如何在两个 Python 日期时间范围内计算属于给定月份的天数？

Question

I have two Python datetime and I want to count the days between those dates, counting ONLY the days belonging to the month I choose. The range might overlap multiple months/years.

Example: If I have 2017-10-29 & 2017-11-04 and I chose to count the days in October, I get 3 (29, 30 & 31 Oct.).

I can't find a straightforward way to do this so I think I'm going to iterate over the days using datetime.timedelta(days=1) , and increment a count each time the day belongs to the month I chose.

Do you know a more performant method?

I'm using Python 2.7.10 with the Django framework.

Answer 1

You can get difference between two datetime objects by simply subtracting them.

So, we start by getting the difference between the two dates. And then we generate all the dates between the two using

gen = (start_date + datetime.timedelta(days = e) for e in range(diff + 1))

And since we only want the dates between the specified ones, we apply a filter.

filter(lambda x : x==10 , gen) Then we will sum them over. And the final code is this:

diff = start_date - end_date
gen = (start_date + datetime.timedelta(days = e) for e in range(diff + 1))
filtered_dates = filter(
    lambda x : x.month == 10  , 
    gen
)
count = sum(1 for e in filtered_dates)

You can also use reduce but sum() is a lot more readable.

Answer 2

Iterating over the days would be the most straightforward way to do it. Otherwise, you would need to know how many days are in a given month and you would need different code for different scenarios:

1. The given month is the month of the first date
2. The given month is the month of the second date
3. The given month is between the first and the second date (if dates span more than two months)

If you want to support dates spanning more than one year then you would need the input to include month and year.

Your example fits scenario #1, which I guess you could do like this:

>>> from datetime import datetime, timedelta
>>>
>>> first_date = datetime(2017, 10, 29)
>>>
>>> first_day_of_next_month = first_date.replace(month=first_date.month + 1, day=1)
>>> last_day_of_this_month = first_day_of_next_month - timedelta(1)
>>> number_of_days_in_this_month = last_day_of_this_month.day
>>> number_of_days_in_this_month - first_date.day + 1
3

This is why I would suggest implementing it the way you originally intended and only turning to this if there's a performance concern.

Answer 3

A potential method of achieving this is to first compare whether your start or end dates you are comparing have the same month that you want to choose.

For example:

start = datetime(2017, 10, 29)
end = datetime(2017, 11, 4)

We create a function to compare the dates like so:

def daysofmonth(start, end, monthsel):
    if start.month == monthsel:
        days = (datetime(start.year, monthsel+1, 1) - start).days
    elif end.month == monthsel:
        days = (end - datetime(end.year, monthsel, 1)).days
    elif not (monthsel > start.month) & (end.month > monthsel):
        return 0
    else:
        days = (datetime(start.year, monthsel+1, 1) - datetime(start.year, monthsel, 1)).days

    return days

So, in our example setting monthsel gives:

 >>> daysofmonth(start, end, 10)
 >>> 3

Answer 4

Using pandas whit your dates:

import pandas as pd
from datetime import datetime

first_date = datetime(2017, 10, 29)
second_date = datetime(2017, 11, 4)
days_count = (second_date - first_date).days
month_date = first_date.strftime("%Y-%m")

values = pd.date_range(start=first_date,periods=days_count,freq='D').to_period('M').value_counts()
print(values)
print(values[month_date])

outputs

2017-10    3
2017-11    3
3