [8,2,3]

Is there a process where i can find the count of the common term for- each of the index,

Expected output: intersection_0, it will compare 0 of list1 with each of list2 and give output, intersection_1 which will compare 1 of list1 with each of list2

Expected_output:

Intersection_0   intersection_1    intersection_2     intersection_3
1                 2                      1                     1
1                 0                      1                     1
0                 0                      0                     0
1                 2                      0                     1

For intersection i was trying:

df['intersection'] = [len(set(a).intersection(b)) for a, b in zip(df1.list1, df1.list2)]

Is there a better way or faster way to achieve this? Thank you in advance

CodePudding user response：

The double loop would go like this:

intersections = []
for l2 in df['list2']:
    intersection = []
    for l1 in df['list1']:
        try:
            i = len(np.intersect1d(l1,l2))
        except:
            i = 0
        intersection.append(i)
    intersections.append(intersection)
    
out = (pd.DataFrame(intersections))

Output:

   0  1  2  3
0  2  2  1  1
1  1  0  1  1
2  0  0  0  0
3  1  2  1  1