Python 单元二维表邻域的确定
我有一个列表，类似于Python 单元二维表邻域的确定,python,matrix,Python,Matrix,我有一个列表，类似于 [[1,2,3，]，[4,5,6，]，[7,8,9]] 以图形方式表示为： 1 2 3 4 5 6 7 8 9 我正在寻找一种优雅的方法来检查单元格相邻区域的值，水平、垂直和对角。例如，[0][2]的邻居是[0][1]、[1][1]和[1][2]或数字2、5、6 现在我意识到，我可以做一次暴力攻击来检查每一个值： [i-1][j] [i][j-1] [i-1][j-1] [i+1][j] [i][j+1] [i+1][j+1] [i+1][j-1] [i-1][j+1]
[[1,2,3，]，[4,5,6，]，[7,8,9]]1 2 3
4 5 6
7 8 9
[i-1][j]
[i][j-1]
[i-1][j-1]
[i+1][j]
[i][j+1]
[i+1][j+1]
[i+1][j-1]
[i-1][j+1]
(x - 1, y - 1) (x, y - 1) (x + 1, y - 1)
(x - 1, y) (x, y) (x + 1, y)
(x - 1, y + 1) (x, y + 1) (x + 1, y + 1)
为了确保邻域适合有限矩阵，如果初始值（x，y）在边上，只需对邻域的坐标施加一个限制-矩阵大小。没有更干净的方法可以做到这一点。如果确实需要，可以创建一个函数：
def top(matrix, x, y):
try:
return matrix[x][y - 1];
except IndexError:
return None
我不知道它在您看来有多优雅，但它似乎在没有任何硬编码的情况下工作。这会生成所有索引：
for x_ in range(max(0,x-1),min(height,x+2)):
for y_ in range(max(0,y-1),min(width,y+2)):
if (x,y)==(x_,y_): continue
# do stuff with the neighbours
>>> a=[[1, 2, 3], [4, 5, 6], [7, 8, 9]]
>>> width=height=3
>>> x,y=0,2
>>> for x_ in range(max(0,x-1),min(height,x+2)):
... for y_ in range(max(0,y-1),min(width,y+2)):
... if (x,y)==(x_,y_): continue
... print a[x_][y_]
...
2
5
6
def neighboring( array ):
nn,mm = len(array), len(array[0])
offset = (0,-1,1) # 0 first so the current cell is the first in the gen
indices = ( (i,j) for i in range(nn) for j in range(mm) )
for i,j in indices:
all_neigh = ( (i+x,j+y) for x in offset for y in offset )
valid = ( (i,j) for i,j in all_neigh if (0<=i<nn) and (0<=j<mm) ) # -1 is a valid index in normal lists, but not here so throw it out
yield valid.next(), valid ## first is the current cell, next are the neightbors
for (x,y), neigh in neighboring( l ):
print l[x][y], [l[x][y] for x,y in neigh]
def相邻（数组）：
nn，mm=len（数组），len（数组[0]）
offset=（0，-1,1）#0 first，因此当前单元格是gen中的第一个单元格
指数=（（i，j）范围内的i（nn）范围内的j（mm））
对于指数中的i，j：
所有_neigh=（（i+x，j+y）表示偏移量中的x表示偏移量中的y）
如果（0可能您正在检查一个数独框，如果框为nxn，当前单元格为（x，y），则开始检查：
startingRow = x / n * n;
startingCol = y/ n * n
#“板”的大小
X=10
Y=10
邻域=λx，y:[（x2，y2）表示范围（x-1，x+2）内的x2
适用于范围内的y2（y-1，y+2）
如果（-1
假设您有一个正方形矩阵：
from itertools import product
size = 3
def neighbours(cell):
for c in product(*(range(n-1, n+2) for n in cell)):
if c != cell and all(0 <= n < size for n in c):
yield c
函数的作用可以可视化如下：
如果有人对选择直接（非对角）邻居的替代方法感到好奇，请看：
neighbors = [(x+a[0], y+a[1]) for a in
[(-1,0), (1,0), (0,-1), (0,1)]
if ( (0 <= x+a[0] < w) and (0 <= y+a[1] < h))]
neighbories=[（x+a[0]，y+a[1]），用于in
[(-1,0), (1,0), (0,-1), (0,1)]
if（（0感谢您提供有关邻居的重要提示。以下是此问题的运行代码：
def findNeighbours(l,elem):
#This try is for escaping from unbound error that happens
#when we try to iterate through indices that are not in array
try:
#Iterate through each item of multidimensional array using enumerate
for row,i in enumerate(l):
try:
#Identifying the column index of the givem element
column=i.index(elem)
except ValueError:
continue
x,y=row,column
# hn=list(((x,y+1),(x,y-1))) #horizontal neighbours=(x,y+/-1)
# vn=list(((x+1,y),(x-1,y))) #vertical neighbours=(x+/-1,y)
# dn=list(((x+1,y+1),(x-1,y-1),(x+1,y-1),(x-1,y+1))) #diagonal neighbours=(x+/-1,y+/-1)
#Creating a list with values that are actual neighbors for the extracted index of array
