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Python - Get Yesterday's date as a string in YYYY-MM-DD format

As an input to an API request I need to get yesterday's date as a string in the format YYYY-MM-DD. I have a working version which is:
yesterday = datetime.date.fromordinal(datetime.date.today().toordinal()-1)
report_date = str(yesterday.year) + \
('-' if len(str(yesterday.month)) == 2 else '-0') + str(yesterday.month) + \
('-' if len(str(yesterday.day)) == 2 else '-0') + str(yesterday.day)
There must be a more elegant way to do this, interested for educational purposes as much as anything else!

You Just need to subtract one day from today's date. In Python datetime.timedelta object lets you create specific spans of time as a timedelta object.
datetime.timedelta(1) gives you the duration of "one day" and is subtractable from a datetime object. After you subtracted the objects you can use datetime.strftime in order to convert the result --which is a date object-- to string format based on your format of choice:
>>> from datetime import datetime, timedelta
>>> yesterday = datetime.now() - timedelta(1)
>>> type(yesterday)
>>> datetime.datetime
>>> datetime.strftime(yesterday, '%Y-%m-%d')
'2015-05-26'
Note that instead of calling the datetime.strftime function, you can also directly use strftime method of datetime objects:
>>> (datetime.now() - timedelta(1)).strftime('%Y-%m-%d')
'2015-05-26'
As a function:
from datetime import datetime, timedelta
def yesterday(frmt='%Y-%m-%d', string=True):
yesterday = datetime.now() - timedelta(1)
if string:
return yesterday.strftime(frmt)
return yesterday
example:
In [10]: yesterday()
Out[10]: '2022-05-13'
In [11]: yesterday(string=False)
Out[11]: datetime.datetime(2022, 5, 13, 12, 34, 31, 701270)

An alternative answer that uses today() method to calculate current date and then subtracts one using timedelta(). Rest of the steps remain the same.
https://docs.python.org/3.7/library/datetime.html#timedelta-objects
from datetime import date, timedelta
today = date.today()
yesterday = today - timedelta(days = 1)
print(today)
print(yesterday)
Output:
2019-06-14
2019-06-13

>>> import datetime
>>> datetime.date.fromordinal(datetime.date.today().toordinal()-1).strftime("%F")
'2015-05-26'

Calling .isoformat() on a date object will give you YYYY-MM-DD
from datetime import date, timedelta
(date.today() - timedelta(1)).isoformat()

I'm trying to use only import datetime based on this answer.
import datetime
oneday = datetime.timedelta(days=1)
yesterday = datetime.date.today() - oneday

How to convert python timestamp string to epoch?

I have the following string:
mytime = "2009-03-08T00:27:31.807Z"
How do I convert it to epoch in python?
I tried:
import time
p = '%Y-%m-%dT%H:%M:%S'
int(time.mktime(time.strptime(s, p)))
But it does not work with the 31.807Z.

There are two parts:
Convert the time string into a broken-down time. See How to parse ISO formatted date in python?
Convert the UTC time to "seconds since the Epoch" (POSIX timestamp).
#!/usr/bin/env python
from datetime import datetime
utc_time = datetime.strptime("2009-03-08T00:27:31.807Z", "%Y-%m-%dT%H:%M:%S.%fZ")
epoch_time = (utc_time - datetime(1970, 1, 1)).total_seconds()
# -> 1236472051.807
If you are sure that you want to ignore fractions of a second and to get an integer result:
#!/usr/bin/env python
import time
from calendar import timegm
utc_time = time.strptime("2009-03-08T00:27:31.807Z", "%Y-%m-%dT%H:%M:%S.%fZ")
epoch_time = timegm(utc_time)
# -> 1236472051
To support timestamps that correspond to a leap second such as Wed July 1 2:59:60 MSK 2015, you could use a combination of time.strptime() and datetime (if you care about leap seconds you should take into account the microseconds too).

