Using only Python (no NumPy):
import itertools as IT
L = [ [1, 2, 3],
[4, 5, 6],
[7, 8, 9] ]
N = len(L)
d = dict()
for i,j in IT.product(range(N), repeat=2):
d.setdefault(j-i, []).append((i,j))
print([[L[i][j] for i,j in d[k]] for k in range(-N+1, N)])
# [[7], [4, 8], [1, 5, 9], [2, 6], [3]]
Or, even better, use Nemo's transformation (generalized to h x w shaped matrices:
L = [ [1, 2, 3,],
[4, 5, 6,],
[7, 8, 9,], ]
h, w = len(L), len(L[0])
print([[L[h-1-q][p-q]
for q in range(min(p, h-1), max(0, p-w+1)-1, -1)]
for p in range(h+w-1)])
# [[7], [4, 8], [1, 5, 9], [2, 6], [3]]
We can also put this code in a function for easier usage:
def diagonals(L):
"""
https://stackoverflow.com/a/31373955/190597 (unutbu)
>>> L = array([[ 0, 1, 2],
[ 3, 4, 5],
[ 6, 7, 8],
[ 9, 10, 11]])
>>> diagonals(L)
[[9], [6, 10], [3, 7, 11], [0, 4, 8], [1, 5], [2]]
"""
h, w = len(L), len(L[0])
return [[L[h - p + q - 1][q]
for q in range(max(p-h+1, 0), min(p+1, w))]
for p in range(h + w - 1)]
def antidiagonals(L):
"""
>>> L = array([[ 0, 1, 2],
[ 3, 4, 5],
[ 6, 7, 8],
[ 9, 10, 11]])
>>> antidiagonals(L)
[[0], [3, 1], [6, 4, 2], [9, 7, 5], [10, 8], [11]]
"""
h, w = len(L), len(L[0])
return [[L[p - q][q]
for q in range(max(p-h+1,0), min(p+1, w))]
for p in range(h + w - 1)]
