Question

我正在尝试使用循环填充字典，问题是在完成条件后循环停止而不检查是否有更多可用条件要实现，

sec=[
     [1, 2, 4.0, 100.0], 
     [2, 3, 1.5, 500.0], 
     [2, 4, 10.0, 700.0], 
     [2, 5, 5.75, 1500.0], 
     [3, 4, 11.4, 200.0], 
     [4, 5, 10.5, 750.0], 
     [4, 6, 6.75, 550.0]]

我制作了这个列表，还有这本词典

graph={1: [1], 2: [2], 3: [3], 4: [4], 5: [5], 6: [6]} 我想要完成的是

graph = {1: ['2'],
         2: ['3', '4','5'],
         3: ['4'],
         4: ['5','6'],
         5: [],
         6: []}

它应该如何工作是它收集所有sec[x][0]作为字典的键，sec[n][1]作为字典中的值如果sec [x] [0]的值为1且sec [x] [1]的值为2，则将数字2作为键1的值添加到字典中我得到的代码就是这个

def magic(numList): #turns string into int
   s = ''.join(map(str, numList))
   return int(s)
for i in range(1,len(loc)+1): #len of loc is 6
    for n in range(0,len(loc)+1):
        if magic(graph[i])==sec[n][0]:
            graph[i].append(sec[n][1])

但它只会添加第一个值，然后索引n将停止计数，然后索引i继续运行，因此它不会检查键中的更多值

Answer 1

graph的初始定义对预期结果没有帮助。使用空列表初始化值，然后附加到一个简单的循环中：

>>> sec=[
     [1, 2, 4.0, 100.0], 
     [2, 3, 1.5, 500.0], 
     [2, 4, 10.0, 700.0], 
     [2, 5, 5.75, 1500.0], 
     [3, 4, 11.4, 200.0], 
     [4, 5, 10.5, 750.0], 
     [4, 6, 6.75, 550.0]]
>>> graph = {i:[] for i in range(1,7)}
>>> for x,y,*z in sec: graph[x].append(str(y))

>>> graph
{1: ['2'], 2: ['3', '4', '5'], 3: ['4'], 4: ['5', '6'], 5: [], 6: []}

Answer 2

使用collections模块中的defaultdict。见下文：

from collections import defaultdict

graph_dd = defaultdict(list)

for s in sec:
    from_, to_ = s[:2]
    graph_dd[from_].append(to_)
    # do-nothing reference to to_ creates empty list, so that all nodes are represented
    # in graph_dd; else 5 and 6 would not be created
    graph_dd[to_]

# convert to regular dict (not really necessary, but OP wanted a dict)
graph = dict(graph_dd.items())

for k,v in graph.items():
    print(k,':', v)

打印：

1 : [2]
2 : [3, 4, 5]
3 : [4]
4 : [5, 6]
5 : []
6 : []

Python for循环迭代

2 个答案: