Question about finding the number of times an element appears in an array

Can someone explain why/how this code works in creating an array B which corresponds to the number of each integer from 1-9 in array A?

c=[1:9]
B=sum(A(:)==c, 1)

[Lewis]

E.g. for A=[1 2 4] it will create B = [1 1 0 1]

[Rik]

Combining unique with histc (or histcounts) might be either safer, or more performant for larger arrays than this setup.

Answer 1

[Chris]

A = randi(9,2)

A = 2×2

8 8 5 9

A(:) formats A in a column vector.

A(:)

ans = 4×1

8 5 8 9

A(:)==c compares the column to each element of c (possible because the dimensions of a and c are orthogonal), returning an nx9 matrix with ones where the comparison is true.

A(:)==1:9

ans = 4×9 logical array
   0   0   0   0   0   0   0   1   0
   0   0   0   0   1   0   0   0   0
   0   0   0   0   0   0   0   1   0
   0   0   0   0   0   0   0   0   1

Summing across dimension 1 (the default):

B=sum(A(:)==1:9)

B = 1×9

0 0 0 0 1 0 0 2 1

[Lewis]

Thank you !