转自：http://www.cnblogs.com/lkprof/p/3179850.html，感谢分享~

问题1：如果日期中有千年以前的情况(没法用格式化函数)，如('2010-11-23','1989-3-7','2010-1-5','978-12-1','2010-2-4')参照方法1

问题2：如果日期中没有千年以前的情况，做法就很多了。参照方法2和方法3

1 # -*- coding: utf-8 -*-

2 import time

3 from operator import itemgetter

4 arr=('2010-11-23','1989-3-7','2010-1-5','978-12-1','2010-2-4')

5 def date_sort1(x):

6 ls=list(x)

7 #用了冒泡排序来排序，其他方法效果一样

8 for j in range(len(ls)-1):

9 for i in range(len(ls)-j-1):

10 lower=ls[i].split('-')

11 upper=ls[i+1].split('-')

12 for s in range(3):

13 if int(lower[s])>int(upper[s]):

14 ls[i],ls[i+1]=ls[i+1],ls[i]

15 break

16 elif int(lower[s])

17 break

18 ar=tuple(ls)

19 return ar

20 ar=('2010-11-23','1989-3-7','2010-2-4','2010-1-5')

21 def date_sort2(x):

22 ls=list(x)

23 dic={}

24 for l in ls:

25 #返回用秒数来表示时间的浮点数

26 dic[l]=time.mktime(time.strptime(l, '%Y-%m-%d'))

27

28 dic=sorted(dic.iteritems(), key=itemgetter(1))

29 sorted_items=[keys[0] for keys in dic]

30 '''

31 items=dic.items()

32 backitems=[[v[1],v[0]] for v in items]

33 backitems.sort()

34 sorted_items=[keys[1] for keys in backitems]

35 '''

36 '''

37 items=dic.items()

38 backitems=[[v[0],v[1]] for v in items]

39 backitems=sorted(backitems, key=lambda x : x[1])

40 sorted_items=[keys[0] for keys in backitems]

41 '''

42 return tuple(sorted_items)

43 import datetime

44 def date_sort3(x):

45 ls=list(x)

46 #用了冒泡排序来排序，其他方法效果一样

47 for j in range(len(ls)-1):

48 for i in range(len(ls)-j-1):

49 lower=datetime.datetime.strptime(ls[i], '%Y-%m-%d')

50 upper=datetime.datetime.strptime(ls[i+1], '%Y-%m-%d')

51 if lower>upper:

52 ls[i],ls[i+1]=ls[i+1],ls[i]

53 return tuple(ls)

54 print date_sort1(arr)

55 print date_sort2(ar)

56 print date_sort3(ar)

运行结果：

('978-12-1', '1989-3-7', '2010-1-5', '2010-2-4', '2010-11-23')

('1989-3-7', '2010-1-5', '2010-2-4', '2010-11-23')

('1989-3-7', '2010-1-5', '2010-2-4', '2010-11-23')

正则表达式同样可以处理这类问题,下面是正则表达式的解决方案。

1 #利用正则表达式

2 import re

3

4 data = ['2010-11-23','1989-3-7','2010-1-5','978-12-1','2010-2-4']

5 patt = '(\d+)-(\d+)-(\d+)'

6 #交换排序

7 for i in range(len(data)-1):

8 for x in range(i+1, len(data)):

9 j = 1

10 while j<4:

11 lower = re.match(patt, data[i]).group(j)

12 upper = re.match(patt, data[x]).group(j)

13 #print lower,upper

14 if int(lower) < int(upper):

15 j = 4

16 elif int(lower) == int(upper):

17 j += 1

18 else:

19 data[i],data[x] = data[x],data[i]

20 j = 4

21 print data