按另一个列表对嵌套列表进行排序[重复]

Question

我有以下双重嵌套列表：

records = [[['Jack', 'male', 1],['Jack', 'male', 2],['Jack', 'male', 3]],[['Sally', 'female', 1],['Sally', 'female', 2],['Sally', 'female', 3]]]

我想根据这个列表对这个列表进行排序...

list_order = ['female', 'male']

...结果是这样的：

records 
[[['Sally', 'female', 1],['Sally', 'female', 2],['Sally', 'female', 3]],[['Jack', 'male', 1],['Jack', 'male', 2],['Jack', 'male', 3]]]

Answer 1

这可能比您要求的要复杂一些。我假设：

1. 您不能保证每个组将仅包含来自一个“list_order”类别的项目

2. 您希望最终列表是双重嵌套的，所有“list_order”类别组合在一起

3. 除了与“list_order”匹配的一列之外，没有其他排序
4. 项目应该按照它们在“list_order”中列出的确切顺序

因此，您可以使用以下代码：

records = [[['Jack', 'male', 1],['Jack', 'male', 2],['Jack', 'male', 3]],[['Sally', 'female', 1],['Sally', 'female', 2],['Sally', 'female', 3]]]
list_order = ['female', 'male']

# "flatten" list into singly-nested list
records = [leaf for branch in records for leaf in branch]

# sort the items in the list by list_order 
# sorted passes each item in the list to the lambda function
# record[1] is the position of "male"/"female"
records = sorted(records, key=lambda record: list_order.index(record[1]))

# group the list by list_order again, creating doubly-nested list
records = [ [record for record in records if record[1] == item ] for item in list_order ]

print records

Try it online (with debugging printouts)!

Answer 2

你可以使用sum:

records = [[['Jack', 'male', 1],['Jack', 'male', 2],['Jack', 'male', 3]],[['Sally', 'female', 1],['Sally', 'female', 2],['Sally', 'female', 3]]]
list_order = ['female', 'male']
new_records = sorted(records, key=lambda x:sum(list_order.index(i[1]) for i in x))

输出：

[[['Sally', 'female', 1], ['Sally', 'female', 2], ['Sally', 'female', 3]], [['Jack', 'male', 1], ['Jack', 'male', 2], ['Jack', 'male', 3]]]