Question

我需要在Python中更新一个列表，

data = [{' Customers ','null,blank '},{' CustomersName ','max=50,null,blank '},{' CustomersAddress ','max=150,blank '},{' CustomersActive ','Active '}]

我想编写一个Lambda表达式以在列表中存储客户，CustomersName，并删除空白。我绝对是Python新手，没有任何知识！

Answer 1

正如我所看到的，您已经在列表中声明了字典，但是Dict是错误的，它应该是{“ key”：“ value”}，所以我认为您需要这样将其更改为List：

  data = [[' Customers ','null,blank '],[' CustomersName ','max=50,null,blank '],[' CustomersAddress ','max=150,blank '],[' CustomersActive ','Active ']]

然后，以下内容将使您获得您的期望！

data_NameExtracted = [x[0].strip() for x in data]

Answer 2

您不能将其放在lambda表达式中，但可以使用如下生成器对象：

# please note that i have used tuples instead of sets, 
# because sets are unordered
data = [
    (' Customers ','null,blank '),
    (' CustomersName ','max=50,null,blank '), 
    (' CustomersAddress ','max=150,blank '),
    (' CustomersActive ','Active ')
]

# Indexing is not allowed for set objects
values = [item[0].strip() for item in data]

请参阅：

https://wiki.python.org/moin/Generators

https://docs.python.org/3/tutorial/datastructures.html#sets

编辑：

如果您不想使用字典，则可以使用以下内容：

data = [
    {' Customers ': 'null,blank '},
    {' CustomersName ': 'max=50,null,blank '}, 
    {' CustomersAddress ': 'max=150,blank '},
    {' CustomersActive ': 'Active '}
]
# expecting a single value in the dicts
values = [item.values()[0].strip() for item in data]

列表的Python Django Lambda表达式

2 个答案: