Question

我有一个很大的字典列表，我想要分组，如下所示：

first_list = [{'name':'James','gender':'M','address':'California'},{'name':'Tom','gender':'M','address':'California'},
{'name':'Jane','gender':'F','address':'Utah'},
{'name':'Kim','gender':'F','address':'Wisconsin'},
{'name':'Ron','gender':'M','address':'Montana'}]

我有另一个名单：

second_list = ['James', 'Tom']

我希望列出名称＆＃39;在第一个列表中不是第二个列表的一部分，它只是删除了James和Tom词典。

[{'name':'Jane','gender':'F','address':'Utah'},
{'name':'Kim','gender':'F','address':'Wisconsin'},
{'name':'Ron','gender':'M','address':'Montana'}]

我尝试使用列表理解，但我不认为这适用于不同的列表：

third_list = [x for x in first_list if x['name'] != (y for y in second_list)]

这不会起作用，将返回与第一个列表相同的列表。我的语法错了吗？

Answer 1

使用适合您的not in。

third_list = [i for i in first_list if i['name'] not in second_list]

<强>结果

[{'address': 'Utah', 'gender': 'F', 'name': 'Jane'},
 {'address': 'Wisconsin', 'gender': 'F', 'name': 'Kim'},
 {'address': 'Montana', 'gender': 'M', 'name': 'Ron'}]

Answer 2

您的代码不起作用，因为(y for y in second_list)是一个生成器，x['name']是一个字符串，这意味着它将一直返回False，我想你想要的是这个：

>>> first_list = [{'name':'James','gender':'M','address':'California'},{'name':'Tom','gender':'M','address':'California'},
... {'name':'Jane','gender':'F','address':'Utah'},
... {'name':'Kim','gender':'F','address':'Wisconsin'},
... {'name':'Ron','gender':'M','address':'Montana'}]
>>>
>>>
>>> second_list = ['James', 'Tom']
>>>
>>> [x for x in first_list if x['name'] not in second_list]
[{'gender': 'F', 'name': 'Jane', 'address': 'Utah'}, {'gender': 'F', 'name': 'Kim', 'address': 'Wisconsin'}, {'gender': 'M', 'name': 'Ron', 'address': 'Montana'}]

Pythonic方法，尝试使用filter方法：

>>> filter(lambda x:x["name"] not in second_list,first_list)
[{'gender': 'F', 'name': 'Jane', 'address': 'Utah'}, {'gender': 'F', 'name': 'Kim', 'address': 'Wisconsin'}, {'gender': 'M', 'name': 'Ron', 'address': 'Montana'}]

Answer 3

使用

third_list = [x for x in first_list if x['name'] not in second_list]

如何使用另一个列表使用列表推导对列表进行子集化

3 个答案: