How to Calculate Number of Days Between Two Given Dates

How to calculate number of days between two given dates

If you have two date objects, you can just subtract them, which computes a timedelta object.

from datetime import date

d0 = date(2008, 8, 18)
d1 = date(2008, 9, 26)
delta = d1 - d0
print(delta.days)

The relevant section of the docs:
https://docs.python.org/library/datetime.html.

See this answer for another example.

How to calculate number of days between two dates

http://momentjs.com/ or https://date-fns.org/

From Moment docs:

var a = moment([2007, 0, 29]);
var b = moment([2007, 0, 28]);
a.diff(b, 'days')   // =1

or to include the start:

a.diff(b, 'days')+1   // =2

Beats messing with timestamps and time zones manually.

Depending on your specific use case, you can either

Use a/b.startOf('day') and/or a/b.endOf('day') to force the diff to be inclusive or exclusive at the "ends" (as suggested by @kotpal in the comments).
Set third argument true to get a floating point diff which you can then Math.floor, Math.ceil or Math.round as needed.
Option 2 can also be accomplished by getting 'seconds' instead of 'days' and then dividing by 24*60*60.

How to calculate number of days between two dates?

Here is a quick and dirty implementation of datediff, as a proof of concept to solve the problem as presented in the question. It relies on the fact that you can get the elapsed milliseconds between two dates by subtracting them, which coerces them into their primitive number value (milliseconds since the start of 1970).

// new Date("dateString") is browser-dependent and discouraged, so we'll write// a simple parse function for U.S. date format (which does no error checking)function parseDate(str) {    var mdy = str.split('/');    return new Date(mdy[2], mdy[0]-1, mdy[1]);}
function datediff(first, second) {    // Take the difference between the dates and divide by milliseconds per day.    // Round to nearest whole number to deal with DST.    return Math.round((second-first)/(1000*60*60*24));}
alert(datediff(parseDate(first.value), parseDate(second.value)));

<input id="first" value="1/1/2000"/><input id="second" value="1/1/2001"/>

Finding the number of days between two dates

$now = time(); // or your date as well
$your_date = strtotime("2010-01-31");
$datediff = $now - $your_date;

echo round($datediff / (60 * 60 * 24));

count the number of days between two dates per year

Here's one approach using tidyverse and lubridate.

First, separate the rows by calendar year, to use to measure the number of days for each year. Each row will include dates to be counted in each calendar year, starting with January 1st and ending with December 31st if overlapping multiple years. Then, it is easy to calculate the number of days in a given year.

The results from this example are slightly different than what I have. Year 2016 is a leap year and has 366 days. If the number of days are not inclusive of either start or end dates, you would get a different answer.

library(tidyverse)
library(lubridate)

df %>%
  mutate(date_int = interval(start, end),
         year = map2(year(start), year(end), seq)) %>%
  unnest(year) %>%
  mutate(year_int = interval(as.Date(paste0(year, '-01-01')), as.Date(paste0(year, '-12-31'))),
         year_sect = intersect(date_int, year_int),
         start_new = as.Date(int_start(year_sect)),
         end_new = as.Date(int_end(year_sect))) %>%
  select(id, start_new, end_new) %>%
  mutate(year = year(start_new),
         days = as.numeric(end_new - start_new)) %>%
  right_join(df) %>%
  pivot_wider(id_cols = c(id, start, end), names_from = year, values_from = days, names_prefix = "year_", values_fill = list(days = 0)) %>%
  mutate(days_number = reduce(select(., starts_with("year_")), `+`))

Output

     id start      end        year_2015 year_2016 days_number
  <dbl> <date>     <date>         <dbl>     <dbl>       <dbl>
1     1 2015-01-01 2016-12-31       364       365         729
2     2 2016-01-01 2016-12-31         0       365         365
3     3 2015-07-01 2016-12-31       183       365         548

Calculate days between two dates given specific values

def dayDiff(groupby):
    if (not (groupby.Yes == 1).any()) or (not (groupby.Value > 0).any()):
        return np.zeros(groupby.Date.count())

    min_date = groupby[groupby.Yes == 1].Date.iloc[0]    
    max_date = groupby[groupby.Value > 0].Date.iloc[0]
    delta = max_date - min_date
    return np.where(groupby.Value > 0 , delta.days, 0)


df1.Date = pd.to_datetime(df1.Date, dayfirst=True)
DateDiff = df1.groupby('UserId').apply(dayDiff).explode().rename('DateDiff').reset_index(drop=True)
pd.concat([df1, DateDiff], axis=1)

Returns:


Date    UserId  Value     Yes       DateDiff
0   2017-01-02  1   0      1          0
1   2017-01-03  1   0      0          0
2   2017-01-04  1   0      0          0
3   2017-01-05  1   100    0          3
4   2017-01-01  2   0      1          0
5   2017-01-02  2   1000   0          1
6   2017-01-03  2   0      0          0

Although this answers your question, the date diff logic is hard to follow, especially when it comes to the placement of the DateDiff values.

Update

pd.Series.explode() was only introduced in pandas version 0.25, for those using previous versions:

df1.Date = pd.to_datetime(df1.Date, dayfirst=True)
DateDiff = (df1
            .groupby('UserId')
            .apply(dayDiff)
            .to_frame()
            .explode(0)
            .reset_index(drop=True)
            .rename(columns={0: 'DateDiff'}))
pd.concat([df1, DateDiff], axis=1)

This will yield the same results.

calculate number of 'Real' days between two dates when given a number of business days as input

For business days over 5 days you can take the business day, divide by 5 and make at an integer (floor), multiply by 7, and add the modulus of business days and 5. E.g for 26/5 => 5 (weeks) * 7 = 35 + 1 (modulus of 26 and 5) = 36

For under 5 days you will need some logic to check the day of the week for your current and to see if it does cross a weekend or not, and add 2 if it does.

How to Calculate Number of Days Between Two Given Dates