如何在 python 中逐行比較兩個矩陣？

Question

我有兩個列數相同但行數不同的矩陣，其中一個大很多。 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]]

都是numpy矩陣

我試圖找出 matA 的每一行出現在 matB 中的次數並將其放入數組中，因此在這種情況下數組將變為 arr = [1,2,1]因為 matA 的第一行出現在 mat 中一次，第二行出現兩次，最后一行只出現一次

Answer 1

在 numpy.array 中查找唯一行

在 numpy 中獲取唯一行位置的更快方法是什么

這是一個解決方案：

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...