You are missing .%fZ from your format string.
p = '%Y-%m-%dT%H:%M:%S.%fZ'
The correct way to convert to epoch is to use datetime:
from datetime import datetime
p = '%Y-%m-%dT%H:%M:%S.%fZ'
mytime = "2009-03-08T00:27:31.807Z"
epoch = datetime(1970, 1, 1)
print((datetime.strptime(mytime, p) - epoch).total_seconds())
Or call int if you want to ignore fractions.

dateutil has recently been added back to python packages, it's an easy one liner that handles formatting on its own.
from dateutil import parser
strtime = '2009-03-08T00:27:31.807Z'
epoch = parser.parse(strtime).timestamp()

dateutil is the only library i have found that correctly deals with the timezone offset identitifier (Z)
pip install python-dateutil
then
from dateutil.parser import parse as date_parse
print date_parse("2009-03-08T00:27:31.807Z")
#get timestamp
import calendar
dt = date_parse("2009-03-08T00:27:31.807Z")
timestamp1 = calendar.timegm(dt.timetuple())

Code:
import datetime
epoch = datetime.datetime(1970, 1, 1)
mytime = "2009-03-08T00:27:31.807Z"
myformat = "%Y-%m-%dT%H:%M:%S.%fZ"
mydt = datetime.datetime.strptime(mytime, myformat)
val = (mydt - epoch).total_seconds()
print(val)
> 1236472051.81
repr(val)
> '1236472051.807'
Notes:
When using time.strptime(), the returned time.struct_time does not support sub-second precision.
The %f format is for microseconds. When parsing it need not be the full 6 digits.

Python 3.7+ The string format in question can be parsed by strptime:
from datetime import datetime
datetime.strptime("2009-03-08T00:27:31.807Z", '%Y-%m-%dT%H:%M:%S.%f%z')
>>> datetime.datetime(2009, 3, 8, 0, 27, 31, 807000, tzinfo=datetime.timezone.utc)
Another option using the built-in datetime.fromisoformat(): As mentioned in this thread linked by #jfs, fromisoformat() doesn't parse the 'Z' character to UTC although this is part of the RFC3339 definitions. A little work-around can make it work - some will consider this nasty but it's efficient after all.
from datetime import datetime
mytime = "2009-03-08T00:27:31.807Z"
datetime.fromisoformat(mytime.replace("Z", "+00:00")).timestamp()
>>> 1236472051.807

This code works in Python 3.6 to convert a datetime string to epoch in UTC or local timezone.
from datetime import datetime, timedelta
from dateutil.tz import tzutc, tzlocal
mydate = '2020-09-25'
mytime = '06:00:00'
epoch1970 = datetime(1970, 1, 1, 0, 0, 0, tzinfo=tzutc())
myepochutc = int((datetime.strptime(mydate + ' ' + mytime, "%Y-%m-%d %H:%M:%S").replace(tzinfo=tzutc()) - epoch1970).total_seconds()*1000)
myepochlocal = int((datetime.strptime(mydate + ' ' + mytime, "%Y-%m-%d %H:%M:%S").replace(tzinfo=tzlocal()) - epoch1970).total_seconds()*1000)
#epoch will be in milliseconds
print(myepochutc) #if mydate/mytime was in utc
print(myepochlocal) #if mydate/mytime was in local timezone

DateTime Conversion to Epoch Time

I need to convert time stored in a variable in the format using Python <=2.7
07/20-10:38:04.360700
to epoch time (since Midnight, Jan 1st, 1970) like this
1405852684.360700
Is it best to import the time module, or just split and use some math calculations?

If the string date is with respect to UTC, then:
In [31]: import datetime as DT
In [32]: text = '07/20-10:38:04.360700'
In [33]: date = DT.datetime.strptime('2014/'+text, '%Y/%m/%d-%H:%M:%S.%f')
In [34]: (date - DT.datetime(1970,1,1)).total_seconds()
Out[34]: 1405852684.3607
If the string date refers to a date with respect to some other timezone, then you could use pytz to make the datetime timezone-aware before doing the calculation. For example,
import pytz
import datetime as DT
text = '07/20-10:38:04.360700'
tz = pytz.timezone('US/Eastern')
date = DT.datetime.strptime('2014/'+text, '%Y/%m/%d-%H:%M:%S.%f')
# interpret the date as coming from US/Eastern
date_tz = tz.localize(date)
epoch = DT.datetime(1970,1,1, tzinfo=pytz.utc)
timestamp = (date_tz - epoch).total_seconds()
print(repr(timestamp))
# 1405867084.3607

You can use python's timetuple() function (and a little math)
>>> import datetime
>>> import time
>>> v = datetime.datetime(2014, 7, 20, 10, 38, 4, 360700)
>>> time.mktime(v.timetuple())
1405870684.0
Now we need your microseconds:
>>> time.mktime(v.timetuple())+(v.microsecond/1000000.)
1405870684.3607

