python - 计算两天之间每个月的天数

Question

我正在寻找一个函数来获取 2 个日期(入院和出院)和一个财政年度，并返回这些日期之间每个月的天数。

财政年度为 4 月 1 日 -> 3 月 31 日

我目前有一个解决方案(如下)，它是 SPSS 和 Python 的一团糟，最终需要将其实现回 SPSS，但作为一个更整洁的 Python 函数，不幸的是，这意味着它只能使用标准库(不是 Pandas ).

例如

+-----------------+-----------------+------+--+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+
|    Admission    |    Discharge    |  FY  |  | Apr | May | Jun | Jul | Aug | Sep | Oct | Nov | Dec | Jan | Feb | Mar |
+-----------------+-----------------+------+--+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+
| 01 January 2017 | 05 January 2017 | 1617 |  |   0 |   0 |   0 |   0 |   0 |   0 |   0 |   0 |   0 |   4 |   0 |   0 |
| 01 January 2017 | 05 June 2017    | 1617 |  |   0 |   0 |   0 |   0 |   0 |   0 |   0 |   0 |   0 |  31 |  28 |  31 |
| 01 January 2017 | 05 June 2017    | 1718 |  |  30 |  31 |   4 |   0 |   0 |   0 |   0 |   0 |   0 |   0 |   0 |   0 |
| 01 January 2017 | 01 January 2019 | 1718 |  |  30 |  31 |  30 |  31 |  31 |  30 |  31 |  30 |  31 |  31 |  28 |  31 |
+-----------------+-----------------+------+--+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+

Related - How to calculate number of days between two given dates?

Current solution (SPSS code)

 * Count the beddays.
 * Similar method to that used in Care homes.
 * 1) Declare an SPSS macro which will set the beddays for each month.
 * 2) Use python to run the macro with the correct parameters.
 * This means that different month lengths and leap years are handled correctly.
Define !BedDaysPerMonth (Month = !Tokens(1) 
   /MonthNum = !Tokens(1) 
   /DaysInMonth = !Tokens(1) 
   /Year = !Tokens(1))

 * Store the start and end date of the given month.
Compute #StartOfMonth = Date.DMY(1, !MonthNum, !Year).
Compute #EndOfMonth = Date.DMY(!DaysInMonth, !MonthNum, !Year).

 * Create the names of the variables e.g. April_beddays and April_cost.
!Let !BedDays = !Concat(!Month, "_beddays").

 * Create variables for the month.
Numeric !BedDays (F2.0).

 * Go through all possibilities to decide how many days to be allocated.
Do if keydate1_dateformat LE #StartOfMonth.
   Do if keydate2_dateformat GE #EndOfMonth.
      Compute !BedDays = !DaysInMonth.
   Else.
      Compute !BedDays = DateDiff(keydate2_dateformat, #StartOfMonth, "days").
   End If.
Else if keydate1_dateformat LE #EndOfMonth.
   Do if keydate2_dateformat GT #EndOfMonth.
      Compute !BedDays = DateDiff(#EndOfMonth, keydate1_dateformat, "days") + 1.
   Else.
      Compute !BedDays = DateDiff(keydate2_dateformat, keydate1_dateformat, "days").
   End If.
Else.
   Compute !BedDays = 0.
End If.

 * Months after the discharge date will end up with negatives.
If !BedDays < 0 !BedDays = 0.
!EndDefine.

 * This python program will call the macro for each month with the right variables.
 * They will also be in FY order.
Begin Program.
from calendar import month_name, monthrange
from datetime import date
import spss

#Set the financial year, this line reads the first variable ('year')
fin_year = int((int(spss.Cursor().fetchone()[0]) // 100) + 2000)

#This line generates a 'dictionary' which will hold all the info we need for each month
#month_name is a list of all the month names and just needs the number of the month
#(m < 4) + 2015 - This will set the year to be 2015 for April onwards and 2016 other wise
#monthrange takes a year and a month number and returns 2 numbers, the first and last day of the month, we only need the second.
months = {m: [month_name[m], (m < 4) + fin_year, monthrange((m < 4) + fin_year, m)[1]]  for m in range(1,13)}
print(months) #Print to the output window so you can see how it works

#This will make the output look a bit nicer
print("\n\n***This is the syntax that will be run:***")

#This loops over the months above but first sorts them by year, meaning they are in correct FY order
for month in sorted(months.items(), key=lambda x: x[1][1]):
   syntax = "!BedDaysPerMonth Month = " + month[1][0][:3]
   syntax += " MonthNum = " + str(month[0])
   syntax += " DaysInMonth = " + str(month[1][2])
   syntax += " Year = " + str(month[1][1]) + "."

   print(syntax)
   spss.Submit(syntax)
End Program.

Answer 1

我能想到的唯一方法是遍历每一天并解析它所属的月份:

import time, collections
SECONDS_PER_DAY = 24 * 60 * 60
def monthlyBedDays(admission, discharge, fy=None):

    start = time.mktime(time.strptime(admission, '%d-%b-%Y'))
    end = time.mktime(time.strptime( discharge, '%d-%b-%Y'))
    if fy is not None:
        fy = str(fy)
        start = max(start, time.mktime(time.strptime('01-Apr-'+fy[:2], '%d-%b-%y')))
        end   = min(end,   time.mktime(time.strptime('31-Mar-'+fy[2:], '%d-%b-%y')))
    days = collections.defaultdict(int)
    for day in range(int(start), int(end) + SECONDS_PER_DAY, SECONDS_PER_DAY):
        day = time.localtime(day)
        key = time.strftime('%Y-%m', day)  # use '%b' to answer the question exactly, but that's not such a good idea
        days[ key ] += 1
    return days

output = monthlyBedDays(admission="01-Jan-2018", discharge="25-Apr-2018")
print(output)
# Prints:
# defaultdict(<class 'int'>, {'2018-01': 31, '2018-02': 28, '2018-03': 31, '2018-04': 25})

print(monthlyBedDays(admission="01-Jan-2018", discharge="25-Apr-2018", fy=1718))
# Prints:
# defaultdict(<class 'int'>, {'2018-01': 31, '2018-02': 28, '2018-03': 31})

print(monthlyBedDays(admission="01-Jan-2018", discharge="25-Apr-2018", fy=1819))
# Prints:
# defaultdict(<class 'int'>, {'2018-04': 25})

请注意，输出是一个 defaultdict，这样，如果您要求它提供任何月份(或任何键)未记录的天数(例如 output['1999-12']) 它将返回 0。另请注意，我对输出键使用了 '%Y-%m' 格式。与使用最初要求的键类型 ('%b' -> 'Jan').