operator.itemgetter 和 sort() 在 Python 中是如何工作的？
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How does operator.itemgetter and sort() work in Python?
提问by Nickl
I have the following code:
我有以下代码：
# initialize
a = []
# create the table (name, age, job)
a.append(["Nick", 30, "Doctor"])
a.append(["John", 8, "Student"])
a.append(["Paul", 22, "Car Dealer"])
a.append(["Mark", 66, "Retired"])
# sort the table by age
import operator
a.sort(key=operator.itemgetter(1))
# print the table
print(a)
It creates a 4x3 table and then it sorts it by age. My question is, what exactly key=operator.itemgetter(1)does? Does the operator.itemgetterfunction return the item's value? Why can't I just type something like key=a[x][1]there? Or can I? How could with operator print a certain value of the form like 3x2which is 22?
它创建一个 4x3 表，然后按年龄对其进行排序。我的问题是，究竟是什么key=operator.itemgetter(1)？该operator.itemgetter函数是否返回项目的值？为什么我不能在key=a[x][1]那里输入类似的东西？或者我可以吗？操作员如何打印某个值，例如3x2which is 22？
How does exactly Python sort the table? Can I reverse-sort it?
How can I sort it based on two columns like first age, and then if age is the same b name?
How could I do it without
operator?
Python究竟是如何对表格进行排序的？我可以反向排序吗？
如何根据两列（如第一个年龄）对其进行排序，然后如果年龄与 b 名称相同？
没有我怎么办
operator？
采纳答案by J0HN
Looks like you're a little bit confused about all that stuff.
看起来你对所有这些东西有点困惑。
operatoris a built-in module providing a set of convenient operators. In two words operator.itemgetter(n)constructs a callable that assumes an iterable object (e.g. list, tuple, set) as input, and fetches the n-th element out of it.
operator是一个内置模块，提供了一组方便的操作符。用两个词operator.itemgetter(n)构造一个可调用对象，它假定一个可迭代对象（例如列表、元组、集合）作为输入，并从中取出第 n 个元素。
So, you can't use key=a[x][1]there, because python has no idea what xis. Instead, you could use a lambdafunction (elemis just a variable name, no magic there):
所以，你不能key=a[x][1]在那里使用，因为 python 不知道是什么x。相反，你可以使用一个lambda函数（elem只是一个变量名，没有魔法）：
a.sort(key=lambda elem: elem[1])
Or just an ordinary function:
或者只是一个普通的函数：
def get_second_elem(iterable):
return iterable[1]
a.sort(key=get_second_elem)
So, here's an important note: in python functions are first-class citizens, so you can pass them to other functions as a parameter.
所以，这里有一个重要的注意事项：在 python 中，函数是一等公民，因此您可以将它们作为参数传递给其他函数。
Other questions:
其他问题：
- Yes, you can reverse sort, just add
reverse=True:a.sort(key=..., reverse=True) - To sort by more than one column you can use
itemgetterwith multiple indices:operator.itemgetter(1,2), or with lambda:lambda elem: (elem[1], elem[2]). This way, iterables are constructed on the fly for each item in list, which are than compared against each other in lexicographic(?) order (first elements compared, if equal - second elements compared, etc) - You can fetch value at [3,2] using
a[2,1](indices are zero-based). Using operator... It's possible, but not as clean as just indexing.
- 是的，您可以反向排序，只需添加
reverse=True：a.sort(key=..., reverse=True) - 要按多列排序，您可以使用
itemgetter多个索引：operator.itemgetter(1,2)或 lambda:lambda elem: (elem[1], elem[2])。这样，可以为列表中的每个项目动态构建可迭代对象，然后按字典（？）顺序相互比较（比较第一个元素，如果相等 - 比较第二个元素等） - 您可以使用 [3,2] 获取值
a[2,1]（索引从零开始）。使用运算符...这是可能的，但不像索引那样干净。
Refer to the documentation for details:
有关详细信息，请参阅文档：
回答by Paulo Almeida
You are asking a lot of questions that you could answer yourself by reading the documentation, so I'll give you a general advice: read it and experiment in the python shell. You'll see that itemgetterreturns a callable:
你问了很多问题，你可以通过阅读文档来回答自己，所以我会给你一个一般性的建议：阅读它并在 python shell 中进行实验。你会看到itemgetter返回一个可调用的：
>>> func = operator.itemgetter(1)
>>> func(a)
['Paul', 22, 'Car Dealer']
>>> func(a[0])
8
To do it in a different way, you can use lambda:
要以不同的方式做到这一点，您可以使用lambda：
a.sort(key=lambda x: x[1])
And reverse it:
并反转它：
a.sort(key=operator.itemgetter(1), reverse=True)
Sort by more than one column:
按多列排序：
a.sort(key=operator.itemgetter(1,2))
See the sorting How To.
请参阅排序方法。
回答by Lutz Prechelt
Answer for Python beginners
Python初学者的答案
In simpler words:
简单来说：
- The
key=parameter ofsortrequires a key function(to be applied to be objects to be sorted) rather than a single key valueand - that is just what
operator.itemgetter(1)will give you: A functionthat grabs the first item from a list-like object.
- 所述
key=的参数sort需要密钥函数（要施加到要进行排序的对象），而不是单个密钥值和 - 这正是
operator.itemgetter(1)会给你：一个功能是从列表状物体抓住的第一个项目。
(More precisely those are callables, not functions, but that is a difference that can often be ignored.)
（更准确地说，那些是callables，而不是函数，但这是一个经常可以忽略的差异。）
回答by user2329366
#sorting first by age then profession,you can change it in function "fun".
a = []
def fun(v):
return (v[1],v[2])
# create the table (name, age, job)
a.append(["Nick", 30, "Doctor"])
a.append(["John", 8, "Student"])
a.append(["Paul", 8,"Car Dealer"])
a.append(["Mark", 66, "Retired"])
a.sort(key=fun)
print a
回答by shubham sinha
a = []
a.append(["Nick", 30, "Doctor"])
a.append(["John", 8, "Student"])
a.append(["Paul", 8,"Car Dealer"])
a.append(["Mark", 66, "Retired"])
print a
[['Nick', 30, 'Doctor'], ['John', 8, 'Student'], ['Paul', 8, 'Car Dealer'], ['Mark', 66, 'Retired']]
def _cmp(a,b):
if a[1]<b[1]:
return -1
elif a[1]>b[1]:
return 1
else:
return 0
sorted(a,cmp=_cmp)
[['John', 8, 'Student'], ['Paul', 8, 'Car Dealer'], ['Nick', 30, 'Doctor'], ['Mark', 66, 'Retired']]
def _key(list_ele):
return list_ele[1]
sorted(a,key=_key)
[['John', 8, 'Student'], ['Paul', 8, 'Car Dealer'], ['Nick', 30, 'Doctor'], ['Mark', 66, 'Retired']]
>>>
