问题:按两个字段对Python列表进行排序

我有一个从排序的csv创建的以下列表

list1 = sorted(csv1, key=operator.itemgetter(1))

我实际上想按两个条件对列表进行排序:首先按字段1中的值,然后按字段2中的值。我该怎么做?

I have the following list created from a sorted csv

list1 = sorted(csv1, key=operator.itemgetter(1))

I would actually like to sort the list by two criteria: first by the value in field 1 and then by the value in field 2. How do I do this?


回答 0

像这样:

import operator
list1 = sorted(csv1, key=operator.itemgetter(1, 2))

like this:

import operator
list1 = sorted(csv1, key=operator.itemgetter(1, 2))

回答 1

使用lambda函数时无需导入任何内容。
以下list按第一个元素排序,然后按第二个元素排序。

sorted(list, key=lambda x: (x[0], -x[1]))

No need to import anything when using lambda functions.
The following sorts list by the first element, then by the second element.

sorted(list, key=lambda x: (x[0], -x[1]))

回答 2

Python具有稳定的排序方式,因此,只要性能不成问题,最简单的方法就是按字段2对其进行排序,然后再次按字段1对其进行排序。

这将为您提供所需的结果,唯一的陷阱是,如果列表很大(或者您希望经常对其进行排序),则两次调用sort可能是不可接受的开销。

list1 = sorted(csv1, key=operator.itemgetter(2))
list1 = sorted(list1, key=operator.itemgetter(1))

这样一来,还可以轻松处理需要对某些列进行反向排序的情况,只需在必要时添加’reverse = True’参数即可。

否则,您可以将多个参数传递给itemgetter或手动构建一个元组。这可能会更快一些,但是有一个问题,就是如果某些列想要反向排序,它不能很好地推广(数字列仍然可以通过取反来反转,但是这会使排序保持稳定)。

因此,如果您不需要对任何列进行反向排序,则可以向itemgetter输入多个参数(如果可能),并且这些列不是数字的,或者您希望保持排序稳定以进行多个连续排序。

编辑:对于在理解此答案的原始方式时遇到问题的评论者,以下示例准确显示了排序的稳定性,从而确保了我们可以对每个键进行单独的排序并最终对多个条件下的数据进行排序:

DATA = [
    ('Jones', 'Jane', 58),
    ('Smith', 'Anne', 30),
    ('Jones', 'Fred', 30),
    ('Smith', 'John', 60),
    ('Smith', 'Fred', 30),
    ('Jones', 'Anne', 30),
    ('Smith', 'Jane', 58),
    ('Smith', 'Twin2', 3),
    ('Jones', 'John', 60),
    ('Smith', 'Twin1', 3),
    ('Jones', 'Twin1', 3),
    ('Jones', 'Twin2', 3)
]

# Sort by Surname, Age DESCENDING, Firstname
print("Initial data in random order")
for d in DATA:
    print("{:10s} {:10s} {}".format(*d))

print('''
First we sort by first name, after this pass all
Twin1 come before Twin2 and Anne comes before Fred''')
DATA.sort(key=lambda row: row[1])

for d in DATA:
    print("{:10s} {:10s} {}".format(*d))

print('''
Second pass: sort by age in descending order.
Note that after this pass rows are sorted by age but
Twin1/Twin2 and Anne/Fred pairs are still in correct
firstname order.''')
DATA.sort(key=lambda row: row[2], reverse=True)
for d in DATA:
    print("{:10s} {:10s} {}".format(*d))

print('''
Final pass sorts the Jones from the Smiths.
Within each family members are sorted by age but equal
age members are sorted by first name.
''')
DATA.sort(key=lambda row: row[0])
for d in DATA:
    print("{:10s} {:10s} {}".format(*d))

这是一个可运行的示例,但是为了节省运行它的人员,输出为:

Initial data in random order
Jones      Jane       58
Smith      Anne       30
Jones      Fred       30
Smith      John       60
Smith      Fred       30
Jones      Anne       30
Smith      Jane       58
Smith      Twin2      3
Jones      John       60
Smith      Twin1      3
Jones      Twin1      3
Jones      Twin2      3

