给定坐标列表，将值添加到这些坐标，直到最短路径发生变化
Given a list of coordinates, add values to those coordinates until shortest path changes
所以我几天前提出了一个问题（这里是 ) basically what I'm trying to do is given a 3D array, and a list of coordinates, add values to those coordinates until shortest path changes, the user Mohammed Kashif 帮我解决了我的问题，但那是一个二维坐标列表，现在我想做的是做同样的事情但是使用 3D 坐标列表，第一个坐标列表与前 5 个数组和第二个坐标列表与接下来的 5 个数组。这就是我尝试的：
（这只是一个例子，我的坐标列表，3D数组可以有更多的元素）
import numpy as np
import networkx as nx
from copy import deepcopy
arr = np.array ([[[ 0., 303., 43., 26., 20.],
[ 0., 0., 0., 0., 0.],
[ 0., 21., 0., 8., 0.],
[ 0., 47., 5., 0., 4.],
[ 0., 35., 0., 1., 0.]],
[[ 0., 356., 40., 32., 49.],
[ 0., 0., 0., 0., 0.],
[ 0., 36., 0., 1., 0.],
[ 0., 25., 1., 0., 1.],
[ 0., 40., 0., 3., 0.]],
[[ 0., 372., 27., 34., 44.],
[ 0., 0., 0., 0., 0.],
[ 0., 37., 0., 3., 0.],
[ 0., 41., 8., 0., 1.],
[ 0., 34., 0., 6., 0.]],
[[ 0., 300., 46., 37., 46.],
[ 0., 0., 0., 0., 0.],
[ 0., 40., 0., 1., 0.],
[ 0., 48., 2., 0., 5.],
[ 0., 43., 0., 2., 0.]],
[[ 0., 321., 42., 22., 22.],
[ 0., 0., 0., 0., 0.],
[ 0., 42., 0., 3., 0.],
[ 0., 20., 3., 0., 5.],
[ 0., 20., 0., 9., 0.]],
[[ 0., 319., 48., 21., 39.],
[ 0., 0., 0., 0., 0.],
[ 0., 29., 0., 0., 1.],
[ 0., 38., 0., 0., 7.],
[ 0., 40., 1., 5., 0.]],
[[ 0., 374., 46., 25., 28.],
[ 0., 0., 0., 0., 0.],
[ 0., 25., 0., 0., 2.],
[ 0., 44., 0., 0., 6.],
[ 0., 44., 2., 9., 0.]],
[[ 0., 341., 34., 21., 49.],
[ 0., 0., 0., 0., 0.],
[ 0., 27., 0., 0., 9.],
[ 0., 25., 0., 0., 8.],
[ 0., 49., 1., 1., 0.]],
[[ 0., 310., 30., 44., 47.],
[ 0., 0., 0., 0., 0.],
[ 0., 34., 0., 0., 2.],
[ 0., 21., 0., 0., 8.],
[ 0., 37., 9., 8., 0.]],
[[ 0., 321., 27., 44., 31.],
[ 0., 0., 0., 0., 0.],
[ 0., 21., 0., 0., 5.],
[ 0., 41., 0., 0., 1.],
[ 0., 41., 1., 5., 0.]]])
n = 5 #Number of array for each path list and coordinates list
graphs = []
pathaux = []
for i in arr:
graphs.append(nx.from_numpy_array(i, create_using = nx.DiGraph)) #Create graphs from numpy array
for graph in graphs:
pathaux.append(nx.shortest_path(graph, 0, 1, weight = 'weight')) #Find the shortest path
path = [pathaux[i: i+n] for i in range(0, len(pathaux), n)] #Creates a 3D list "separating" for each 5 arrays
print(path)
#path = [[[0, 4, 3, 2, 1], [0, 3, 1], [0, 2, 1], [0, 3, 2, 1], [0, 3, 1]], [[0, 3, 4, 2, 1], [0, 4, 2, 1], [0, 3, 1], [0, 2, 4, 3, 1], [0, 2, 1]]]
#path[0] = [[0, 4, 3, 2, 1], [0, 3, 1], [0, 2, 1], [0, 3, 2, 1], [0, 3, 1]] #Shortest path for the first 5 arrays
#path[1] = [[0, 3, 4, 2, 1], [0, 4, 2, 1], [0, 3, 1], [0, 2, 4, 3, 1], [0, 2, 1]] #Shortest path for the next 5 arrays
#Here are the 3D list of coordinates, the first two pair is for the first 5 array, the next four pairs are for the other 5 arrays
coordinates = [[[4, 3], [3, 2]], [[3, 4], [4, 2], [2, 4], [4, 3]]]
for i in coordinates:
for x,y in i:
# Make deepcopies of path and arr
# For the first iteration, set newpath = path
new_path = deepcopy(pathaux)
temp_arr = deepcopy(arr)
# Set counter for each coordinate to zero
cnt = 0
# Iterate till a change in path is observed
while pathaux == new_path:
# Add 1 to x,y
temp_arr[:, x, y] = temp_arr[:, x, y] + 1
# Increment the counter
cnt += 1
# Reconstruct the graph and shortest path
temp_graph = []
new_path = []
for i in temp_arr:
temp_graph.append(nx.from_numpy_array(i, create_using = nx.DiGraph))
for graph in temp_graph:
new_path.append(nx.shortest_path(graph, 0, 1, weight = 'weight'))
#If we are out of the loop, this means that
# the shortest path has changed. Print the details.
print("For coordinates X={} and Y={} the change is at {}".format(x, y, cnt))
这是这段代码的输出
#For the first 5 arrays
For coordinates X=4 and Y=3 the change is at 3
For coordinates X=3 and Y=2 the change is at 1
#For the next 5 arrays
For coordinates X=3 and Y=4 the change is at 1
For coordinates X=4 and Y=2 the change is at 1
For coordinates X=2 and Y=4 the change is at 1
For coordinates X=4 and Y=3 the change is at 3
如您所见，坐标是正确的，但更改的值是错误的，如果我手动执行，将值添加到坐标直到我的列表更改这是我的真实输出：
#For the first 5 arrays
For coordinates X=4 and Y=3 the change is at 5
For coordinates X=3 and Y=2 the change is at 6
#For the next 5 arrays
For coordinates X=3 and Y=4 the change is at 1
For coordinates X=4 and Y=2 the change is at 1
For coordinates X=2 and Y=4 the change is at 3
For coordinates X=4 and Y=3 the change is at 3
我考虑过将 arr 从 3D 数组重塑为形状为 (2, 5, 5, 5) 的 4D 数组，但我相信我必须做一个双 for-loop 所以它更慢，我知道这是一个很长的问题，但我不知道如何使具有第一个最短路径和第一个坐标列表的相应前 5 个数组一起工作，并为接下来的五个具有相应最短路径和坐标的数组做同样的事情。再次抱歉，问题很长，如有任何帮助，我们将不胜感激，谢谢！
你应该单独考虑分区。对现有代码应用以下更改，它应该可以工作：
...
arr_partition = [arr[i: i+n] for i in range(0, len(arr), n)]
j = 0
for i in coordinates:
print("The tree is", j)
for x, y in i:
new_path = deepcopy(path[j])
temp_arr = deepcopy(arr_partition[j])
cnt = 0
while path[j] == new_path:
...
print()
j += 1
...
输出：
The tree is 0
For coordinates X=4 and Y=3 the change is at 5
For coordinates X=3 and Y=2 the change is at 6
The tree is 1
For coordinates X=3 and Y=4 the change is at 1
For coordinates X=4 and Y=2 the change is at 1
For coordinates X=2 and Y=4 the change is at 3
For coordinates X=4 and Y=3 the change is at 3
所以我几天前提出了一个问题（这里是
import numpy as np
import networkx as nx
from copy import deepcopy
arr = np.array ([[[ 0., 303., 43., 26., 20.],
[ 0., 0., 0., 0., 0.],
[ 0., 21., 0., 8., 0.],
[ 0., 47., 5., 0., 4.],
[ 0., 35., 0., 1., 0.]],
[[ 0., 356., 40., 32., 49.],
[ 0., 0., 0., 0., 0.],
[ 0., 36., 0., 1., 0.],
[ 0., 25., 1., 0., 1.],
[ 0., 40., 0., 3., 0.]],
[[ 0., 372., 27., 34., 44.],
[ 0., 0., 0., 0., 0.],
[ 0., 37., 0., 3., 0.],
[ 0., 41., 8., 0., 1.],
[ 0., 34., 0., 6., 0.]],
[[ 0., 300., 46., 37., 46.],
[ 0., 0., 0., 0., 0.],
[ 0., 40., 0., 1., 0.],
[ 0., 48., 2., 0., 5.],
[ 0., 43., 0., 2., 0.]],
[[ 0., 321., 42., 22., 22.],
[ 0., 0., 0., 0., 0.],
[ 0., 42., 0., 3., 0.],
[ 0., 20., 3., 0., 5.],
[ 0., 20., 0., 9., 0.]],
[[ 0., 319., 48., 21., 39.],
[ 0., 0., 0., 0., 0.],
[ 0., 29., 0., 0., 1.],
[ 0., 38., 0., 0., 7.],
[ 0., 40., 1., 5., 0.]],
[[ 0., 374., 46., 25., 28.],
[ 0., 0., 0., 0., 0.],
[ 0., 25., 0., 0., 2.],
[ 0., 44., 0., 0., 6.],
[ 0., 44., 2., 9., 0.]],
[[ 0., 341., 34., 21., 49.],
[ 0., 0., 0., 0., 0.],
[ 0., 27., 0., 0., 9.],
[ 0., 25., 0., 0., 8.],
[ 0., 49., 1., 1., 0.]],
[[ 0., 310., 30., 44., 47.],
[ 0., 0., 0., 0., 0.],
[ 0., 34., 0., 0., 2.],
[ 0., 21., 0., 0., 8.],
[ 0., 37., 9., 8., 0.]],
[[ 0., 321., 27., 44., 31.],
[ 0., 0., 0., 0., 0.],
[ 0., 21., 0., 0., 5.],
[ 0., 41., 0., 0., 1.],
[ 0., 41., 1., 5., 0.]]])
n = 5 #Number of array for each path list and coordinates list
graphs = []
pathaux = []
for i in arr:
graphs.append(nx.from_numpy_array(i, create_using = nx.DiGraph)) #Create graphs from numpy array
for graph in graphs:
pathaux.append(nx.shortest_path(graph, 0, 1, weight = 'weight')) #Find the shortest path
path = [pathaux[i: i+n] for i in range(0, len(pathaux), n)] #Creates a 3D list "separating" for each 5 arrays
print(path)
#path = [[[0, 4, 3, 2, 1], [0, 3, 1], [0, 2, 1], [0, 3, 2, 1], [0, 3, 1]], [[0, 3, 4, 2, 1], [0, 4, 2, 1], [0, 3, 1], [0, 2, 4, 3, 1], [0, 2, 1]]]
#path[0] = [[0, 4, 3, 2, 1], [0, 3, 1], [0, 2, 1], [0, 3, 2, 1], [0, 3, 1]] #Shortest path for the first 5 arrays
#path[1] = [[0, 3, 4, 2, 1], [0, 4, 2, 1], [0, 3, 1], [0, 2, 4, 3, 1], [0, 2, 1]] #Shortest path for the next 5 arrays
#Here are the 3D list of coordinates, the first two pair is for the first 5 array, the next four pairs are for the other 5 arrays
coordinates = [[[4, 3], [3, 2]], [[3, 4], [4, 2], [2, 4], [4, 3]]]
for i in coordinates:
for x,y in i:
# Make deepcopies of path and arr
# For the first iteration, set newpath = path
new_path = deepcopy(pathaux)
temp_arr = deepcopy(arr)
# Set counter for each coordinate to zero
cnt = 0
# Iterate till a change in path is observed
while pathaux == new_path:
# Add 1 to x,y
temp_arr[:, x, y] = temp_arr[:, x, y] + 1
# Increment the counter
cnt += 1
# Reconstruct the graph and shortest path
temp_graph = []
new_path = []
for i in temp_arr:
temp_graph.append(nx.from_numpy_array(i, create_using = nx.DiGraph))
for graph in temp_graph:
new_path.append(nx.shortest_path(graph, 0, 1, weight = 'weight'))
#If we are out of the loop, this means that
# the shortest path has changed. Print the details.
print("For coordinates X={} and Y={} the change is at {}".format(x, y, cnt))
这是这段代码的输出
#For the first 5 arrays
For coordinates X=4 and Y=3 the change is at 3
For coordinates X=3 and Y=2 the change is at 1
#For the next 5 arrays
For coordinates X=3 and Y=4 the change is at 1
For coordinates X=4 and Y=2 the change is at 1
For coordinates X=2 and Y=4 the change is at 1
For coordinates X=4 and Y=3 the change is at 3
如您所见，坐标是正确的，但更改的值是错误的，如果我手动执行，将值添加到坐标直到我的列表更改这是我的真实输出：
#For the first 5 arrays
For coordinates X=4 and Y=3 the change is at 5
For coordinates X=3 and Y=2 the change is at 6
#For the next 5 arrays
For coordinates X=3 and Y=4 the change is at 1
For coordinates X=4 and Y=2 the change is at 1
For coordinates X=2 and Y=4 the change is at 3
For coordinates X=4 and Y=3 the change is at 3
我考虑过将 arr 从 3D 数组重塑为形状为 (2, 5, 5, 5) 的 4D 数组，但我相信我必须做一个双 for-loop 所以它更慢，我知道这是一个很长的问题，但我不知道如何使具有第一个最短路径和第一个坐标列表的相应前 5 个数组一起工作，并为接下来的五个具有相应最短路径和坐标的数组做同样的事情。再次抱歉，问题很长，如有任何帮助，我们将不胜感激，谢谢！
你应该单独考虑分区。对现有代码应用以下更改，它应该可以工作：
...
arr_partition = [arr[i: i+n] for i in range(0, len(arr), n)]
j = 0
for i in coordinates:
print("The tree is", j)
for x, y in i:
new_path = deepcopy(path[j])
temp_arr = deepcopy(arr_partition[j])
cnt = 0
while path[j] == new_path:
...
print()
j += 1
...
输出：
The tree is 0
For coordinates X=4 and Y=3 the change is at 5
For coordinates X=3 and Y=2 the change is at 6
The tree is 1
For coordinates X=3 and Y=4 the change is at 1
For coordinates X=4 and Y=2 the change is at 1
For coordinates X=2 and Y=4 the change is at 3
For coordinates X=4 and Y=3 the change is at 3