"不使用numpy或zip找到矩阵的子矩阵"

Question

我试图找到一种方法来找到数组X的子矩阵，其中用户提供了输入I和j，如下所示：

def submatrix(X, i, j):

预期的输出应该是矩阵X，没有行i和列j

示例：

X = [[1, 2, 3],
     [4, 5, 6],
     [7, 8, 9]]

子矩阵（X，1，1）

    [[1, 3],
     [7, 9]]

我已经试着自己解决它，但显然没有设法做到这一点，不知道从哪里开始。因此请求帮助

Answer 1

怎么样

def submatrix(X, i, j):
    return [[elem for x, elem in enumerate(row) if x != i]
            for y, row in enumerate(X) if y != j]

X = [[1, 2, 3],
     [4, 5, 6],
     [7, 8, 9]]

print(submatrix(X, 1, 1))

Answer 2

使用for loop：

X = [[1, 2, 3],
     [4, 5, 6],
     [7, 8, 9]]

def submatrix(X, i, j):
    X=X[:]
    del(X[i]) # delete the row
    for n in range(len(X)):
        del(X[n][j])  # delete the column elements of the rows
    return X

X_new = submatrix(X, 1, 1)
[[1, 3], [7, 9]]

Answer 3

def submatrix(x, i, j):
    matrix = []
    for row in x[:i] + x[i+1:]:
        matrix.append(row[:j] + row[j+1:])
    return matrix

示例

>>> submatrix([[1, 2, 3],
               [4, 5, 6],
               [7, 8, 9]], 1, 1)
[[1, 3],
 [7, 9]]

>>> submatrix([[1, 2, 3, 4],
               [5, 6, 7, 8],
               [2, 4, 5, 9],
               [9, 6, 4, 3]], 2, 1)
[[1, 3, 4],
 [5, 7, 8],
 [9, 4, 3]]