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How can I format timedelta microseconds to 2 decimal digits?

Running this
import time
import datetime
timenow = time.time()
timedifference = time.time() - timenow
timedifference = datetime.timedelta( seconds=timedifference )
print( "%s" % timedifference )
I got this:
0:00:00.000004
How can I format trimming the microseconds to 2 decimal digits using the deltatime object?
0:00:00.00
Related questions:
Timedelta in hours,minutes,seconds,microseconds format
Formatting microseconds to two decimal places (in fact converting microseconds into tens of microseconds)

Convert the timedifference to a string with str(), then separate on either side of the decimal place with .split('.'). Then keep the first portion before the decimal place with [0]:
Your example with the only difference on the last line:
import time
import datetime
timenow = time.time()
timedifference = time.time() - timenow
timedifference = datetime.timedelta( seconds=timedifference )
print( "%s" % str(timedifference).split('.')[0] )
generates:
0:00:00

Another solution is to split the fractional part numerically and format it separately:
>>> seconds = 123.995
>>> isec, fsec = divmod(round(seconds*100), 100)
>>> "{}.{:02.0f}".format(timedelta(seconds=isec), fsec)
'0:02:04.00'
As you can see, this takes care of the rounding. It is also easy to adjust the output precision by changing 100 above to another power of 10 (and adjusting the format string):
def format_td(seconds, digits=2):
isec, fsec = divmod(round(seconds*10**digits), 10**digits)
return ("{}.{:0%d.0f}" % digits).format(timedelta(seconds=isec), fsec)

You'll have to format it yourself. A timedelta object contains days, seconds and microseconds so you'll have to do the math to convert to days/hours/min/sec/microsec and then format using python string.format. For your microsec, you'll want ((microsec+5000)/10000) to get the top two digits (the +5000 is for rounding).

A bit late, but here's a 2021 answer with f-strings (modified from #Seb's original answer):
def format_td(seconds, digits=3):
isec, fsec = divmod(round(seconds*10**digits), 10**digits)
return f'{timedelta(seconds=isec)}.{fsec:0{digits}.0f}'

parsing of seconds and millisecond using python

I have time stamp in format of "10:24:00.744" I want to fetch the seconds value (here -- 00) and millisecond value (here -- 744) in different variables using python. Hope someone can help me with this.

You can use datetime.datetime module to parse the time, and then get the seconds and milliseconds from it -
>>> from datetime import datetime
>>> h = datetime.strptime('10:24:00.744','%H:%M:%S.%f')
>>> seconds = h.second
>>> milliseconds = h.microsecond/1000
>>> print(seconds)
0
>>> print(milliseconds)
744.0
strptime() function is used to parse datetime from string, the directives mean -
%H - Hour
%M - Minutes
%S - Seconds
%f - Microseconds
More details about support directives can be found here.

Try the following:
a = "10:24:00.744"
seconds = a[6:8]
milliseconds = a[9:]
print "Seconds", seconds
print "Milliseconds", milliseconds
You will get the following output:
Seconds '00'
Milliseconds '744'

Parse time string in python [duplicate]

This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
How to construct a timedelta object from a simple string
I have a string that is in the format hours:minutes:seconds but it is not a time of day but a duration. For example, 100:00:00 means 100 hours.
I am trying to find the time that is offset from the current time by the amount of time specified in the string. I could use regular expressions to manually pull apart the time string and convert it to seconds and add it to the floating point returned by time.time(), but is there a time function to do this?
The time.strptime() function formatting seems to work on time of day/date strings and not arbitrary strings.

import datetime
dur_str = "100:00:00"
h, m, s = map(int, dur_str.split(':'))
dur = datetime.timedelta(hours=h, minutes=m, seconds=s)

not using re but sometimes it's more work to understand the regex than write the python.
>>> import datetime
>>> time_str = "100:00:00"
>>> hours, minutes, seconds = [int(i) for i in time_str.split(":")]
>>> time_in_seconds = hours * 60 * 60 + minutes * 60 + seconds
>>> time_in_seconds
360000
>>> now = datetime.datetime.now()
>>> now
datetime.datetime(2012, 10, 2, 10, 24, 6, 639000)
>>> new_time = now + datetime.timedelta(seconds=time_in_seconds)
>>> new_time
datetime.datetime(2012, 10, 6, 14, 24, 6, 639000)
As nneonneo pointed out datetime.timedelta() accepts the hours, minutes, and seconds as arguments. So you can even do something silly like this (not recommended):
>>> datetime.timedelta(**{k:v for k,v in zip(["hours", "minutes", "seconds"], [int(i) for i in "100:00:00".split(":")])})
datetime.timedelta(4, 14400)

datetime: Round/trim number of digits in microseconds

Currently I am logging stuff and I am using my own formatter with a custom formatTime():
def formatTime(self, _record, _datefmt):
t = datetime.datetime.now()
return t.strftime('%Y-%m-%d %H:%M:%S.%f')
My issue is that the microseconds, %f, are six digits. Is there anyway to spit out less digits, like the first three digits of the microseconds?

The simplest way would be to use slicing to just chop off the last three digits of the microseconds:
def format_time():
t = datetime.datetime.now()
s = t.strftime('%Y-%m-%d %H:%M:%S.%f')
return s[:-3]
I strongly recommend just chopping. I once wrote some logging code that rounded the timestamps rather than chopping, and I found it actually kind of confusing when the rounding changed the last digit. There was timed code that stopped running at a certain timestamp yet there were log events with that timestamp due to the rounding. Simpler and more predictable to just chop.
If you want to actually round the number rather than just chopping, it's a little more work but not horrible:
def format_time():
t = datetime.datetime.now()
s = t.strftime('%Y-%m-%d %H:%M:%S.%f')
head = s[:-7] # everything up to the '.'
tail = s[-7:] # the '.' and the 6 digits after it
f = float(tail)
temp = "{:.03f}".format(f) # for Python 2.x: temp = "%.3f" % f
new_tail = temp[1:] # temp[0] is always '0'; get rid of it
return head + new_tail
Obviously you can simplify the above with fewer variables; I just wanted it to be very easy to follow.

As of Python 3.6 the language has this feature built in:
def format_time():
t = datetime.datetime.now()
s = t.isoformat(timespec='milliseconds')
return s

This method should always return a timestamp that looks exactly like this (with or without the timezone depending on whether the input dt object contains one):
2016-08-05T18:18:54.776+0000
It takes a datetime object as input (which you can produce with datetime.datetime.now()). To get the time zone like in my example output you'll need to import pytz and pass datetime.datetime.now(pytz.utc).
import pytz, datetime
time_format(datetime.datetime.now(pytz.utc))
def time_format(dt):
return "%s:%.3f%s" % (
dt.strftime('%Y-%m-%dT%H:%M'),
float("%.3f" % (dt.second + dt.microsecond / 1e6)),
dt.strftime('%z')
)
I noticed that some of the other methods above would omit the trailing zero if there was one (e.g. 0.870 became 0.87) and this was causing problems for the parser I was feeding these timestamps into. This method does not have that problem.

An easy solution that should work in all cases:
def format_time():
t = datetime.datetime.now()
if t.microsecond % 1000 >= 500: # check if there will be rounding up
t = t + datetime.timedelta(milliseconds=1) # manually round up
return t.strftime('%Y-%m-%d %H:%M:%S.%f')[:-3]
Basically you do manual rounding on the date object itself first, then you can safely trim the microseconds.

Edit: As some pointed out in the comments below, the rounding of this solution (and the one above) introduces problems when the microsecond value reaches 999500, as 999.5 is rounded to 1000 (overflow).
Short of reimplementing strftime to support the format we want (the potential overflow caused by the rounding would need to be propagated up to seconds, then minutes, etc.), it is much simpler to just truncate to the first 3 digits as outlined in the accepted answer, or using something like:
'{:03}'.format(int(999999/1000))
-- Original answer preserved below --
In my case, I was trying to format a datestamp with milliseconds formatted as 'ddd'. The solution I ended up using to get milliseconds was to use the microsecond attribute of the datetime object, divide it by 1000.0, pad it with zeros if necessary, and round it with format. It looks like this:
'{:03.0f}'.format(datetime.now().microsecond / 1000.0)
# Produces: '033', '499', etc.

You can subtract the current datetime from the microseconds.
d = datetime.datetime.now()
current_time = d - datetime.timedelta(microseconds=d.microsecond)
This will turn 2021-05-14 16:11:21.916229 into 2021-05-14 16:11:21

This method allows flexible precision and will consume the entire microsecond value if you specify too great a precision.
def formatTime(self, _record, _datefmt, precision=3):
dt = datetime.datetime.now()
us = str(dt.microsecond)
f = us[:precision] if len(us) > precision else us
return "%d-%d-%d %d:%d:%d.%d" % (dt.year, dt.month, dt.day, dt.hour, dt.minute, dt.second, int(f))
This method implements rounding to 3 decimal places:
import datetime
from decimal import *
def formatTime(self, _record, _datefmt, precision='0.001'):
dt = datetime.datetime.now()
seconds = float("%d.%d" % (dt.second, dt.microsecond))
return "%d-%d-%d %d:%d:%s" % (dt.year, dt.month, dt.day, dt.hour, dt.minute,
float(Decimal(seconds).quantize(Decimal(precision), rounding=ROUND_HALF_UP)))
I avoided using the strftime method purposely because I would prefer not to modify a fully serialized datetime object without revalidating it. This way also shows the date internals in case you want to modify it further.
In the rounding example, note that the precision is string-based for the Decimal module.

Here is my solution using regexp:
import re
# Capture 6 digits after dot in a group.
regexp = re.compile(r'\.(\d{6})')
def to_splunk_iso(dt):
"""Converts the datetime object to Splunk isoformat string."""
# 6-digits string.
microseconds = regexp.search(dt.isoformat()).group(1)
return regexp.sub('.%d' % round(float(microseconds) / 1000), dt.isoformat())

Fixing the proposed solution based on Pablojim Comments:
from datetime import datetime
dt = datetime.now()
dt_round_microsec = round(dt.microsecond/1000) #number of zeroes to round
dt = dt.replace(microsecond=dt_round_microsec)

If once want to get the day of the week (i.e, 'Sunday)' along with the result, then by slicing '[:-3]' will not work. At that time you may go with,
dt = datetime.datetime.now()
print("{}.{:03d} {}".format(dt.strftime('%Y-%m-%d %I:%M:%S'), dt.microsecond//1000, dt.strftime("%A")))
#Output: '2019-05-05 03:11:22.211 Sunday'
%H - for 24 Hour format
%I - for 12 Hour format
Thanks,

Adding my two cents here as this method will allow you to write your microsecond format as you would a float in c-style. It takes advantage that they both use %f.
import datetime
import re
def format_datetime(date, format):
"""Format a ``datetime`` object with microsecond precision.
Pass your microsecond as you would format a c-string float.
e.g "%.3f"
Args:
date (datetime.datetime): You input ``datetime`` obj.
format (str): Your strftime format string.
Returns:
str: Your formatted datetime string.
"""
# We need to check if formatted_str contains "%.xf" (x = a number)
float_format = r"(%\.\d+f)"
has_float_format = re.search(float_format, format)
if has_float_format:
# make microseconds be decimal place. Might be a better way to do this
microseconds = date.microsecond
while int(microseconds): # quit once it's 0
microseconds /= 10
ms_str = has_float_format.group(1) % microseconds
format = re.sub(float_format, ms_str[2:], format)
return date.strftime(format)
print(datetime.datetime.now(), "%H:%M:%S.%.3f")
# '17:58:54.424'

Using %f with strftime() in Python to get microseconds

I'm trying to use strftime() to microsecond precision, which seems possible using %f (as stated here). However when I try the following code:
import time
import strftime from time
print strftime("%H:%M:%S.%f")
...I get the hour, the minutes and the seconds, but %f prints as %f, with no sign of the microseconds. I'm running Python 2.6.5 on Ubuntu, so it should be fine and %f should be supported (it's supported for 2.6 and above, as far as I know.)

