Question

出于某种原因，我真的很难接受这个（相对论）问题。

如何计算两个日期之间的差异。我想在不使用模块的情况下这样做。但由于某种原因，我的代码没有输出正确的答案。

这是我的思考过程：

如果被要求计算2014年12月10日到2015年2月2日之间的天数。

首先从10日（31 - 10）找到12月剩余的天数 = 21天
查找12月到2月（又名Jan）之间的月数添加该月的天数= 31天
添加12月（21）剩余的天数+月份（31）之间的天数+上个月（2）= 54天的天数。
然后检查异常，即闰年等。

这是我的功能：

def Calculate_Date (year1, month1, day1, year2, month2, day2):
"""
This function takes to dates (year/month/day) and returned the
difference between the dates
"""

#Create a dict for the # of days in each month
month_days = {1:31,2:28,3:31,4:30,5:31,6:30,7:31,8:31,9:30,10:31,11:30,12:31}

days_left_in_month1 = month_days[month1] - day1
days_left_in_year1 =0 
days_into_year2 =0
days_between_year1_and_year2= 0
difference_in_days = 0

# Find the number days left in year one
i = month1
days_left_in_year = []
while i <= 12:
    days = month_days[i]
    days_left_in_year.append(days)
    i = i + 1

days_left_in_year1 = (sum(days_left_in_year)) - day1

# Find the number days into year two
i = 1
days_into_year = []
while i <= month2:
    days = month_days[i]
    days_into_year.append(days)
    i = i + 1
days_into_year2 = sum(days_into_year) - day2

#find the differernce in years 
days_between_year1_and_year2 = (year2 - year1) * 365


#Check if its a leap year
leap_year = False
while True:
    if float(year1 % 4) == 0:
        if float(year1 % 100) != 0:
                leap_year = True
                break
        if float(year1 % 100) == 0:
            if float(year1 % 400) ==0:
                leap_year = True
                break               
    else:
        break

#test output
print "The number of days left in the year One are %r " % days_left_in_year1
print "The number of days into the year Two are %r "    % days_into_year2
print "The number of days between the years are %r "    % days_between_year1_and_year2

#add an increment if leap year was true
if leap_year == True:
    difference_in_days = days_left_in_year1 + days_into_year2 + days_between_year1_and_year2 + 1

else:
    difference_in_days = days_left_in_year1 + days_into_year2 + days_between_year1_and_year2

return difference_in_days


print Calculate_Date(2011,6,30,2012,06,30)

Answer 1

而不是date2 - date1，您可能会发现(date2 - x) - (date1 - x)更容易，其中x是一个易于处理的日期，即year1的“Jan 0”

让我们定义几个函数：

def days_in_month(year, month):
    """
    Return number of days in the specified month (28, 29, 30, or 31)
    """
    if month == 2:    # February
        if not (year % 400):
            return 29
        elif not (year % 100):
            return 28
        elif not (year % 4):
            return 29
        else:
            return 28
    elif month in {1, 3, 5, 7, 8, 10, 12}:
        return 31
    else:
        return 30

def days_in_year(year):
    """
    Return the number of days in the specified year (365 or 366)
    """
    return 337 + days_in_month(year, 2)

def days_this_year(year, month, day):
    """
    Return the number of days so far this year
    """
    return sum(days_in_month(year, m) for m in range(1, month)) + day

def year_days_since(base_year, this_year):
    """
    Return the number of days from the start of base_year to the start of this_year
    """
    if base_year > this_year:
        raise ValueError("base_year must be <= this_year")
    elif base_year == this_year:
        return 0
    else:
        return sum(days_in_year(y) for y in range(base_year, this_year))

然后两个日期之间的差异变为：

def date_diff(y1, m1, d1, y2, m2, d2):
    x = min(y1, y2)   # base date
    days1 = year_days_since(x, y1) + days_this_year(y1, m1, d1)
    days2 = year_days_since(x, y2) + days_this_year(y2, m2, d2)
    return days2 - days1

由于这个答案的对称性，它也会很乐意做出负面的差异：

date_diff(2001, 1, 3, 2002, 2, 5)   # =>  398 == 365 + 31 + 2
date_diff(2002, 2, 5, 2001, 1, 3)   # => -398

Answer 2

如果这是真正的代码，而不是学校作业，这就是我的方式：

from datetime import date

def date_diff(y1, m1, d1, y2, m2, d2):
    return (date(y2, m2, d2) - date(y1, m1, d1)).days

两个日期之间的Python差异

2 个答案: