Compare a 2d array to an array Python

Question

I have an 2D array:

2d_arr = [[a,b,c,d],[a,b],[c,d],[a,f]]

And another array:

arr = [a,w,b,x]

I want to compare every element in the 2D array (2d_arr) with the array (arr) and get the output as a new 2D array like this: (if the 2D array elements match the array put 1 else 0)

[a,w,b,x]
[1,0,1,0]
[1,0,1,0]
[0,0,0,0]
[1,0,0,0]

I have tried the follow:

for i in range(len(2d_arr)):
   for j in range(len(2d_arr[i])):
      if 2d_arr[i][j] == arr[i]
         .....

I know the arr[i] from the last line is wrong but how would i iterate??

Dima Tisnek · Accepted Answer · 2018-05-08 02:52:48Z

2

You can use a list comprehension:

arr_2d = [['a','b','c','d'],['a','b'],['c','d'],['a','f']]
arr = ['a','w','b','x']

[[int(x in a) for x in arr] for a in arr_2d]

[[1, 0, 1, 0], [1, 0, 1, 0], [0, 0, 0, 0], [1, 0, 0, 0]]

edited May 8, 2018 at 2:52

Dima Tisnek

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answered Apr 26, 2018 at 4:33

Julien

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