You don't specify the year so here I assume that it is the current year. The following assumes that all times are UTC, i.e. not local time.
from datetime import datetime
time_string = '07/20-10:38:04.360700'
dt = datetime.strptime(time_string, '%m/%d-%H:%M:%S.%f')
dt = dt.replace(year=datetime.today().year)
>>> (dt - datetime.utcfromtimestamp(0)).total_seconds()
1405852684.3607

Specify which timezone a datetime is in, in python

I have a datetime that i get from a database, this datetime is a UTC datetime. But when i pull it from the DB, it is unaware of the timezone. What i need to do, is convert this datetime to a "seconds from epoch" time for another function. The problem with this, is that the system's time is in PST and i am not able to change it for specific reasons.
So, what i want to do is, take this datetime that i get from the database, and tell python that this datetime is a UTC datetime. Every way that i have done that, results in it losing time or gaining time due to timezone conversions. Again, not trying to convert the time, just trying to specify that it is UTC.
If anyone can help with this that would be great.
Thanks!
Example
Assume database_function() returns a datetime data type that is '2013-06-01 01:06:18'
datetime = database_function()
epoch = datetime.strftime('%s')
pytz.utc.localize(database_function()).datetime.strftime('%s')
datetime.replace(tzinfo=pytz.utc).datetime.strftime('%s')
Both of these return a epoch timestamp of 1370077578
But, it SHOULD return a timestamp of 1370048778 per http://www.epochconverter.com/
Remember, this timestamp is a utc timestamp

Using the fabolous pytz:
import datetime, pytz
dt = datetime.datetime(...)
utc_dt = pytz.utc.localize(dt)
This creates a tz-aware datetime object, in UTC.

How about Setting timezone in Python This appears to reset the timezone within your python script. You are changing the time zone that your system sees given the specified time, not changing the specified time into the specified time zone. You probably want to set it to 'UTC'
time.tzset()
Resets the time conversion rules used by the library routines.
The environment variable TZ specifies how this is done.
New in version 2.3.
Availability: Unix.
I do not have this available on my home platform so I could not test it. I had to get this from the previous answer.
The answer marked best on the question is:
>>> import os, time
>>> time.strftime('%X %x %Z')
'12:45:20 08/19/09 CDT'
>>> os.environ['TZ'] = 'Europe/London'
>>> time.tzset()
>>> time.strftime('%X %x %Z')
'18:45:39 08/19/09 BST'
To get the specific values you've listed:
>>> year = time.strftime('%Y')
>>> month = time.strftime('%m')
>>> day = time.strftime('%d')
>>> hour = time.strftime('%H')
>>> minute = time.strftime('%M')
See here for a complete list of directives. Keep in mind that the strftime() function will always return a string, not an integer or other type.

You can Use pytz, which is a time zone definitions package.
from datetime import datetime
from pytz import timezone
fmt = "%Y-%m-%d %H:%M:%S %Z%z"
# Current time in UTC
now_utc = datetime.now(timezone('UTC'))
print now_utc.strftime(fmt)
# Convert to US/Pacific time zone
now_pacific = now_utc.astimezone(timezone('US/Pacific'))
print now_pacific.strftime(fmt)
# Convert to Europe/Berlin time zone
now_berlin = now_pacific.astimezone(timezone('Europe/Berlin'))
print now_berlin.strftime(fmt)
output:
2014-04-04 21:50:55 UTC+0000
2014-04-04 14:50:55 PDT-0700
2014-04-04 23:50:55 CEST+0200
or may be it helps
>> import pytz
>>> import datetime
>>>
>>> now_utc = datetime.datetime.utcnow() #Our UTC naive time from DB,
for the time being here I'm taking it as current system UTC time..
>>> now_utc
datetime.datetime(2011, 5, 9, 6, 36, 39, 883479) # UTC time in Naive
form.
>>>
>>> local_tz = pytz.timezone('Europe/Paris') #Our Local timezone, to
which we want to convert the UTC time.
>>>
>>> now_utc = pytz.utc.localize(now_utc) #Add Timezone information to
UTC time.
>>>
>>> now_utc
datetime.datetime(2011, 5, 9, 6, 36, 39, 883479, tzinfo=<UTC>) # The
full datetime tuple
>>>
>>> local_time = now_utc.astimezone(local\_tz) # Convert to local
time.
>>>
>>> local_time #Current local time in Paris
datetime.datetime(2011, 5, 9, 8, 36, 39, 883479, tzinfo=<DstTzInfo
'Europe/Paris' CEST+2:00:00 DST>)

