Question

我有以下列表

list_of_dict = [{'flat': ['103'], 'wing': u'C'}, {'flat': ['102', '104'], 'wing': u'B'}, {'flat': ['105'], 'wing': u'D'}]

我希望转换成

list_of_dict = [{'flat': [{'103'}], 'wing': u'C'}, {'flat': [{'102'}, {'104'}], 'wing': u'B'}, {'flat': [{'105'}], 'wing': u'D'}]

Flat应该是“{}”

中包含的数字列表

Answer 1

假设您需要单元素集列表，可以使用for循环：

for d in list_of_dict:
     if 'flat' in d:
         d['flat'] = list(map(lambda x: set([x]), d['flat']))

此处的if检查可提供更高级别的安全性，以防您的词典不包含flat作为密钥。或者，使用EAFP方法，您可以使用try-except大括号：

for d in list_of_dict:
     try:
         d['flat'] = list(map(lambda x: set([x]), d['flat']))
     except KeyError: 
         pass

>>> list_of_dict
[{'flat': [{'103'}], 'wing': 'C'},
 {'flat': [{'102'}, {'104'}], 'wing': 'B'},
 {'flat': [{'105'}], 'wing': 'D'}]

Answer 2

以下内容完全符合您的要求，

list_of_dict = [{'flat': ['103'], 'wing': u'C'}, {'flat': ['102', '104'], 'wing': u'B'}, {'flat': ['105'], 'wing': u'D'}]

for e in list_of_dict:
    e['flat'] = [{x} for x in e['flat']]

print(list_of_dict)

<强>更新以下代码应根据您在下面的评论工作

list_of_dict = [{'flat': ['103'], 'wing': u'C'}, {'flat': ['102', '104'], 'wing': u'B'}, {'flat': ['105'], 'wing': u'D'}]

for e in list_of_dict:
    e['flat'] = ['{'+x+'}' for x in e['flat']]

print(list_of_dict)

Answer 3

{}表示set。假设这是你想要的，你可以这样做：

for x in list_of_dict:
    x['flat'] = Set(x['flat'])

这会将list_of_dict编辑为：

[{'flat': Set(['103']), 'wing': u'C'}, {'flat': Set(['102', '104']), 'wing': u'B'}, {'flat': Set(['105']), 'wing': u'D'}]

如果你在Python解释器中输入{'103'}，它会回复你：

set(['103'])

这就是为什么我认为你需要套装。

Answer 4

理解能够满足您的需求：

list_of_dict = [{'flat': ['103'], 'wing': u'C'}, {'flat': ['102', '104'], 'wing': u'B'}, {'flat': ['105'], 'wing': u'D'}]
result  = [{k: ['{}{}{}'.format('{',item,'}') for item in v] if k=='flat' else v for k, v in d.items()} for d in list_of_dict]
#[{'flat': ['{103}'], 'wing': u'C'}, {'flat': ['{102}', '{104}'], 'wing': u'B'}, {'flat': ['{105}'], 'wing': u'D'}]

将字符串列表转换为单元素集列表

4 个答案: