如何在 python 中逐行比较两个矩阵？

Question

I have two matrices with the same number of columns but a different number of rows, one is a lot larger. matA = [[1,0,1],[0,0,0],[1,1,0]] , matB = [[0,0,0],[1,0,1],[0,0,0],[1,1,1],[1,1,0]]

Both of them are numpy matrices

I am trying to find how many times each row of matA appears in matB and put that in an array so the array in this case will become arr = [1,2,1] because the first row of matA appeared one time in mat, the second row appeared two times and the last row only one time

Answer 1

Find unique rows in numpy.array

What is a faster way to get the location of unique rows in numpy

Here is a solution:

import numpy as np

A = np.array([[1,0,1],[0,0,0],[1,1,0]])

B = np.array([[0,0,0],[1,0,1],[0,0,0],[1,1,1],[1,1,0]])

# stack the rows, A has to be first
combined = np.concatenate((A, B), axis=0) #or np.vstack

unique, unique_indices, unique_counts = np.unique(combined,
                                                  return_index=True,
                                                  return_counts=True,
                                                  axis=0) 

print(unique)
print(unique_indices)
print(unique_counts)

# now we need to derive your desired result from the unique
# indices and counts

# we know the number of rows in A
n_rows_in_A = A.shape[0]

# so we know that the indices from 0 to (n_rows_in_A - 1)
# in unique_indices are rows that appear first or only in A

indices_A = np.nonzero(unique_indices < n_rows_in_A)[0] #first 
#indices_A1 = np.argwhere(unique_indices < n_rows_in_A) 
print(indices_A)
#print(indices_A1)

unique_indices_A = unique_indices[indices_A]
unique_counts_A = unique_counts[indices_A]

print(unique_indices_A)
print(unique_counts_A)

# now we need to subtract one count from the unique_counts
# that's the one occurence in A that we are not interested in.

unique_counts_A -= 1
print(unique_indices_A)
print(unique_counts_A)

# this is nearly the result we want
# now we need to sort it and account for rows that are not
# appearing in A but in B

# will do that later...