Here is one way, using the pytz module:
import pytz
utc_datetime = (datetime.datetime(1970, 1, 1, tzinfo=pytz.utc)
+ datetime.timedelta(seconds=seconds_since_epoch)
If you don't want to install the pytz module, you can copy the example UTC class from the datetime documentation (search for "class UTC"):
https://docs.python.org/2/library/datetime.html#tzinfo-objects

Here's stdlib only solution without 3-party modules.
.., this datetime is a UTC datetime. But when i pull it from the DB, it is unaware of the timezone. What i need to do, is convert this datetime to a "seconds from epoch" time for another function.emphasize is mine
To convert an unaware (naive) datetime object that represents time in UTC to POSIX timestamp:
from datetime import datetime
timestamp = (dt_from_db - datetime(1970, 1, 1)).total_seconds()
Example:
>>> from datetime import datetime
>>> dt = datetime.strptime('2013-06-01 01:06:18', '%Y-%m-%d %H:%M:%S')
>>> (dt - datetime(1970, 1, 1)).total_seconds()
1370048778.0
See Converting datetime.date to UTC timestamp in Python that provides solutions for various Python versions.
To answer the question from the title: In general you need pytz library to handle timezones in Python. In particular, you should use .localize method to convert an unaware datetime object into timezone-aware one.
import pytz # $ pip install pytz
from tzlocal import get_localzone # $ pip install tzlocal
tz = get_localzone() # local timezone whatever it is (just an example)
aware_dt = tz.localize(naive_dt_in_local_timezone, is_dst=None)
is_dst=None asserts that naive_dt_in_local_timezone exists and unambiguous.
Though you don't need it for UTC timezone because it always has the same UTC offset (zero) around the year and in all past years:
import pytz
aware_dt = utc_dt.replace(tzinfo=pytz.utc)
See Python - simplest and most coherent way to get timezone-aware current time in UTC? (it provides a stdlib-only solution):
aware_dt = utc_dt.replace(tzinfo=timezone.utc)

How to truncate the time on a datetime object?

What is a classy way to way truncate a python datetime object?
In this particular case, to the day. So basically setting hour, minute, seconds, and microseconds to 0.
I would like the output to also be a datetime object, not a string.

I think this is what you're looking for...
>>> import datetime
>>> dt = datetime.datetime.now()
>>> dt = dt.replace(hour=0, minute=0, second=0, microsecond=0) # Returns a copy
>>> dt
datetime.datetime(2011, 3, 29, 0, 0)
But if you really don't care about the time aspect of things, then you should really only be passing around date objects...
>>> d_truncated = datetime.date(dt.year, dt.month, dt.day)
>>> d_truncated
datetime.date(2011, 3, 29)

Use a date not a datetime if you dont care about the time.
>>> now = datetime.now()
>>> now.date()
datetime.date(2011, 3, 29)
You can update a datetime like this:
>>> now.replace(minute=0, hour=0, second=0, microsecond=0)
datetime.datetime(2011, 3, 29, 0, 0)

Four years later: another way, avoiding replace
I know the accepted answer from four years ago works, but this seems a tad lighter than using replace:
dt = datetime.date.today()
dt = datetime.datetime(dt.year, dt.month, dt.day)
Notes
When you create a datetime object without passing time properties to the constructor, you get midnight.
As others have noted, this assumes you want a datetime object for later use with timedeltas.
You can, of course, substitute this for the first line: dt = datetime.datetime.now()

To get a midnight corresponding to a given datetime object, you could use datetime.combine() method:
>>> from datetime import datetime, time
>>> dt = datetime.utcnow()
>>> dt.date()
datetime.date(2015, 2, 3)
>>> datetime.combine(dt, time.min)
datetime.datetime(2015, 2, 3, 0, 0)
The advantage compared to the .replace() method is that datetime.combine()-based solution will continue to work even if datetime module introduces the nanoseconds support.
tzinfo can be preserved if necessary but the utc offset may be different at midnight e.g., due to a DST transition and therefore a naive solution (setting tzinfo time attribute) may fail. See How do I get the UTC time of “midnight” for a given timezone?

