Question

我目前有以下子列表：

[['jjj', '3', 'bbb', '0', 'ddd', '9', 'ggg', '8', 'hhh', '2'],
 ['ccc', '2', 'ddd', '0', 'aaa', '3', 'hhh', '9'],
 ['ddd', '2', 'ggg', '5', 'ccc', '6', 'jjj', '1'],
 ['hhh', '9', 'iii', '5', 'eee', '7', 'bbb', '1'],
 ['iii', '6', 'ddd', '5', 'eee', '4', 'jjj', '3']]

我想做的是将这些列表转换为单个字典，并为列表中的每个键指定特定的值

例如

{'jjj': ['3','1','3'],
 'bbb': ['0','1'],
 'ddd': ['9','0','2,'5']}

Answer 1

尝试一下：

import collections

D=collections.defaultdict(list)

for sublist in main_list:
    for k in range(len(sublist)//2):
        D[sublist[2*k]].append(sublist[2*k+1])

或者，如果您只想使用简单的字典：

D=dict()

for sublist in main_list:
    for k in range(len(sublist)//2):
        key=sublist[2*k]
        value=sublist[2*k+1]
        current=D.get(key)
        if current is None:
            D[key]=[value]
        else:
            current.append(value)

Answer 2

您可以尝试创建一个简单的循环以遍历嵌套列表，并根据键值创建字典。这是您的操作方法。

x = [['jjj', '3', 'bbb', '0', 'ddd', '9', 'ggg', '8', 'hhh', '2'],
     ['ccc', '2', 'ddd', '0', 'aaa', '3', 'hhh', '9'],
     ['ddd', '2', 'ggg', '5', 'ccc', '6', 'jjj', '1'],
     ['hhh', '9', 'iii', '5', 'eee', '7', 'bbb', '1'],
     ['iii', '6', 'ddd', '5', 'eee', '4', 'jjj', '3']]

y = {}
for i in x:
    for j in range (0, len(i),2):
        if i[j] in y.keys(): #if key already exists, add to list
            y[i[j]].append(i[j+1])
        else: #if key does not exist, create a list
            y[i[j]] = list(i[j+1])
print (y)

您的输出将是：

{'jjj': ['3', '1', '3'], 'bbb': ['0', '1'], 'ddd': ['9', '0', '2', '5'], 'ggg': ['8', '5'], 'hhh': ['2', '9', '9'], 'ccc': ['2', '6'], 'aaa': ['3'], 'iii': ['5', '6'], 'eee': ['7', '4']}

将子列表转换成字典

2 个答案: