Numpy zeros 二维数组：替换特定索引处的元素

Question

For a function I have to write again for CodeSignal, I create an 'empty' matrix with numpy called 'result'. During the course of a for loop, I want to add 1s to certain elements of this zeros matrix:

matrix = [[True, False, False],
          [False, True, False],
          [False, False, False]]

matrix = np.array(matrix)                        ## input matrix 
(row, col) = matrix.shape
result = np.zeros((row,col), dtype=int)         ## made empty matrix of same size 

for i in range(0, row): 
    for j in range(0, col):
        mine = matrix[i,j],[i,j]
        if mine[0] == True:                     ##for indices in input matrix where element is called True..
            result[i+1,j+1][i+1,j+1] = 1        ##..replace neighbouring elements with 1 (under construction ;) ) 
print(result)

My very first problem comes with the last part, substituting elements at given indices with another value. Eg result[1,1][1,1] = 1 I always get the error

TypeError: object does not support item assignment

and this happened after setting np.zeros to various object types - int32, int8, complex, float64...

If I try:

Eg result[1,1][1,1] == 1

I get:

IndexError: invalid index to scalar variable.

So what is the way to change or add elements to 2d np arrays at specific locations?

Answer 1

It makes no sense t write:

matrix[i,j][i,j]

The matrix is a 2d array, so that means that matrix[i,j] is a scalar, not an array. Applying 0[i,j] is non-sensical.

You can implement this as:

for i in range(row): 
    for j in range(col):
        if matrix[i,j]:
            result[i+1,j+1] = 1

here you thus will "shift" the values of matrix one to the right, and one down. But then you better perform this with:

result[1:,1:] = matrix[:-1,:-1]

This then gives us:

>>> result
array([[0., 0., 0.],
       [0., 1., 0.],
       [0., 0., 1.]])