用于列表的Python Django Lambda表达式

我需要用Python更新一个列表，它是：

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

我想编写一个Lambda表达式来存储列表中的Customers和CustomersName，并删除空格。

---

我是Python的新手，没有任何知识

在我看来，您已经在一个列表中声明了字典，但是Dict是错误的，它应该是{“key”：“value”}，所以我假设您需要将其更改为List，如下所示：

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

然后下面的内容会让你得到你想要的

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

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不能将其放入lambda表达式中，但可以使用以下生成器对象：

# 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]

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]

见：

编辑：

如果您不想使用字典，您可以使用以下内容：

# 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]

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]