You can use datetime's strftime function to get this. The problem is that time's strftime accepts a timetuple that does not carry microsecond information.
from datetime import datetime
datetime.now().strftime("%H:%M:%S.%f")
Should do the trick!

You are looking at the wrong documentation. The time module has different documentation.
You can use the datetime module strftime like this:
>>> from datetime import datetime
>>>
>>> now = datetime.now()
>>> now.strftime("%H:%M:%S.%f")
'12:19:40.948000'

With Python's time module you can't get microseconds with %f.
For those who still want to go with time module only, here is a workaround:
now = time.time()
mlsec = repr(now).split('.')[1][:3]
print time.strftime("%Y-%m-%d %H:%M:%S.{} %Z".format(mlsec), time.localtime(now))
You should get something like 2017-01-16 16:42:34.625 EET (yes, I use milliseconds as it's fairly enough).
To break the code into details, paste the below code into a Python console:
import time
# Get current timestamp
now = time.time()
# Debug now
now
print now
type(now)
# Debug strf time
struct_now = time.localtime(now)
print struct_now
type(struct_now)
# Print nicely formatted date
print time.strftime("%Y-%m-%d %H:%M:%S %Z", struct_now)
# Get miliseconds
mlsec = repr(now).split('.')[1][:3]
print mlsec
# Get your required timestamp string
timestamp = time.strftime("%Y-%m-%d %H:%M:%S.{} %Z".format(mlsec), struct_now)
print timestamp
For clarification purposes, I also paste my Python 2.7.12 result here:
>>> import time
>>> # get current timestamp
... now = time.time()
>>> # debug now
... now
1484578293.519106
>>> print now
1484578293.52
>>> type(now)
<type 'float'>
>>> # debug strf time
... struct_now = time.localtime(now)
>>> print struct_now
time.struct_time(tm_year=2017, tm_mon=1, tm_mday=16, tm_hour=16, tm_min=51, tm_sec=33, tm_wday=0, tm_yday=16, tm_isdst=0)
>>> type(struct_now)
<type 'time.struct_time'>
>>> # print nicely formatted date
... print time.strftime("%Y-%m-%d %H:%M:%S %Z", struct_now)
2017-01-16 16:51:33 EET
>>> # get miliseconds
... mlsec = repr(now).split('.')[1][:3]
>>> print mlsec
519
>>> # get your required timestamp string
... timestamp = time.strftime("%Y-%m-%d %H:%M:%S.{} %Z".format(mlsec), struct_now)
>>> print timestamp
2017-01-16 16:51:33.519 EET
>>>

This should do the work
import datetime
datetime.datetime.now().strftime("%H:%M:%S.%f")
It will print
HH:MM:SS.microseconds like this e.g 14:38:19.425961

You can also get microsecond precision from the time module using its time() function.
(time.time() returns the time in seconds since epoch. Its fractional part is the time in microseconds, which is what you want.)
>>> from time import time
>>> time()
... 1310554308.287459 # the fractional part is what you want.
# comparision with strftime -
>>> from datetime import datetime
>>> from time import time
>>> datetime.now().strftime("%f"), time()
... ('287389', 1310554310.287459)

When the "%f" for micro seconds isn't working, please use the following method:
import datetime
def getTimeStamp():
dt = datetime.datetime.now()
return dt.strftime("%Y%j%H%M%S") + str(dt.microsecond)

If you want an integer, try this code:
import datetime
print(datetime.datetime.now().strftime("%s%f")[:13])
Output:
1545474382803

If you want speed, try this:
def _timestamp(prec=0):
t = time.time()
s = time.strftime("%H:%M:%S", time.localtime(t))
if prec > 0:
s += ("%.9f" % (t % 1,))[1:2+prec]
return s
Where prec is precision -- how many decimal places you want.
Please note that the function does not have issues with leading zeros in fractional part like some other solutions presented here.