Question

我有一个列表l1，如下所示：

l1 = [[(1,"INC1"), (1, "INC2")],[(3, "INC4"),(4, "INC6")]]

我有一个列表names，如下所示：

names = ['John', 'Marcia']

我有另一个列表label_issue_list，如下所示：

label_issue_list = [["Problem1","Problem1"],["Problem3","Problem4"]]

所有三个列表的索引都连接起来，如下面的输出所示。

我试图使用所有这三个列表并获得一个字典输出，如下所示：

{
"John" : {
            "Problem1" : 2,
          "tick_info" : {"1": ["INC1", "INC2"]}

            },
"Marcia" : {
            "Problem3" : 1,
            "Problem4" : 1,
            "tick_info" : {"3":["INC4"], "4":["INC6"]}
           }
}

我尝试了以下代码，但这似乎不起作用：

clust_stack = {}
for i in range(len(l1)):
      fq= defaultdict( list )
      for n,v in l1[i]:
           fq[str(n)].append(v)
      for name_, data in zip(names, label_issue_list)[i]:
           clust_stack[name_] = Counter(map(str,data))
           clust_stack[name_]["tick_info"] = {}
           clust_stack[name_]["tick_info"] = dict(fq)

我在for name_, data in zip(names, label_issue_list)[i]:行中收到以下错误：

ValueError: too many values to unpack

我需要做什么才能获得所需的输出？

Answer 1

从压缩所有3个列表开始：

clust_stack = {}
for name, data, fq in zip(names, label_issue_list, l1):
    clust_stack[name] = dict(Counter(data))  # turn back into regular dictionary
    tick_info = clust_stack[name]['tick_info'] = {}
    for num, tick in fq:
        tick_info.setdefault(num, []).append(tick)

使用Counter计算每个l1元素中的信息非常好，但您需要将该结果转回常规字典，以便为其添加其他键。

然后，您可以将tick_info密钥添加到该密钥，以每个密钥的列表值收集这些列表中的信息。

演示：

>>> from collections import Counter
>>> from pprint import pprint
>>> l1 = [[(1,"INC1"), (1, "INC2")],[(3, "INC4"),(4, "INC6")]]
>>> names = ['John', 'Marcia']
>>> label_issue_list = [["Problem1","Problem1"],["Problem3","Problem4"]]
>>> clust_stack = {}
>>> for name, data, fq in zip(names, label_issue_list, l1):
...     clust_stack[name] = dict(Counter(data))  # turn back into regular dictionary
...     tick_info = clust_stack[name]['tick_info'] = {}
...     for num, tick in fq:
...         tick_info.setdefault(num, []).append(tick)
... 
>>> pprint(clust_stack)
{'John': {'Problem1': 2, 'tick_info': {1: ['INC1', 'INC2']}},
 'Marcia': {'Problem3': 1,
            'Problem4': 1,
            'tick_info': {3: ['INC4'], 4: ['INC6']}}}

Python：使用列表列表，元组列表和列表创建字典

1 个答案: