Question

我有一个锥体之间的连接点的嵌套列表。

a = [0,1]
b = [2,4]
c = [2,0]
d = [4,3]

e=[a,b,c,d]

我想编写一个程序，列出锥0的每个邻居，锥1的每个邻居...... 即我希望输出为[[1,2]，[0]，[0,4]，[4]，[2,3]]

我做了以下事情：

neighbour_list =[]
cone = 0

for element in e:

   while cone in element:

      if cone == element[0]:
        neighbour = element[1]
        neighbour_list.append(neighbour)

      else:
        neighbour = element[0]
        neighbour_list.append(neighbour)


      cone = cone + 1
 print(neighbour_list)

我的想法是确定包含锥0，然后是锥1，然后是锥2等的列表，并为每个列提取邻居（即元素[1]或元素[0]）并将其附加到neighbour_list。

我收到错误消息：“while element in element” - 类型（int）的参数不可迭代

怎么了？

Answer 1

我建议改用词典。

a = [0, 1]
b = [2, 4]
c = [2, 0]
d = [4, 3]

e = [a, b, c, d]
neighbour_list = {}

for x, y in e:
    neighbour_list.setdefault(x, [])
    neighbour_list[x].append(y)
    neighbour_list.setdefault(y, [])
    neighbour_list[y].append(x)
neighbour_list = list(neighbour_list.values())
print(neighbour_list)

运行它给出：

[[1, 2], [0], [4, 0], [4], [2, 3]]

Answer 2

使用描述性名称有助于使您的代码更易于阅读。

a = [0,1]
b = [2,4]
c = [2,0]
f = [4,6]
junctions=[a,b,c,d,f]
neighbour_list =[]

如果我们事先并不知道锥体是什么，我们需要找到路口的所有锥体

# find all the cones in the junctions
cones = list()
for junction in junctions:
    for cone in junction:
        cones.append(cone)
# The previous could be replaced with a list comprehension
# cones = [cone for junction in junctions for cone in junction]

如果您只想考虑使用结点的锥体

##cones = set(cones)

如果您想考虑结点中所示范围内的所有锥体，请使用

cones = range(min(cones), max(cones) + 1)

如果 cone 循环是外循环，那么构建列表会容易一些，因为我们试图找到每个圆锥的邻居：

# for each cone look through the junctions to find neighbors
for cone in sorted(cones):    #search for neighbors from the smallest to the largest cone
    neighbors = list()    #make a new list for each cone
    for junction in junctions:
        if junction[0] == cone:
            neighbors.append(junction[1])
        elif junction[1] == cone:
            neighbors.append(junction[0])
    neighbour_list.append(neighbors)

>>> print neighbour_list
[[1, 2], [0], [4, 0], [4], [2, 3, 6], [], [4]]
>>>

neighbour_list似乎缺少一些信息 - 你不能轻易分辨出邻居所在的锥形区域。您可以使用zip添加信息：

>>> print zip(sorted(cones), neighbour_list)
[(0, [1, 2]), (1, [0]), (2, [4, 0]), (3, [4]), (4, [2, 3, 6]), (5, []), (6, [4])]
>>>

或者程序中的细微更改可以添加信息neighbour_list

neighbour_list.append((cone,neighbors))

难以组合每个锥体的邻居

2 个答案: