CODEWITHSUNDEEP
pythonarrayslistnumpyscipy
Sayantan Mukherjee
Sayantan Mukherjee

Reputation: 205

Comparing rows of a multidimesional array in python

I have two numpy array

a= np.array([[2,2],[3,2],[4,2],[3,3],[5,3]])
b= np.array([[1,1],[1,3],[5,3]])

I want to compare a with b and return a-b such that :

a-b = array([[2,2],
             [3,2],
             [4,2],
             [3,3]])

I have tried doing :

[x for x in a if x not in b]

and it resulted in 

[array([2, 2]), array([3, 2]), array([4, 2])] # where clearly [3,3] is missing

I even tried comparing each row of both a and b within loop where it gave me an error 

The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()

Can anyone please help me to solve this problem ???

Upvotes: 1

Views: 82

Answers (2)

Divakar
Divakar

Reputation: 221754

Broadcasting based vectorized approach -

a[~((b[:,None,:] == a).all(2)).any(0)]

Using cdist from scipy.spatial.distance -

from scipy.spatial.distance import cdist

a[~(cdist(a,b)==0).any(1)]

Sample run -

In [89]: a
Out[89]: 
array([[2, 2],
       [3, 2],
       [4, 2],
       [3, 3],
       [5, 3]])

In [90]: b
Out[90]: 
array([[1, 1],
       [1, 3],
       [5, 3]])

In [91]: a[~((b[:,None,:] == a).all(2)).any(0)]
Out[91]: 
array([[2, 2],
       [3, 2],
       [4, 2],
       [3, 3]])

In [92]: a[~(cdist(a,b)==0).any(1)]
Out[92]: 
array([[2, 2],
       [3, 2],
       [4, 2],
       [3, 3]])

Upvotes: 2

Anand S Kumar
Anand S Kumar

Reputation: 91009

One way to do this would be to convert the numpy arrays into list of tuples and b to set of tuples then do the same list comprehension you used on them. Example -

In [1]: import numpy as np

In [2]: a= np.array([[2,2],[3,2],[4,2],[3,3],[5,3]])

In [3]: b= np.array([[1,1],[1,3],[5,3]])

In [18]: alist = list(map(tuple, a))

In [19]: bset = set(map(tuple, b))

In [20]: np.array([x for x in alist if x not in bset])
Out[20]:
array([[2, 2],
       [3, 2],
       [4, 2],
       [3, 3]])

Upvotes: 1

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