How to round a datetime variable in Python?-CodePudding

How do I round a datetime variable like this one:

x = datetime.now()

which produces something like this:

2022-05-04 15:36:01.055696

to something like this?

2022-05-04 15:36:01.057

CodePudding user response：

In case you want to modify a datetime object, you can round its .microsecond attribute:

x = datetime.now()
print(x.isoformat(' '))
# 2022-05-04 15:36:01.055696

x = x.replace(microsecond=round(x.microsecond / 1000) * 1000)
print(x.isoformat(' '))
# 2022-05-04 15:36:01.056

In any case (even without modifying the object and rounding the microseconds), to get the representation you want you can format it with the appropriate method:

x.isoformat(' ', timespec='milliseconds')
# 2022-05-04 15:36:01.056

Note that modifying the object using round(x.microsecond / 1000) * 1000 for the microseconds will round them to the closest millisecond. If you just want to truncate the microseconds instead, use x.microsecond // 1000 * 1000.

CodePudding user response：

If you just want to remove the last few characters you could do:

print(x.strftime('%Y-%m-%d %H:%M:%S.%f')[:-3])

The formatting str is the default which is what you are seeing in the output

CodePudding user response：

You can convert it to a unix timestamp and then round the number to 3 decimal places.

from datetime import datetime

x = '2022-05-04 15:36:01.055696'

# Convert to datetime object
d = datetime.strptime(x, '%Y-%m-%d %H:%M:%S.%f')

# Convert to rounded unix timestamp
ts = datetime.fromtimestamp(round(d.timestamp(), 3))

# You know have rounded datetime object, you can convert back to string with
ts.strftime('%Y-%m-%d %H:%M:%S.%f')[:-3]

CodePudding user response：

Pandas timestamp has a built in-function for this. Make the timestamp a timestamp object then use round.

import pandas as pd

x = '2022-05-04 15:36:01.055696'
ts = pd.Timestamp(x)
ts_rounded = ts.round(freq='L')
print(ts_rounded) 2022-05-04 15:36:01.056000