You cannot truncate a datetime object because it is immutable.
However, here is one way to construct a new datetime with 0 hour, minute, second, and microsecond fields, without throwing away the original date or tzinfo:
newdatetime = now.replace(hour=0, minute=0, second=0, microsecond=0)

See more at https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.Series.dt.floor.html
It's now 2019, I think the most efficient way to do it is:
df['truncate_date'] = df['timestamp'].dt.floor('d')

You could use pandas for that (although it could be overhead for that task). You could use round, floor and ceil like for usual numbers and any pandas frequency from offset-aliases:
import pandas as pd
import datetime as dt
now = dt.datetime.now()
pd_now = pd.Timestamp(now)
freq = '1d'
pd_round = pd_now.round(freq)
dt_round = pd_round.to_pydatetime()
print(now)
print(dt_round)
"""
2018-06-15 09:33:44.102292
2018-06-15 00:00:00
"""

There is a great library used to manipulate dates: Delorean
import datetime
from delorean import Delorean
now = datetime.datetime.now()
d = Delorean(now, timezone='US/Pacific')
>>> now
datetime.datetime(2015, 3, 26, 19, 46, 40, 525703)
>>> d.truncate('second')
Delorean(datetime=2015-03-26 19:46:40-07:00, timezone='US/Pacific')
>>> d.truncate('minute')
Delorean(datetime=2015-03-26 19:46:00-07:00, timezone='US/Pacific')
>>> d.truncate('hour')
Delorean(datetime=2015-03-26 19:00:00-07:00, timezone='US/Pacific')
>>> d.truncate('day')
Delorean(datetime=2015-03-26 00:00:00-07:00, timezone='US/Pacific')
>>> d.truncate('month')
Delorean(datetime=2015-03-01 00:00:00-07:00, timezone='US/Pacific')
>>> d.truncate('year')
Delorean(datetime=2015-01-01 00:00:00-07:00, timezone='US/Pacific')
and if you want to get datetime value back:
>>> d.truncate('year').datetime
datetime.datetime(2015, 1, 1, 0, 0, tzinfo=<DstTzInfo 'US/Pacific' PDT-1 day, 17:00:00 DST>)

You can use datetime.strftime to extract the day, the month, the year...
Example :
from datetime import datetime
d = datetime.today()
# Retrieves the day and the year
print d.strftime("%d-%Y")
Output (for today):
29-2011
If you just want to retrieve the day, you can use day attribute like :
from datetime import datetime
d = datetime.today()
# Retrieves the day
print d.day
Ouput (for today):
29

If you are dealing with a Series of type DateTime there is a more efficient way to truncate them, specially when the Series object has a lot of rows.
You can use the floor function
For example, if you want to truncate it to hours:
Generate a range of dates
times = pd.Series(pd.date_range(start='1/1/2018 04:00:00', end='1/1/2018 22:00:00', freq='s'))
We can check it comparing the running time between the replace and the floor functions.
%timeit times.apply(lambda x : x.replace(minute=0, second=0, microsecond=0))
>>> 341 ms ± 18.2 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%timeit times.dt.floor('h')
>>>>2.26 ms ± 451 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

>>> import datetime
>>> dt = datetime.datetime.now()
>>> datetime.datetime.date(dt)
datetime.date(2019, 4, 2)

Here is yet another way which fits in one line but is not particularly elegant:
dt = datetime.datetime.fromordinal(datetime.date.today().toordinal())

There is a module datetime_truncate which handlers this for you. It just calls datetime.replace.

6 years later... I found this post and I liked more the numpy aproach:
import numpy as np
dates_array = np.array(['2013-01-01', '2013-01-15', '2013-01-30']).astype('datetime64[ns]')
truncated_dates = dates_array.astype('datetime64[D]')
cheers

If you want to truncate to an arbitrary timedelta:
from datetime import datetime, timedelta
truncate = lambda t, d: t + (datetime.min - t) % - d
# 2022-05-04 15:54:19.979349
now = datetime.now()
# truncates to the last 15 secondes
print(truncate(now, timedelta(seconds=15)))
# truncates to the last minute
print(truncate(now, timedelta(minutes=1)))
# truncates to the last 2 hours
print(truncate(now, timedelta(hours=2)))
# ...
"""
2022-05-04 15:54:15
2022-05-04 15:54:00
2022-05-04 14:00:00
"""
PS: This is for python3

You can just use
datetime.date.today()
It's light and returns exactly what you want.

You could do it by specifying isoformat
>>> import datetime
>>> datetime.datetime.now().isoformat(timespec='seconds', sep=' ')
2022-11-24 12:42:05
The documentation offers more details about the isoformat() usage.
https://docs.python.org/3/library/datetime.html#datetime.datetime.isoformat

What does truncate mean?
You have full control over the formatting by using the strftime() method and using an appropriate format string.
http://docs.python.org/library/datetime.html#strftime-strptime-behavior