sorted（）排序详解

http://wiki.python.org/moin/HowTo/Sorting?highlight=%28howto%29#The_Old_Way_Using_the_cmp_Parameter

一个列表存储了由数字组成的字符串，比如 '21'
,如何对他们按照数字大小排序得到新列表呢？

l=['1','3','2','32','1','453','232']

l.sort(key = int) # key = int,只按照数值大小进行比较

from operator import itemgetter,attrgetter

s = [('john', 'A', 15), ('jane', 'B', 12), ('dave', 'B', 10),]

print itemgetter(0,2)

print sorted(s,key=itemgetter(2))

from operator import itemgetter,attrgetter

d = {'data1':3,'data2':1,'data3':2,'data4':4}

print d.items()

print sorted(d.iteritems(),key = itemgetter(0)) #根据元组的第一项进行排序，此处即字典的键

print sorted(d.iteritems(),key = itemgetter(1)) #根据元组的第二项进行排序，此处即字典的值

python列表排序

简单记一下python中List的sort方法（或者sorted内建函数）的用法。

关键字：
python列表排序 python字典排序 sorted

List的元素可以是各种东西，字符串，字典，自己定义的类等。

sorted函数用法如下：

sorted(data, cmp=None, key=None, reverse=False)

其中，data是待排序数据，可以使List或者iterator, cmp和key都是函数，这两个函数作用与data的元素上产生一个结果，sorted方法根据这个结果来排序。
cmp(e1, e2) 是带两个参数的比较函数, 返回值: 负数: e1 < e2, 0: e1 == e2, 正数: e1 > e2. 默认为 None, 即用内建的比较函数.
key 是带一个参数的函数, 用来为每个元素提取比较值. 默认为 None, 即直接比较每个元素.
通常, key 和 reverse 比 cmp 快很多, 因为对每个元素它们只处理一次; 而 cmp 会处理多次.

通过例子来说明sorted的用法：

1. 对由tuple组成的List排序

>>> students = [('john', 'A', 15), ('jane', 'B', 12), ('dave', 'B', 10),]

用key函数排序(lambda的用法见注释1)

>>> sorted(students, key=lambda s : s[2]) # sort by age
[('dave', 'B', 10), ('jane', 'B', 12), ('john', 'A', 15)]

用cmp函数排序

>>> sorted(students, cmp=lambda x,y : cmp(x[2], y[2])) # sort by age asc
[('dave', 'B', 10), ('jane', 'B', 12), ('john', 'A', 15)]

年龄倒序（详见末尾）：

>>> sorted(students, cmp=lambda x,y : cmp(y[2], x[2])) # sort by age desc，交换x/y的位置
[('john', 'A', 15), ('jane', 'B', 12),('dave', 'B', 10)]

用 operator 函数来加快速度, 上面排序等价于:(itemgetter的用法见注释2)

>>> from operator import itemgetter, attrgetter
>>> sorted(students, key=itemgetter(2))

用 operator 函数进行多级排序

>>> sorted(students, key=itemgetter(1,2)) # sort by grade then by age
[('john', 'A', 15), ('dave', 'B', 10), ('jane', 'B', 12)]

2. 对由字典排序

>>> d = {'data1':3, 'data2':1, 'data3':2, 'data4':4}
>>> sorted(d.iteritems(), key=itemgetter(1), reverse=True)
[('data4', 4), ('data1', 3), ('data3', 2), ('data2', 1)]

from operator import itemgetter,attrgetter
d = {'data1':3,'data2':1,'data3':2,'data4':4}

print d.items()
print sorted(d.iteritems(),key = itemgetter(0)) #根据字典的键进行排序
print sorted(d.iteritems(),key = itemgetter(1)) #根据字典的值进行排序

注释1
参考：http://jasonwu.me/2011/10/29/introduce-to-python-lambda.html

注释2
参考:http://ar.newsmth.net/thread-90745710c90cf1.html
class itemgetter(__builtin__.object)
| itemgetter(item, ...) --> itemgetter object
|
| Return a callable object that fetches the given item(s) from its operand.
| After, f=itemgetter(2), the call f(r) returns r[2].
| After, g=itemgetter(2,5,3), the call g(r) returns (r[2], r[5], r[3])

相当于

def itemgetter(i,*a):
def func(obj):
r = obj[i]
if a:
r = (r,) + tuple(obj[i] for i in a)
return r
return func
>>> a = [1,2,3]
>>> b=operator.itemgetter(1)
>>> b(a)
2
>>> b=operator.itemgetter(1,0)
>>> b(a)
(2, 1)
>>> b=itemgetter(1)
>>> b(a)
2
>>> b=itemgetter(1,0)
>>> b(a)
(2, 1)

In Py2.x, sort allowed an optional function which can be called for doing thecomparisons. That function should take two arguments to be compared andthen return a negative value for less-than, return zero if they are equal,or return a positive value for greater-than. For example, we can do:

>>> def numeric_compare(x, y): return x - y>>> sorted([5, 2, 4, 1, 3], cmp=numeric_compare)[1, 2, 3, 4, 5]

Or you can reverse the order of comparison with:

>>> def reverse_numeric(x, y): return y - x>>> sorted([5, 2, 4, 1, 3], cmp=reverse_numeric)[5, 4, 3, 2, 1]

list.sort() method is only defined for lists. In contrast, the sorted() function accepts anyiterable.

>>> sorted({1: 'D', 2: 'B', 3: 'B', 4: 'E', 5: 'A'})[1, 2, 3, 4, 5]