neighbours=[(x,y+1),(x,y-1),(x+1,y),(x-1,y),(x+1,y+1),(x-1,y-1),(x+1,y-1),(x-1,y+1)]
#Creating a universe of indices from given array
index_list=[(i,j) for i in range(len(l)) for j in range(len(l[i]))]
#Looping through index_list and nested loop for neighbours but filter for matched ones
# and extract the value of respective index
return_values=[l[index[0]][index[1]] for index in index_list for neighbour in neighbours if index==neighbour]
return return_values,neighbours
except UnboundLocalError:
return []
如果lambdas吓唬了你，你就在这里。但是lambdas让你的代码看起来很干净。@johniek_comp有一个非常干净的解决方案
k,l=(2,3)
x = (0,-1,+1)
y = (0,-1,+1)
cell_u = ((k+a,l+b) for a in x for b in y)
print(list(cell_u))
灵感来自前面的一个答案
可以使用min（）和max（）函数缩短计算时间：
width = 3
height = 3
[(x2, y2) for x2 in range(max(0, x-1), min(width, x+2))
for y2 in range(max(0, y-1), min(height, y+2))
if (x2, y2) != (x, y)]
您想要获取索引还是值？您想要一个可以对每个索引进行随机访问的函数，还是一个返回（val，neights，of，of，val）列表的函数结对？--仅仅获取索引对于一个优雅的解决方案来说太简单了，但是你真正想做的可能更有趣或者-我故意把这个问题留得相当笼统，这样人们就不会感到拘束了。@Kaizer.se:谢谢，我不确定它是哪一个，也懒得去查找或尝试它。+1这不是一个完整的解决方案，但有趣的是，这可能是搜索邻居的一种非常好的方法——这实际上不起作用。我正在做一个类似的解决方案，并意识到了这一点。负索引实际上用于按相反顺序查找索引。例如-1表示最后一个元素。我不明白这个答案，我将其与问题一起运行在Python2.7中的示例中，如何添加检查以确保坐标不超过矩阵的边界？如果“board”is10201x10201
？@CF84:那么常数X
和Y
需要为它们指定不同的值。您对此有什么问题吗？您的观点是什么？@CDspace编辑如何偏离帖子的原始意图？如果您指的是一行，它仍然可以用一个普通字符写在一行上函数。此外，自上次编辑以来，它还不是一行。至少大写的X和Y变量应该更改，因为它们很难识别。您好，这看起来不错。但是对于行数和列数不等的矩阵，如2*4数组，又如何呢？
neighbors = [(x+a[0], y+a[1]) for a in
[(-1,0), (1,0), (0,-1), (0,1)]
if ( (0 <= x+a[0] < w) and (0 <= y+a[1] < h))]
def findNeighbours(l,elem):
#This try is for escaping from unbound error that happens
#when we try to iterate through indices that are not in array
try:
#Iterate through each item of multidimensional array using enumerate
for row,i in enumerate(l):
try:
#Identifying the column index of the givem element
column=i.index(elem)
except ValueError:
continue
x,y=row,column
# hn=list(((x,y+1),(x,y-1))) #horizontal neighbours=(x,y+/-1)
# vn=list(((x+1,y),(x-1,y))) #vertical neighbours=(x+/-1,y)
# dn=list(((x+1,y+1),(x-1,y-1),(x+1,y-1),(x-1,y+1))) #diagonal neighbours=(x+/-1,y+/-1)
#Creating a list with values that are actual neighbors for the extracted index of array
neighbours=[(x,y+1),(x,y-1),(x+1,y),(x-1,y),(x+1,y+1),(x-1,y-1),(x+1,y-1),(x-1,y+1)]
#Creating a universe of indices from given array
index_list=[(i,j) for i in range(len(l)) for j in range(len(l[i]))]
#Looping through index_list and nested loop for neighbours but filter for matched ones
# and extract the value of respective index
return_values=[l[index[0]][index[1]] for index in index_list for neighbour in neighbours if index==neighbour]
return return_values,neighbours
except UnboundLocalError:
return []
k,l=(2,3)
x = (0,-1,+1)
y = (0,-1,+1)
cell_u = ((k+a,l+b) for a in x for b in y)
print(list(cell_u))
width = 3
height = 3
[(x2, y2) for x2 in range(max(0, x-1), min(width, x+2))
for y2 in range(max(0, y-1), min(height, y+2))
if (x2, y2) != (x, y)]