你如何在二维列表中找到一个项目的邻居？

Question

[
[a,b,c,d],
[e,f,g,h],
[i,j,k,l]
]

例如。 g 正在接触 b,c,d,f,h,j,k,l，a 正在接触 e,f,b。我尝试了一种方法，我把它列成一个大清单并做了一些垃圾，但它崩溃了，tbh。如果有兴趣，这是代码：

import math

big_list = []
x= [
["a", "b", "c", "d", "e", "f"],
["g", "h", "i", "j", "k", "l"],
["m", "n", "o", "p", "q", "r"],
["s", "t", "u", "v", "w", "x"],
["y", "z", "1", "2", "3", "4"]
]

#def find_neighbours(num):
for y in x:
    for element in y:
        big_list.append(element)

def find_row(element):
    columns = len(x)
    loc = big_list.index(element)
    exact_row = (loc)/columns
    return(math.floor(exact_row))

def find_column(element):
    row = find_row(element)
    exact_column = x[row]
    return(exact_column)

print(find_column("x"))

列表可以是任意大小，并不总是正方形。

Answer 1

找到元素的行索引和列索引。然后，您可以定义要检查的行/列差异列表，以生成邻居的索引。最后，您可以遍历这些差异来找到邻居。

from itertools import product

board = [
    ['a','b','c','d'],
    ['e','f','g','h'],
    ['i','j','k','l']
]

target = 'a'

DELTAS = [x for x in product((-1, 0, 1), repeat=2) if x != (0, 0)]

for row_idx in range(len(board)):
    for col_idx in range(len(board[0])):
        if board[row_idx][col_idx] == target:
            row = row_idx
            col = col_idx
            break

neighbors = []
for row_delta, col_delta in DELTAS:
    if 0 <= row + row_delta < len(board) and 0 <= col + col_delta < len(board[0]):
        neighbors.append(board[row + row_delta][col + col_delta])

print(neighbors)

这输出：

['b', 'e', 'f']

Answer 2

基本上涉及两个步骤：

查找目标项的索引（如果有的话）。
确定其周围有效项的索引。

这是一个文字实现：

m = [['a', 'b', 'c', 'd', 'e', 'f'],
     ['g', 'h', 'i', 'j', 'k', 'l'],
     ['m', 'n', 'o', 'p', 'q', 'r'],
     ['s', 't', 'u', 'v', 'w', 'x'],
     ['y', 'z', '1', '2', '3', '4']]

width = len(m[0])
height = len(m)

target = 'a'
x = None
for y in range(height):
    try:
        x = m[y].index(target)
    except ValueError:
        continue
    else:
        break  # Found

if x is not None:
    print(f'Target {target!r} found in position [{x}, {y}]')
else:
    raise RuntimeError(f'Target {target!r} not found.')

neighbours = []
for j in range(y-1, y+2):
    if -1 < j < height:
        for i in range(x-1, x+2):
            if -1 < i < width and (i, j) != (x, y):
                neighbours.append((i, j))

print('neighbours:', neighbours)

Answer 3

# How do you find the neighbours of an item in a 2D list?

import numpy as np


def neighbors(array, item):
    item_loc = tuple(np.asarray(np.where(array == item)).T[0].tolist())
    s = tuple(np.array(array.shape) - 1)
    n = np.array([-1, 0, +1])
    rows = item_loc[0] + n
    cols = item_loc[1] + n
    neighbor_loc = [
        (x, y) for x in rows for y in cols if (0 <= x <= s[0]) & (0 <= y <= s[1])
    ]
    neighbor_loc.remove(item_loc)
    print(
        f"The neighbors of {item} are {array[tuple(np.transpose(neighbor_loc))].tolist()}."
    )


array = np.array(
    [
        ["a", "b", "c", "d", "e", "f"],
        ["g", "h", "i", "j", "k", "l"],
        ["m", "n", "o", "p", "q", "r"],
        ["s", "t", "u", "v", "w", "x"],
        ["y", "z", "1", "2", "3", "4"],
    ]
)
item = "r"
neighbors(array, item)

首先找到物品的位置。使用形状来确定阵列的边缘。确定具有邻居的行和列，但删除任何不存在的位置。然后删除项目位置。最后，检索邻居。

此代码假定数组中的所有项目都是唯一的，并且所请求的项目存在于数组中。

你如何在二维列表中找到一个项目的邻居？

How do you find the neighbours of an item in a 2D list?

python

python-3.x