python 容器排序

本文主要是对下面链接的翻译和概括http://docs.python.org/2/howto/sorting.html

通常简单的调用sorted已经可以满足大多数的需求，但是对于复杂的数据结构，以及多层次优先级的组合排序和稳定性的需求，我们就要使用sorted更高级别的设置，sorted提供了自定义关键字和比较函数的方式来满足这些要求。

sorted函数用法如下：

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

排序基础

>>> sorted([5, 2, 3, 1, 4])
[1, 2, 3, 4, 5]

>>> a = [5, 2, 3, 1, 4]
>>> a.sort()
>>> a
[1, 2, 3, 4, 5]

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

自定义关键字排序

注意key是一个接受一个参数，然后将其转化成一个可以比较的简单数据的函数，如下面 str.lower就是一个函数

>>> sorted("This is a test string from Andrew".split(), key=str.lower)
['a', 'Andrew', 'from', 'is', 'string', 'test', 'This']

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

>>> class Student:
        def __init__(self, name, grade, age):
            self.name = name
            self.grade = grade
            self.age = age
        def __repr__(self):
            return repr((self.name, self.grade, self.age))

>>>

>>> student_objects = [
    Student('john', 'A', 15),
    Student('jane', 'B', 12),
    Student('dave', 'B', 10),
]
>>> sorted(student_objects, key=lambda student: student.age)   # sort by age
[('dave', 'B', 10), ('jane', 'B', 12), ('john', 'A', 15)]

上面写lambda可能有点麻烦，python还提供了更简单的办法，另外这个办法还可以按照优先级对不同关键字组合排序

>>> from operator import itemgetter, attrgetter

>>>

>>> sorted(student_tuples, key=itemgetter(2))
[('dave', 'B', 10), ('jane', 'B', 12), ('john', 'A', 15)]

>>>

>>> sorted(student_objects, key=attrgetter('age'))
[('dave', 'B', 10), ('jane', 'B', 12), ('john', 'A', 15)]

>>> sorted(student_tuples, key=itemgetter(1,2))
[('john', 'A', 15), ('dave', 'B', 10), ('jane', 'B', 12)]

>>>

>>> sorted(student_objects, key=attrgetter('grade', 'age'))
[('john', 'A', 15), ('dave', 'B', 10), ('jane', 'B', 12)]

自定义比较函数排序

这个在3.0之后不建议使用

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

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

但是如果还要兼容过去的版本可以使用下面的办法将cmp转换成key

def cmp_to_key(mycmp):
    'Convert a cmp= function into a key= function'
    class K(object):
        def __init__(self, obj, *args):
            self.obj = obj
        def __lt__(self, other):
            return mycmp(self.obj, other.obj) < 0
        def __gt__(self, other):
            return mycmp(self.obj, other.obj) > 0
        def __eq__(self, other):
            return mycmp(self.obj, other.obj) == 0
        def __le__(self, other):
            return mycmp(self.obj, other.obj) <= 0
        def __ge__(self, other):
            return mycmp(self.obj, other.obj) >= 0
        def __ne__(self, other):
            return mycmp(self.obj, other.obj) != 0
    return K

To convert to a key function, just wrap the old comparison function:

>>>

>>> sorted([5, 2, 4, 1, 3], key=cmp_to_key(reverse_numeric))
[5, 4, 3, 2, 1]

升序和降序

>>> sorted(student_tuples, key=itemgetter(2), reverse=True)
[('john', 'A', 15), ('jane', 'B', 12), ('dave', 'B', 10)]

>>>

>>> sorted(student_objects, key=attrgetter('age'), reverse=True)
[('john', 'A', 15), ('jane', 'B', 12), ('dave', 'B', 10)]

排序的稳定性

这个不需要我们额外设置，python所选用的排序算法都是稳定的。