找到Python中基质子集的重新发生

Question

我有一个场景，其中矩阵输入为5*5，

4 4 4 1 1
4 4 4 0 2
4 4 4 5 1
4 4 4 2 3
4 4 4 4 4 

如果我给出任何输入，例如2 3或3 3，则应找到矩阵的重新出现。 一样

像3 * 3

，它在第1行中具有这样的矩阵子集

; -

4 4 4 
4 4 4
4 4 4

同样，如果我指定3 * 3，则矩阵的子集有3个出现。是否可以将其转换为Python？

Answer 1

假设您的原始5x5矩阵称为input_matrix。您可以做到这一点，以从上左上重复序列中找到给定的子序列的次数：

input_size = input_matrix.shape[0]

# However many rows and columns the sub-matrix should have.
num_rows = 3
num_cols = 3

# The sub-matrix to find the number of occurrences of.
orig_matrix = input_matrix[:num_rows, :num_cols]

num_occurrences = 0

# Iterate through all indices at positions where the sub-matrix could still fit 
# inside the input matrix.
for row_num in range(input_size - num_rows + 1):
    for col_num in range(input_size - num_cols + 1):

      # Get the sub-matrix at those indices
      this_matrix = input_matrix[row_num:row_num + num_rows, col_num:col_num + num_cols]

      # Get the difference between the original sub-matrix and the current submatrix.
      # If difference at all points is 0, the submatrices are the same.
      if np.count_nonzero(orig_matrix - this_matrix, axis=None) == 0:
          num_occurrences += 1

num_occurrences包含您想要的数字。我敢肯定，有更有效的方法可以做到这一点，但这就是我所拥有的。