Question

我有2个列表：

edges = [[0,1],[0,2],[0,3],[1,2],[1,3]]
weight = [10,8,7,3,7]

edges代表连接2个节点以及相应权重的边的列表。

对于给定的起始节点，如edges [i] [0]，我想根据给定的权重选择最短的连接点，因此在这种情况下，结果将如下所示：

connect = [[0,3],[1,2]]
weight = [7,3]

因为在连接到0 3的所有节点中，最接近的是1，对于1，2是最接近的。

我无法提出问题，不胜感激！

Answer 1

edges = [[0,1],[0,2],[0,3],[1,2],[1,3]]
weight = [10,8,7,3,7]

connect = []
wght = []

In [8]: for i in set(e[0] for e in edges):
...:     temp = [(a, b) for a, b in zip(edges, weight) if a[0] == i]
...:     temp = min(temp, key=lambda x: x[1])
...:     connect += [temp[0]]
...:     wght += [temp[1]]

In [9]: connect
Out[9]: [[0, 3], [1, 2]]

In [10]: wght
Out[10]: [7, 3]

以防万一：

In [20]: [min([(a, b) for a, b in zip(edges, weight) if a[0] == i], key=lambda x: x[1]) fo
...: r i in set([e[0] for e in edges])]
Out[20]: [([0, 3], 7), ([1, 2], 3)]

Answer 2

使用Pandas的另一种解决方案：

df = pd.DataFrame(edges, columns=['start','end'])
df['weight'] = weight
df.loc[df.groupby('start')['weight'].idxmin()]

结果为：

start   end weight
0       3   7
1       2   3

给定2d列表的单个值来简化列表

2 个答案: