如果我有两个日期（例如'8/18/2008'和'9/26/2008'），获取这两个日期之间天数的最佳方法是什么？
Ray Vega
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793205 次
14 回答
203
使用日期时间的力量：
from datetime import datetime
date_format = "%m/%d/%Y"
a = datetime.strptime('8/18/2008', date_format)
b = datetime.strptime('9/26/2008', date_format)
delta = b - a
print delta.days # that's it
于 2008-09-29T23:41:59.710 回答
48
离圣诞节还有几天：
>>> import datetime
>>> today = datetime.date.today()
>>> someday = datetime.date(2008, 12, 25)
>>> diff = someday - today
>>> diff.days
86
更多的算术在这里。
于 2008-09-29T23:43:01.937 回答
20
你想要 datetime 模块。
>>> from datetime import datetime, timedelta
>>> datetime(2008,08,18) - datetime(2008,09,26)
datetime.timedelta(4)
另一个例子：
>>> import datetime
>>> today = datetime.date.today()
>>> print(today)
2008-09-01
>>> last_year = datetime.date(2007, 9, 1)
>>> print(today - last_year)
366 days, 0:00:00
正如这里所指出的
于 2008-09-29T23:43:10.300 回答
19
每个人都用日期回答得很好，让我尝试用熊猫回答
dt = pd.to_datetime('2008/08/18', format='%Y/%m/%d')
dt1 = pd.to_datetime('2008/09/26', format='%Y/%m/%d')
(dt1-dt).days
这将给出答案。如果输入之一是数据框列。只需使用dt.days代替days
(dt1-dt).dt.days
于 2019-11-13T06:22:05.673 回答
16
from datetime import datetime
start_date = datetime.strptime('8/18/2008', "%m/%d/%Y")
end_date = datetime.strptime('9/26/2008', "%m/%d/%Y")
print abs((end_date-start_date).days)
于 2012-04-26T10:58:02.887 回答
12
它也可以通过以下方式轻松完成arrow：
import arrow
a = arrow.get('2017-05-09')
b = arrow.get('2017-05-11')
delta = (b-a)
print delta.days
于 2017-05-11T18:35:15.533 回答
9
不使用 Lib 只是纯代码：
#Calculate the Days between Two Date
daysOfMonths = [ 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
def isLeapYear(year):
# Pseudo code for this algorithm is found at
# http://en.wikipedia.org/wiki/Leap_year#Algorithm
## if (year is not divisible by 4) then (it is a common Year)
#else if (year is not divisable by 100) then (ut us a leap year)
#else if (year is not disible by 400) then (it is a common year)
#else(it is aleap year)
return (year % 4 == 0 and year % 100 != 0) or year % 400 == 0
def Count_Days(year1, month1, day1):
if month1 ==2:
if isLeapYear(year1):
if day1 < daysOfMonths[month1-1]+1:
return year1, month1, day1+1
else:
if month1 ==12:
return year1+1,1,1
else:
return year1, month1 +1 , 1
else:
if day1 < daysOfMonths[month1-1]:
return year1, month1, day1+1
else:
if month1 ==12:
return year1+1,1,1
else:
return year1, month1 +1 , 1
else:
if day1 < daysOfMonths[month1-1]:
return year1, month1, day1+1
else:
if month1 ==12:
return year1+1,1,1
else:
return year1, month1 +1 , 1
def daysBetweenDates(y1, m1, d1, y2, m2, d2,end_day):
if y1 > y2:
m1,m2 = m2,m1
y1,y2 = y2,y1
d1,d2 = d2,d1
days=0
while(not(m1==m2 and y1==y2 and d1==d2)):
y1,m1,d1 = Count_Days(y1,m1,d1)
days+=1
if end_day:
days+=1
return days
# Test Case
def test():
test_cases = [((2012,1,1,2012,2,28,False), 58),
((2012,1,1,2012,3,1,False), 60),
((2011,6,30,2012,6,30,False), 366),
((2011,1,1,2012,8,8,False), 585 ),
((1994,5,15,2019,8,31,False), 9239),
((1999,3,24,2018,2,4,False), 6892),
((1999,6,24,2018,8,4,False),6981),
((1995,5,24,2018,12,15,False),8606),
((1994,8,24,2019,12,15,True),9245),
((2019,12,15,1994,8,24,True),9245),
((2019,5,15,1994,10,24,True),8970),
((1994,11,24,2019,8,15,True),9031)]
for (args, answer) in test_cases:
result = daysBetweenDates(*args)
if result != answer:
print "Test with data:", args, "failed"
else:
print "Test case passed!"
test()
于 2018-02-05T22:37:00.863 回答
6
还有一种datetime.toordinal()方法还没有提到：
import datetime
print(datetime.date(2008,9,26).toordinal() - datetime.date(2008,8,18).toordinal()) # 39
https://docs.python.org/3/library/datetime.html#datetime.date.toordinal
date.序数（）返回日期的预测公历序数，其中第 1 年的 1 月 1 日的序数为 1。对于任何
date对象d，date.fromordinal(d.toordinal()) == d.
似乎很适合计算天数差异，尽管不如timedelta.days.
于 2019-06-09T14:03:16.917 回答
5
对于计算日期和时间，有几个选项，但我会写一个简单的方法：
from datetime import timedelta, datetime, date
import dateutil.relativedelta
# current time
date_and_time = datetime.now()
date_only = date.today()
time_only = datetime.now().time()
# calculate date and time
result = date_and_time - timedelta(hours=26, minutes=25, seconds=10)
# calculate dates: years (-/+)
result = date_only - dateutil.relativedelta.relativedelta(years=10)
# months
result = date_only - dateutil.relativedelta.relativedelta(months=10)
# days
result = date_only - dateutil.relativedelta.relativedelta(days=10)
# calculate time
result = date_and_time - timedelta(hours=26, minutes=25, seconds=10)
result.time()
希望能帮助到你
于 2018-01-31T10:53:46.613 回答
3
from datetime import date
def d(s):
[month, day, year] = map(int, s.split('/'))
return date(year, month, day)
def days(start, end):
return (d(end) - d(start)).days
print days('8/18/2008', '9/26/2008')
当然，这假设您已经验证您的日期格式为r'\d+/\d+/\d+'.
于 2016-04-15T14:50:41.070 回答
3
以下是解决此问题的三种方法：
from datetime import datetime
Now = datetime.now()
StartDate = datetime.strptime(str(Now.year) +'-01-01', '%Y-%m-%d')
NumberOfDays = (Now - StartDate)
print(NumberOfDays.days) # Starts at 0
print(datetime.now().timetuple().tm_yday) # Starts at 1
print(Now.strftime('%j')) # Starts at 1
于 2018-03-20T13:29:28.427 回答
1
如果您想自己编写计算代码，那么这里有一个函数将返回给定年、月和日的序数：
def ordinal(year, month, day):
return ((year-1)*365 + (year-1)//4 - (year-1)//100 + (year-1)//400
+ [ 0,31,59,90,120,151,181,212,243,273,304,334][month - 1]
+ day
+ int(((year%4==0 and year%100!=0) or year%400==0) and month > 2))
date.toordinal该函数与datetime 模块中的方法兼容。
您可以获得两个日期之间的差异天数，如下所示：
print(ordinal(2021, 5, 10) - ordinal(2001, 9, 11))
于 2021-05-10T16:29:04.207 回答
0
在 python 中不使用 datetime 对象。
# A date has day 'd', month 'm' and year 'y'
class Date:
def __init__(self, d, m, y):
self.d = d
self.m = m
self.y = y
# To store number of days in all months from
# January to Dec.
monthDays = [31, 28, 31, 30, 31, 30,
31, 31, 30, 31, 30, 31 ]
# This function counts number of leap years
# before the given date
def countLeapYears(d):
years = d.y
# Check if the current year needs to be considered
# for the count of leap years or not
if (d.m <= 2) :
years-= 1
# An year is a leap year if it is a multiple of 4,
# multiple of 400 and not a multiple of 100.
return int(years / 4 - years / 100 + years / 400 )
# This function returns number of days between two
# given dates
def getDifference(dt1, dt2) :
# COUNT TOTAL NUMBER OF DAYS BEFORE FIRST DATE 'dt1'
# initialize count using years and day
n1 = dt1.y * 365 + dt1.d
# Add days for months in given date
for i in range(0, dt1.m - 1) :
n1 += monthDays[i]
# Since every leap year is of 366 days,
# Add a day for every leap year
n1 += countLeapYears(dt1)
# SIMILARLY, COUNT TOTAL NUMBER OF DAYS BEFORE 'dt2'
n2 = dt2.y * 365 + dt2.d
for i in range(0, dt2.m - 1) :
n2 += monthDays[i]
n2 += countLeapYears(dt2)
# return difference between two counts
return (n2 - n1)
# Driver program
dt1 = Date(31, 12, 2018 )
dt2 = Date(1, 1, 2019 )
print(getDifference(dt1, dt2), "days")
于 2020-07-26T03:00:11.510 回答