Get diagonal without using numpy?
matrix, python
Solution
To get the leading diagonal you could do
diag = [ mat[i][i] for i in range(len(mat)) ]
or even
diag = [ row[i] for i,row in enumerate(mat) ]
And play similar games for other diagonals. For example, for the counter-diagonal (top-right to bottom-left) you would do something like:
diag = [ row[-i-1] for i,row in enumerate(mat) ]
For other minor diagonals you would have to use `if` conditionals in the list comprehension, e.g.:
diag = [ row[i+offset] for i,row in enumerate(mat) if 0 <= i+offset < len(row)]
Problem
I'm trying to get the diagonal from a matrix in Python without using `numpy` (I really can't use it). Does someone here knows how to do it? Example of what I want to get: ``` get_diagonal ([[1,2,3,4],[5,6,7,8],[9,10,11,12]], 1, 1, 1) Result: [1, 6, 11] ``` Or like: ``` get_diagonal ([[1,2,3,4],[5,6,7,8],[9,10,11,12]], 1, 2, 1) Result: [2, 7, 12] ``` Until know I've tried a lot of stuff but doesn't work. ``` def obter_diagonal(matrix, line, column, direc): d = [] if direc == 1: for i in matrix: for j in i: if all(i == line, j == column): d.extend(matrix[i][j]) else: for i in matrix: for j in i: d.extend[len(matrix)-1-i][j] return d ``` If `direc==1` I need to get the diagonal that goes from left-> right, top-> bottom. If `direc==-1` need to get the diag that goes from right-> left, top->bottom.