python2.7练习小例子（四）
4）：题目：输入某年某月某日，判断这一天是这一年的第几天？
程序分析：以3月5日为例，应该先把前两个月的加起来，然后再加上5天即本年的第几天，特殊情况，闰年且输入月份大于2时需考虑多加一天。
程序源代码：
#!/usr/bin/python
# -*- coding: UTF-8 -*-
year = int(raw_input('year:\n'))
month = int(raw_input('month:\n'))
day = int(raw_input('day:\n'))
months = (0,31,59,90,120,151,181,212,243,273,304,334)
if 0 < month <= 12:
sum = months[month - 1]
else:
print 'data error'
sum += day
leap = 0
if (year % 400 == 0) or ((year % 4 == 0) and (year % 100 != 0)):
leap = 1
if (leap == 1) and (month > 2):
sum += 1
print 'it is the %dth day.' % sum
以上实例输出结果为：
year:
2015
month:
6
day:
7
it is the 158th day.
看另外一个案例：
#!/usr/bin/python
# -*- coding: UTF-8 -*-
year=int(raw_input("年：\n"))
month=int(raw_input("月：\n"))
day=int(raw_input("日：\n"))
months1=[0,31,60,91,121,152,182,213,244,274,305,335,366] #闰年
months2=[0,31,59,90,120,151,181,212,243,273,304,334,365] #平年
if ((year%4==0)and(year%100!=0)) or((year%100==0)and(year%400==0)):
Dth=months1[month-1]+day
else:
Dth=months2[month-1]+day
print "是该年的第%d天"%Dth
闰年需要同时满足以下条件：
- 1、年份能被4整除；
- 2、年份若是 100 的整数倍的话需被400整除，否则是平年。
#!/usr/bin/python
# -*- coding: UTF-8 -*-
# 输入任意年月日，知道是改年第几天
p = [31,28,31,30,31,30,31,31,30,31,30,31] # 平年
w = [31,29,31,30,31,30,31,31,30,31,30,31] # 闰年
year =int(raw_input("请输入年："+'\n'))
month =int(raw_input("请输入月："+'\n'))
day=int(raw_input("请输入日："+'\n'))
arr=[31,28,31,30,31,30,31,31,30,31,30,31]
sum=day
for i in range(0,month-1):
sum+=arr[i]
if year%4==0:
if year%100==0 and year%400!=0:
print "这是今年的第%d天" % sum
else:
sum=sum+1
print "这是今年的第%d天" % sum
else:
print "这是今年的第%d天" % sum
再来参考一个：
#!/usr/bin/python
# -*- coding: UTF-8 -*-
# 输入任意年月日，知道是改年第几天
p = [31,28,31,30,31,30,31,31,30,31,30,31] # 平年
w = [31,29,31,30,31,30,31,31,30,31,30,31] # 闰年
year =int(raw_input("年:\n"))
month =int(raw_input("月:\n"))
day =int(raw_input("日:\n"))
# 判断闰年，平年
if year % 100 == 0:
if year % 400 == 0:
d=w
else:
d=p
else:
if year % 4 == 0:
d=w
else:
d=p
# 计算天数
days = sum(d[0:month-1])+day
print "%d.%d.%d是%d年的第%s天。"%(year,month,day,year,days)
还有哦：
#!/usr/bin/python
# -*- coding: UTF-8 -*-
year = int(input('请输入年份：'))
month = int(input('请输入月份：'))
day = int(input('请输入日期：'))
total = 0
if year%4 == 0:
days = 29
else:
days = 28
if year%4 == 0:
print year, '年是润年，2月有', days, '天！'
else:
print year, '年是平年，2月有', days, '天！'
if month <= 1:
total = day
elif month == 2:
total = 31 + day
elif month == 9 or month == 11:
total = (month - 2) * 30 + days + month/2 + day + 1
else:
total = (month - 2) * 30 + days + month/2 + day
print year, '年',month, '月', day, '日,是这一年的第', total, '天！'
完事再来看一个：
#!/usr/bin/python
def isLeapYear(a):
if (0 == a%4 or 0 == a%400) and 0 != a%100 :
return 1
else:
return 0
dict = {1: 31, 2: 28, 3: 31, 4: 30, 5: 31, 6: 30, 7: 31, 8: 31, 9: 30, 10: 31, 11: 30, 12: 31}
a = int(input("Input year:"))
b = int(input("Input month:"))
c = int(input("Input day:"))
m = 0
k = isLeapYear(a)
for i in range(1,b):
m = m + dict[i]
m = m + isLeapYear(a) + c
print(m)
#!/usr/bin/env python
# -*- coding: utf-8 -*-
year = int(raw_input('year:\n'))
month = int(raw_input('month:\n'))
day = int(raw_input('day:\n'))
days = [31,28,31,30,31,30,31,31,30,31,30,31]
if year % 400 == 0 or (year % 4 == 0 and year % 100 != 0):
days[2] += 1
now = sum(days[0:month-1])+day
print 'it is the %dth day.' %now
#!/usr/bin/env python
#coding:utf-8
import time
import sys
reload(sys)
sys.setdefaultencoding('utf-8') # 设置 'utf-8'
a = raw_input("输入时间（格式如：2017-04-04）:")
t = time.strptime(a,"%Y-%m-%d")
print time.strftime("今年的第%j天",t).decode('utf-8')
#! /usr/bin/env python
# coding:utf-8
import time
D=raw_input("请输入年份，格式如XXXX-XX-XX：")
d=time.strptime( D,'%Y-%m-%d').tm_yday
print "the {} day of this year!" .format(d)
Python3 参考解法：
#!/usr/bin/python3
date = input("输入年月日(yyyy-mm-dd):")
y,m,d = (int(i) for i in date.split('-'))
sum=0
special = (1,3,5,7,8,10)
for i in range(1,int(m)):
if i == 2:
if y%400==0 or (y%100!=0 and y%4==0):
sum+=29
else:
sum+=28
elif(i in special):
sum+=31
else:
sum+=30
sum+=d
print("这一天是一年中的第%d天"%sum)
再来看：
#!/usr/bin/python
# -*- coding: UTF-8 -*-
i= int(input('请输入年份：'))
j= int(input('请输入月份：'))
k= int(input('请输入天数：'))
month = [31,28,31,30,31,30,31,31,30,31,30,31]
day = 0
if i%4 == 0 or i %400 == 0 and i%100 != 0: #闰年
month = month[:j-1]
if j >2: #大于2月份
day = sum(month)+1+k
elif j== 2 and k ==29: #刚好在闰年闰天
day = sum(month)+1+k
else:
day = sum(month)+k
else: #平年
month = month[:j-1]
day = sum(month)+k
print('这一天是这一年的第{}天'.format(day))
#!/usr/bin/env python
#-*- coding:utf-8 -*-
import calendar
year = int(input("year: "))
month = int(input("month: "))
day = int(input("day: "))
days = 0
if 0 < month <= 12:
_, month_day = calendar.monthrange(year, month)
if day <= month_day:
for m in range(1, month):
_, month_day = calendar.monthrange(year, m)
days += month_day
days += day
print days
else:
print "input day error"
else:
print "input month error"
通过计算输入的日期与相应年份1月1日相差的秒数，然后除以每天的秒数3600*24，即可得到输入日期的天数：
#!/usr/bin/env python3
import time
def datecount(date):
date0=date[0:4]+"-01-01"
datet=time.strptime(date,"%Y-%m-%d") #将输入的字符串转化为时间元组
date0t=time.strptime(date0,"%Y-%m-%d")
dates=time.mktime(datet) #将时间元组转化为时间戳
date0s=time.mktime(date0t)
count=(dates-date0s)/(3600*24) #输入日期的时间戳减当前年份0101的时间戳除以每天秒数
