Subtract One Day from Datetime

How to subtract a day from a date?

You can use a timedelta object:

from datetime import datetime, timedelta
    
d = datetime.today() - timedelta(days=days_to_subtract)

Subtract days from a DateTime

That error usually occurs when you try to subtract an interval from DateTime.MinValue or you want to add something to DateTime.MaxValue (or you try to instantiate a date outside this min-max interval). Are you sure you're not assigning MinValue somewhere?

Subtract one day from datetime

Try this

SELECT DATEDIFF(DAY,  DATEADD(day, -1, '2013-03-13 00:00:00.000'), GETDATE())

OR

SELECT DATEDIFF(DAY,  DATEADD(day, -1, @CreatedDate), GETDATE())

Python datetime minus 1 day using timedelta

Some well placed parentheses should suffice:

(datetime.datetime.now() - datetime.timedelta(days=1)).strftime('%Y-%m-%d')

A friendly suggestion: take a look at Python string formatting instead of constructing your string using concatenation. It'll end up being a lot cleaner and less repetitive.

How to subtract days from a plain Date?

Try something like this:

 var d = new Date();
 d.setDate(d.getDate()-5);

Note that this modifies the date object and returns the time value of the updated date.

var d = new Date();
document.write('Today is: ' + d.toLocaleString());
d.setDate(d.getDate() - 5);
document.write('<br>5 days ago was: ' + d.toLocaleString());

Subtract one day from DateTime object

Just use the AddDays method, remembering that it doesn't change the value it's called on - it returns a new DateTime value.

DateTime date = ...;
date = date.AddDays(-1);

subtract one day from a pandas dataframe date column

You can use .dt.hour == 0 with np.where() to identify rows where the hour is 0 and - pd.tseries.offsets.Day() to subtract a day.

df['date'] = pd.to_datetime(df['date'])
df['FinalStartPunch'] = pd.to_datetime(df['FinalStartPunch'], format='%d%b%Y:%H:%M:%S')
df['Date1'] = df['FinalStartPunch'].where((df['FinalStartPunch'].dt.hour != 0),
                                                    df['date'] - pd.tseries.offsets.Day())
df
Out[37]: 
         date     FinalStartPunch               Date1
0  2015-06-27 2015-06-27 14:15:00 2015-06-27 14:15:00
1  2015-07-23 2015-07-23 13:31:00 2015-07-23 13:31:00
2  2015-07-23 2015-07-23 18:43:00 2015-07-23 18:43:00
3  2015-08-15 2015-08-15 18:35:00 2015-08-15 18:35:00
4  2015-08-15 2015-08-15 23:30:00 2015-08-15 23:30:00
5  2015-08-16 2015-08-16 00:00:00 2015-08-15 00:00:00
6  2016-01-30 2016-01-30 18:25:00 2016-01-30 18:25:00
7  2016-01-30 2016-01-30 23:52:00 2016-01-30 23:52:00
8  2016-01-31 2016-01-31 00:00:00 2016-01-30 00:00:00
9  2016-08-13 2016-08-13 18:30:00 2016-08-13 18:30:00
10 2016-08-13 2016-08-13 23:58:00 2016-08-13 23:58:00
11 2016-08-14 2016-08-14 00:00:00 2016-08-13 00:00:00
12 2017-01-28 2017-01-28 18:30:00 2017-01-28 18:30:00
13 2017-01-28 2017-01-28 23:57:00 2017-01-28 23:57:00
14 2017-01-29 2017-01-29 00:00:00 2017-01-28 00:00:00

I have a list of dates and I want to subtract actual date from each of them to know how many day passed. Is there any fast way to do this?

If I'm understanding correctly, you need to find the current date, and then find the number of days between the current date and the dates in your list?

If so, you could try this:

from datetime import datetime, date

dates = ['2019-10-11', '2013-05-16', '2011-06-16', '2000-04-22']

actual_date = date.today()

days = []

for date in dates:
    date_object = datetime.strptime(date, '%Y-%m-%d').date()
    days_difference = (actual_date - date_object).days
    days.append(days_difference)

print(days)

What I am doing here is:

• Converting the individual date strings to a "date" object
• Subtracting the this date from the actual date. This gets you the time as well, so to strip that out we add .days.
• Save the outcome to a list, although of course you could do whatever you wanted with the output.

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