【分享】Python时间排序

本文来源于：http://www.cnblogs.com/lkprof/p/3179850.html

不过原博客代码中是有点小问题，经过我测试运行，没问题才贴出来代码。

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('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。

代码有两段，分别附上：

date_sort_one.py:

#-*- coding:utf-8 _*-  
""" 
@author:yuanpengfei 
@file: date_sort.py 
@time: 2017/12/01 
"""

# -*- 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) > int(upper):
                    ls[i], ls[i + 1] = ls[i + 1], ls[i]
                    break
                elif int(lower) < int(upper):
                    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.items(), 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))

测试结果：

('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')

date_sort_two.py:

#-*- coding:utf-8 _*-  
""" 
@author:yuanpengfei 
@file: date_sort_two.py 
@time: 2017/12/01 
"""

#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)

测试结果：

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

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