First we sort by first name, after this pass all
Twin1 come before Twin2 and Anne comes before Fred
Smith      Anne       30
Jones      Anne       30
Jones      Fred       30
Smith      Fred       30
Jones      Jane       58
Smith      Jane       58
Smith      John       60
Jones      John       60
Smith      Twin1      3
Jones      Twin1      3
Smith      Twin2      3
Jones      Twin2      3

Second pass: sort by age in descending order.
Note that after this pass rows are sorted by age but
Twin1/Twin2 and Anne/Fred pairs are still in correct
firstname order.
Smith      John       60
Jones      John       60
Jones      Jane       58
Smith      Jane       58
Smith      Anne       30
Jones      Anne       30
Jones      Fred       30
Smith      Fred       30
Smith      Twin1      3
Jones      Twin1      3
Smith      Twin2      3
Jones      Twin2      3

Final pass sorts the Jones from the Smiths.
Within each family members are sorted by age but equal
age members are sorted by first name.

Jones      John       60
Jones      Jane       58
Jones      Anne       30
Jones      Fred       30
Jones      Twin1      3
Jones      Twin2      3
Smith      John       60
Smith      Jane       58
Smith      Anne       30
Smith      Fred       30
Smith      Twin1      3
Smith      Twin2      3

特别要注意的是,在第二步中,reverse=True参数如何按顺序保留名字,而仅对列表进行排序然后反转,则会丢失第三个排序键的期望顺序。

Python has a stable sort, so provided that performance isn’t an issue the simplest way is to sort it by field 2 and then sort it again by field 1.

That will give you the result you want, the only catch is that if it is a big list (or you want to sort it often) calling sort twice might be an unacceptable overhead.

list1 = sorted(csv1, key=operator.itemgetter(2))
list1 = sorted(list1, key=operator.itemgetter(1))

Doing it this way also makes it easy to handle the situation where you want some of the columns reverse sorted, just include the ‘reverse=True’ parameter when necessary.

Otherwise you can pass multiple parameters to itemgetter or manually build a tuple. That is probably going to be faster, but has the problem that it doesn’t generalise well if some of the columns want to be reverse sorted (numeric columns can still be reversed by negating them but that stops the sort being stable).

So if you don’t need any columns reverse sorted, go for multiple arguments to itemgetter, if you might, and the columns aren’t numeric or you want to keep the sort stable go for multiple consecutive sorts.

Edit: For the commenters who have problems understanding how this answers the original question, here is an example that shows exactly how the stable nature of the sorting ensures we can do separate sorts on each key and end up with data sorted on multiple criteria:

DATA = [
    ('Jones', 'Jane', 58),
    ('Smith', 'Anne', 30),
    ('Jones', 'Fred', 30),
    ('Smith', 'John', 60),
    ('Smith', 'Fred', 30),
    ('Jones', 'Anne', 30),
    ('Smith', 'Jane', 58),
    ('Smith', 'Twin2', 3),
    ('Jones', 'John', 60),
    ('Smith', 'Twin1', 3),
    ('Jones', 'Twin1', 3),
    ('Jones', 'Twin2', 3)
]

# Sort by Surname, Age DESCENDING, Firstname
print("Initial data in random order")
for d in DATA:
    print("{:10s} {:10s} {}".format(*d))

print('''
First we sort by first name, after this pass all
Twin1 come before Twin2 and Anne comes before Fred''')
DATA.sort(key=lambda row: row[1])

for d in DATA:
    print("{:10s} {:10s} {}".format(*d))

print('''
Second pass: sort by age in descending order.
Note that after this pass rows are sorted by age but
Twin1/Twin2 and Anne/Fred pairs are still in correct
firstname order.''')
DATA.sort(key=lambda row: row[2], reverse=True)
for d in DATA:
    print("{:10s} {:10s} {}".format(*d))

print('''
Final pass sorts the Jones from the Smiths.
Within each family members are sorted by age but equal
age members are sorted by first name.
''')
DATA.sort(key=lambda row: row[0])
for d in DATA:
    print("{:10s} {:10s} {}".format(*d))

