python 日期排序

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

# -*- coding: utf-8 -*-
import time
from operator import itemgetter
arr=('2010-11-23','1989-3-7','2010-1-5','978-12-1','2010-2-4')
def date_sort1(x):
    ls=list(x)
    #用了冒泡排序来排序，其他方法效果一样
    for j in range(len(ls)-1):
        for i in range(len(ls)-j-1):
            lower=ls[i].split('-')
            upper=ls[i+1].split('-')       
            for s in range(3):
                if int(lower[s])>int(upper[s]):                
                    ls[i],ls[i+1]=ls[i+1],ls[i]
                    break
                elif int(lower[s])<int(upper[s]):
                    break
    ar=tuple(ls)
    return ar  
ar=('2010-11-23','1989-3-7','2010-2-4','2010-1-5')
def date_sort2(x):
    ls=list(x)
    dic={}
    for l in ls:
        #返回用秒数来表示时间的浮点数
        dic[l]=time.mktime(time.strptime(l, '%Y-%m-%d'))
    
    dic=sorted(dic.iteritems(), key=itemgetter(1))
    sorted_items=[keys[0] for keys in dic]
    '''
    items=dic.items() 
    backitems=[[v[1],v[0]] for v in items] 
    backitems.sort() 
    sorted_items=[keys[1] for keys in backitems] 
    '''
    '''
    items=dic.items() 
    backitems=[[v[0],v[1]] for v in items] 
    backitems=sorted(backitems, key=lambda x : x[1])   
    sorted_items=[keys[0] for keys in backitems] 
    '''
    return tuple(sorted_items)
import datetime
def date_sort3(x):
    ls=list(x)
    #用了冒泡排序来排序，其他方法效果一样
    for j in range(len(ls)-1):
        for i in range(len(ls)-j-1):
            lower=datetime.datetime.strptime(ls[i], '%Y-%m-%d')
            upper=datetime.datetime.strptime(ls[i+1], '%Y-%m-%d')
            if lower>upper:
                ls[i],ls[i+1]=ls[i+1],ls[i]
    return tuple(ls)
print date_sort1(arr)         
print date_sort2(ar)
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。

#利用正则表达式
import re

data = ['2010-11-23','1989-3-7','2010-1-5','978-12-1','2010-2-4']
patt = '(d+)-(d+)-(d+)'
    #交换排序
for i in range(len(data)-1):
    for x in range(i+1, len(data)):
        j = 1
        while j<4:
            lower = re.match(patt, data[i]).group(j)
            upper = re.match(patt, data[x]).group(j)
            #print lower,upper
            if int(lower) < int(upper):
                j = 4
            elif int(lower) == int(upper):
                j += 1
            else:
                data[i],data[x] = data[x],data[i]
                j = 4
print data