return count+1
a=input("请输入日期：格式如2017-06-16\n")
print("{}是{}年第{}天".format(a,a[0:4],int(datecount(a))))
考虑实际的情况，比如输入月份为13月或输入天数为65天时候报错（日期仅校对0-31天，未按照实际日期校对）：
#!/usr/bin/env python
#-*- coding:utf-8 -*-
print('输入年月日以查看某一日期是当年第几天\n')
year = int(input('请输入年份：\n'))
month = int(input('请输入月份:\n'))
day = int(input('请输入日期:\n'))
months = [31,28,31,30,31,30,31,31,30,31,30,31]
d = 0
if 0<month<=12:
if 0<day<=31:
d = d + day
if month > 2:
if (year % 400 == 0) or ((year % 4 == 0) and (year % 100 != 0)):
d = d + 1
for i in range(12):
if (i+1) < month:
d = d + months[i]
i = i + 1
print(d)
else:
print('输入的日期有错误')
else:
print('输入的月份有错误')
通过输入时间点的unix时间戳和输入年份首日的Unix时间戳之间的差，来计算经过的时间：
#coding=utf-8
import time
print "Please Enter full number just like 02 01 03"
y = int(raw_input('Enter the year:')) #分别输入年月日
m = int(raw_input('Enter the month:'))
d = int(raw_input('Enter the day:'))
a = (y,m,d,00,00,00,00,00,00) #要求长度为9
b = (y,01,01,00,00,00,00,00,00) #输入年份的第一天
timestampa = time.mktime(a) #两个都转换为Unix时间戳，即1970.1.1到现在经过的秒数
timestampb = time.mktime(b)
timec = int((timestampa - timestampb)/(3600*24)) #输入的时间戳减去年份首天的时间戳等于经过的秒数，再换算成天，取整
print("There are {} days goes by!".format(timec))
python3 利用time模块，简洁写法：
import time
print(time.strptime('2017-9-20', '%Y-%m-%d')[7])
Python2.x 与 Python3.x 兼容：
#!/usr/bin/python
# -*- coding: UTF-8 -*-
# 觉得自己的逻辑看起来更顺眼，嘻嘻！
days = [0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334]
yy = int(input('请输入年份：'))
if yy >= 0:
mm = int(input('请输入月份：'))
if 0 < mm < 13:
dd = int(input('请输入日期：'))
if ((yy % 4 == 0) and (yy % 100 != 0)) or (yy % 400 == 0):
if mm <= 2:
sum = days[mm - 1] + dd
else:
sum = days[mm - 1] + 1 + dd
else:
sum = days[mm - 1] + dd
print('您输入的时间在这一年的第%d天' % sum)
else:
print('您输入的月分不正确')
else:
print('请输入正确的公元年')
#!/usr/bin/python
# -*- coding: UTF-8 -*-
from functools import reduce
year = int(input('请输入年（如：2017）：'))
month = int(input('请输入月（如：3）：'))
day = int(input('请输入日（如：16）：'))
mday = [0,31,28 if year%4 else 29 if year%100 else 28 if year%4 else 29,31,30,31,30,31,31,30,31,30,31]
print('{}年{}月{}日是当年的第{}天'.format(year, month, day, reduce(lambda x,y:x+y, mday[:month])+day))
加入异常处理，确保日期输入格式正确：
import time
while 1:
try:
a=input('请输入日期yyyy-mm-dd:')
b=time.strptime(a,'%Y-%m-%d') #按输入格式转化成时间格式 local变量
except ValueError:
print('请输入正确的日期格式！')
else:
b=time.strptime(a,'%Y-%m-%d') #按输入格式转化成时间格式 global变量
dd=time.strftime('%j',b) #返回一年中的第几天
yy=time.strftime('%Y',b) #返回年份
print('输入的日期是%s年的第%s天'%(yy,dd))
break
基本上就是各种各样的案例。。。可见诸位的大才。。。
如果感觉不错的话，请多多点赞支持哦。。。
原文链接：https://blog.csdn.net/luyaran/article/details/80016648