This is a runnable example, but to save people running it the output is:

Initial data in random order
Jones      Jane       58
Smith      Anne       30
Jones      Fred       30
Smith      John       60
Smith      Fred       30
Jones      Anne       30
Smith      Jane       58
Smith      Twin2      3
Jones      John       60
Smith      Twin1      3
Jones      Twin1      3
Jones      Twin2      3

First we sort by first name, after this pass all
Twin1 come before Twin2 and Anne comes before Fred
Smith      Anne       30
Jones      Anne       30
Jones      Fred       30
Smith      Fred       30
Jones      Jane       58
Smith      Jane       58
Smith      John       60
Jones      John       60
Smith      Twin1      3
Jones      Twin1      3
Smith      Twin2      3
Jones      Twin2      3

Second pass: sort by age in descending order.
Note that after this pass rows are sorted by age but
Twin1/Twin2 and Anne/Fred pairs are still in correct
firstname order.
Smith      John       60
Jones      John       60
Jones      Jane       58
Smith      Jane       58
Smith      Anne       30
Jones      Anne       30
Jones      Fred       30
Smith      Fred       30
Smith      Twin1      3
Jones      Twin1      3
Smith      Twin2      3
Jones      Twin2      3

Final pass sorts the Jones from the Smiths.
Within each family members are sorted by age but equal
age members are sorted by first name.

Jones      John       60
Jones      Jane       58
Jones      Anne       30
Jones      Fred       30
Jones      Twin1      3
Jones      Twin2      3
Smith      John       60
Smith      Jane       58
Smith      Anne       30
Smith      Fred       30
Smith      Twin1      3
Smith      Twin2      3

Note in particular how in the second step the reverse=True parameter keeps the firstnames in order whereas simply sorting then reversing the list would lose the desired order for the third sort key.


回答 3

list1 = sorted(csv1, key=lambda x: (x[1], x[2]) )
list1 = sorted(csv1, key=lambda x: (x[1], x[2]) )

回答 4

employees.sort(key = lambda x:x[1])
employees.sort(key = lambda x:x[0])

我们也可以将.sort与lambda一起使用2次,因为python sort到位且稳定。这将首先根据第二个元素x [1]对列表进行排序。然后,它将对第一个元素x [0](最高优先级)进行排序。

employees[0] = Employee's Name
employees[1] = Employee's Salary

这等效于执行以下操作:employee.sort(key = lambda x:(x [0],x [1]))

employees.sort(key = lambda x:x[1])
employees.sort(key = lambda x:x[0])

We can also use .sort with lambda 2 times because python sort is in place and stable. This will first sort the list according to the second element, x[1]. Then, it will sort the first element, x[0] (highest priority).

employees[0] = Employee's Name
employees[1] = Employee's Salary

This is equivalent to doing the following: employees.sort(key = lambda x:(x[0], x[1]))


回答 5

您可以按升序使用:

sorted_data= sorted(non_sorted_data, key=lambda k: (k[1],k[0]))

或按降序使用:

sorted_data= sorted(non_sorted_data, key=lambda k: (k[1],k[0]),reverse=True)

In ascending order you can use:

sorted_data= sorted(non_sorted_data, key=lambda k: (k[1],k[0]))

or in descending order you can use:

sorted_data= sorted(non_sorted_data, key=lambda k: (k[1],k[0]),reverse=True)

回答 6

使用下面的字典排序列表将以降序对列表进行排序,第一列为薪水,第二列为年龄

d=[{'salary':123,'age':23},{'salary':123,'age':25}]
d=sorted(d, key=lambda i: (i['salary'], i['age']),reverse=True)

输出:[{‘salary’:123,’age’:25},{‘salary’:123,’age’:23}]

Sorting list of dicts using below will sort list in descending order on first column as salary and second column as age

d=[{'salary':123,'age':23},{'salary':123,'age':25}]
d=sorted(d, key=lambda i: (i['salary'], i['age']),reverse=True)

Output: [{‘salary’: 123, ‘age’: 25}, {‘salary’: 123, ‘age’: 23}]


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