如何获得通过给定点的嵌套列表的对角线 - Python

Question

我有一个这样的二维列表：

thelist=[[0,0,0,0,0,0],
         [1,0,0,0,0,0],
         [0,1,0,0,0,0],
         [0,0,1,0,0,0], # (3,2) is the 1 in this row
         [0,0,0,1,0,0],
         [0,0,0,0,1,0]]

我正在尝试获取通过给定坐标的 2d 列表的 2 条对角线。在上面的列表中，如果坐标是(3,2)，那么这两个列表就是[1,1,1,1,1] 和[0,0,1,0,0,0]。

我已经看到了可以获取所有对角线或特定对角线的解决方案，但我无法在线找到可以获取通过一个点的对角线的解决方案。这是我尝试过的：

def find_diagonals(thelist,coor):
    twodiag=[[],[]]
    coor1=[coor[0]-min(coor),coor[1]-min(coor)]
    coor2=[coor[0]-min(coor),coor[1]+min(coor)]
    while True:
       try: 
           twodiag[0].append(thelist[coor1[0]][coor1[1]]) 
           coor1[0]+=1
           coor1[1]+=1
       except IndexError:
           break

    while True:
       try: 
           twodiag[1].append(thelist[coor2[0]][coor2[1]]) 
           coor2[0]+=1
           coor2[1]-=1
       except IndexError:
           break
    return twodiag

但是，它只正确返回一个对角线。如何修复此代码或以其他方式解决问题？如果有任何不清楚的地方，我很乐意在 cmets 中回答。谢谢！

Answer 1

使用numpy的解决方案：

import numpy as np
thelist=[[0,0,0,0,0,0],
         [1,0,0,0,0,0],
         [0,1,0,0,0,0],
         [0,0,1,0,0,0], # (3,2) is the 1 in this row
         [0,0,0,1,0,0],
         [0,0,1,0,1,0]]
lst = matrix = np.array(thelist)

i, j = 3, 2 #position of element
major = np.diagonal(lst, offset=(j - i))
print(major)

minor = np.diagonal(np.rot90(lst), offset=-lst.shape[1] + (j + i) + 1)
print(minor)

输出：

[1 1 1 1 1]
[0 0 0 1 0 0]

Answer 2

width, height = len(thelist[0]), len(thelist)
size = max(width, height)
valid_x, valid_y = range(width), range(height)

pos = (3,2)
x, y = pos

diag1 = [thelist[i][i-x+y] for i in range(size) if i-x+y in valid_y]
diag2 = [thelist[i][x+y-i] for i in range(size) if x+y-i in valid_y]
print (diag1, diag2)
# [1, 1, 1, 1, 1] [0, 0, 0, 1, 0, 0]

Answer 3

最好的解决方案是简单的线性函数：

def find_diagonals(input_list, coordinates):
    diagonals = [[], []]
    row_coordinate, column_coordinates = coordinates
    for row_number, row in enumerate(input_list):
        diagonal_point = column_coordinates-row_coordinate+row_number
        if 0 <= diagonal_point <= len(row):
            diagonals[0].append(row[diagonal_point])
        diagonal_point = column_coordinates+row_coordinate-row_number
        if 0 <= diagonal_point <= len(row):
            diagonals[1].append(row[diagonal_point])
    return diagonals

Answer 4

你可以这样计算偏移量：

def get_diagonals(alist, coordinates):
    # Diagonal 1
    start = coordinates
    while start[0] > 0 and start[1] > 0:
        start = (start[0] - 1, start[1] - 1)

    diag1 = []
    index = start
    while index[0] < len(alist) and index[1] < len(alist[0]):
        diag1.append(alist[index[0]][index[1]])
        index = (index[0] + 1, index[1] + 1)

    # Diagonal 2
    start = coordinates
    while start[0] < len(alist) - 1 and start[1] > 0:
        start = (start[0] + 1, start[1] - 1)

    diag2 = []
    index = start
    while index[0] >= 0 and index[1] < len(alist[0]):
        diag2.append(alist[index[0]][index[1]])
        index = (index[0] - 1, index[1] + 1)

    return diag1, diag2


thelist=[[0,0,0,0,0,0],
         [1,0,0,0,0,0],
         [0,1,0,0,0,0],
         [0,0,1,0,0,0], # (3,2) is the 1 in this row
         [0,2,0,1,0,0],
         [3,0,0,0,1,0]]

coord = (3,2)

get_diagonals(thelist, coord)
> ([1, 1, 1, 1, 1], [3, 2, 1, 0, 0, 0])

代码是直截了当的，并逐一计算矩阵中对角线上的位置。

编辑：修正了一个错误