一個NXN矩陣的所有對角線元素沒有蟒蛇

Question

試圖獲得一個NXN矩陣的所有對角線元素，而無需使用numpy的使用numpy的，

這是Get diagonal without using numpy in Python

不同，請不要標記爲重複。

這是我的代碼片斷使用兩次迭代，它將打印從（0,0）到（n，n）的所有對角線元素。有人可以幫助我改進我的迭代過程或任何遞歸方式來做到這一點。

#!/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 

輸出：

1 
2 2 
3 3 3 
4 4 4 4 
5 5 5 
6 6 
7 

Answer 1

>>> 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]] 

Answer 2

這裏的遞歸方法假設你的矩陣是方陣：

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)]