All Diagonal elements of a NXN matrix without using numpy in python

Question

Trying to get all the diagonal elements of a NXN matrix without using numpy,

This is different from Get diagonal without using numpy in Python

Please don't mark as duplicate.

Here is my snippet using two iteration, it will print all the diagonal element from (0,0) to (n,n). Could some one help me in improving my iteration process or any recursive way to do it.

#!/usr/bin/python 

matrix = [[1,2,3,4], 
          [2,3,4,5], 
          [3,4,5,6], 
          [4,5,6,7]] 

def get_matrix_diagonal(m): 
    col = len(m) 
    result = list() 
    row = 0 
    for c in range(col): 
        position = list() 
        position.append([row,c]) 
        if c > 0: 
            i = c 
            while i > 0: 
                x = i - 1 
                y = c - i + 1 
                position.append([y,x]) 
                i -= 1 
        result.append(position) 
    row = row + 1 
    cc = c 
    for i in range(row,col): 
        position = list() 
        y = i 
        x = c 
        position.append([y,x]) 
        j = x - y 
        while j > 0: 
            y = c - j + 1 
            x = cc - y + 1 
            position.append([y,x]) 
            j -= 1 
        cc += 1
        result.append(position)
    return result

for ls in get_matrix_diagonal(matrix):
    for co in ls:
        x,y = co[0],co[1],
        print matrix[x][y],
    print

Output:

I didn't vote for close, but really this would be better asked on codereview I think — Tim B, Dec 20 '13 at 09:34

John La Rooy · Answer 1 · 2013-12-20T05:53:02.907

6

>>> matrix = [[1,2,3,4], 
...           [2,3,4,5], 
...           [3,4,5,6], 
...           [4,5,6,7]] 
>>> N = 4
>>> [[matrix[y-x][x] for x in range(N) if 0<=y-x<N] for y in range(2*N-1)]
[[1], [2, 2], [3, 3, 3], [4, 4, 4, 4], [5, 5, 5], [6, 6], [7]]

edited Dec 20 '13 at 05:53

answered Dec 20 '13 at 05:46

John La Rooy

295,403
53
369
502

WoW, this is sweet... @gnibbler could you help me with recursion or is it possible to do with recursion ? – James Sapam Dec 20 '13 at 05:49
To solve with recursion, you'll need to find a way to break an NxN matrix into a smaller problem using an MxM matrix, so M – John La Rooy Dec 20 '13 at 05:51
Ok I ll try that, but i was not sure about the base case. – James Sapam Dec 20 '13 at 05:52
@yopy, the base case can be either N=1(trivial) or N=0(also trivial) – John La Rooy Dec 20 '13 at 05:53

score 2 · Answer 2 · answered Dec 20 '13 at 06:40

2

Here an recursive approach supposing your matrix is square:

def diagonals(m):
    if len(m) == 1: return m
    d = [[]] + diagonals([r[1:] for r in m[:-1]]) + [[]]
    r = [r[0] for r in m] + m[-1][1:]
    return [[r] + d for r, d in zip(r,d)]

answered Dec 20 '13 at 06:40

Hyperboreus

31,997
9
47
87

AWESOME, thank you so much. it works but i need to read to understand :) – James Sapam Dec 20 '13 at 06:44
1

Basically, each recursion removes the left most column and the bottom most line. – Hyperboreus Dec 20 '13 at 06:50

All Diagonal elements of a NXN matrix without using numpy in python

2 Answers2