Develop Reference

strftime formatting - what am I doing wrong?

I've got a string 3:01 AM - 18 Dec 2017
I've written the following pattern strftime('%-I:%M %p - %-d %b %Y') and I can't seem to get it to work, following this
My notes:
%-I hour 12-hour clock as a decimal (no zero padding)
: separation between hour and minute
%M minute as a zero padded decimal
%p AM/PM
- space-dash-space pattern betwen time and date
%-d date of the month as a decimal (no zero padding)
%b Month abbreviated
%Y Year with century as decimal
df['tweet_date'] = pd.to_datetime(df['tweet_date'], errors='coerce').apply(lambda x: x.strftime('%I:%M %p - %d %b %Y')if not pd.isnull(x) else '')
On another dataframe with a similar column this works:
df2['created_at'] = pd.to_datetime(df2['created_at'], errors='coerce').apply(lambda x: x.strftime('%Y-%m-%d %H:%M:%S')if not pd.isnull(x) else '')
df2['created_at'] = df2['created_at'].astype('datetime64[s]')`
where values before formatting look like this for example 2017-10-03T15:48:10.000Z

Your format is fine but some os's can't use the negative formatting for zero-padded units. datetime should be able to parse both padded and non-padded instances of those just fine:
from datetime import datetime as dt
z_time = '06:48 PM - 03 Jun 2021'
nz_time = '6:48 PM - 3 Jun 2021'
dt.strptime(z_time, '%I:%M %p - %d %b %Y')
[out:] datetime.datetime(2021, 6, 3, 18, 48)
dt.strptime(nz_time, '%I:%M %p - %d %b %Y')
[out:] datetime.datetime(2021, 6, 3, 18, 48)
And since you're getting strings from datetimes, you should look whether your os supports certain formatting rules. Here's one for windows.

from datetime import datetime
str="3:01 AM - 18 Dec 2017"
date=datetime.strptime(str,"%I:%M %p - %d %b %Y")

To turn your string into a time, do this:
>>> import time
>>> s = "3:01 AM - 18 Dec 2017"
>>> time.strptime(s,'%I:%M %p - %d %b %Y')
time.struct_time(tm_year=2017, tm_mon=12, tm_mday=18, tm_hour=3, tm_min=1,
tm_sec=0, tm_wday=0, tm_yday=352, tm_isdst=-1)
No hyphens after %. The are not mentioned in the official Python documentation.

I wanted to figure out how to use strftime in a lambda function or better, using the .dt accessor after the column in my dataframe had been converted to a datetime
I couldn't figure this out so I went for the next fastest method
from datetime import datetime
formatted_dates = []
for item in df.tweet_date:
formatted_dates.append(datetime.strptime(item,"%I:%M %p - %d %b %Y"))
df.tweet_date = formatted_dates

Extracting just the year, month and day from this format

If I have a dict with date formatted as 'Thu May 01 04:19:08 +0000 2014' how do I go about extracting just the Year, Month and Day to a variable so that I can then search via date range. I need it in '%Y %M %D' '%H %M %S' using python.

You can use datetime.datetime.strptime to parse that string into a date object and then use strftime (strftime for format, strptime for parse) to extract it in your desired format.
import datetime
date = "Thu May 01 04:19:08 +0000 2014"
objectified = datetime.datetime.strptime(date, "%a %b %d %H:%M:%S %z %Y")
formatted = objectified.strftime("'%Y %M %D' '%H %M %S'")
>>> formatted
"'2014 19 05/01/14' '04 19 08'"
Placeholders are explained here

How to strip date time in python

From a website I'm getting a date in such format: Sun Jan 22 21:32:58 +0000 2012. I understand that I must get rid of +0000 to convert it to the date, but how exactly I can do it? I read the documentation but my code is not working:
from datetime import datetime
strDate = 'Mon Apr 29 14:30:53 2019'
objDate = datetime.strptime(strDate, '%a %b %H %M %S %Y')
I'm getting an error:
ValueError: time data 'Mon Apr 29 14:30:53 2019' does not match format '%d %m %H %M %S %Y'
And I don't really understand why. Or anyone knows how I can get a date from Sun Jan 22 21:32:58 +0000 2012?

If your object is datetime.datetime you can just simply do date()
from datetime import datetime
datetime1 = datetime.now()
date1 = datetime1.date()

One line solution:
strDate ='Sun Jan 22 21:32:58 +0000 2012'
objDate = datetime.strptime(strDate, '%a %b %d %H:%M:%S +%f %Y')
print(objDate)
#2019-04-29 14:30:53
Details:
You just forgot to use %d in order to capture the date number and the : for the time and you ALSO need to capture +0000.
Proof:

I'm afraid that the currently accepted answer, by seralouk, is incorrect. Using "+%f" turns the numbers into fractions of seconds. It's fine for 0000, but will mess things up if they happen to be anything else.
This is because the "+0000" part is a time zone offset, and the proper way to parse it is by using the "%z" directive, which will handle the "+" sign as well, so remove that from the format string:
>>> date_string = "Sun Jan 22 21:32:58 +0000 2012"
>>> datetime.strptime(date_string, "%a %b %d %H:%M:%S %z %Y")
datetime.datetime(2012, 1, 22, 21, 32, 58, tzinfo=datetime.timezone.utc)

You're missing colons : and the day format string %d. See the official documentation of strptime for a table that shows the different formatting values.
from datetime import datetime
strDate = 'Mon Apr 29 14:30:53 2019'
objDate = datetime.strptime(strDate, '%a %b %d %H:%M:%S %Y')

You can get rid of the '+0000' like this:
from datetime import datetime
strDate ='Sun Jan 22 21:32:58 +0000 2012'
objDate = datetime.strptime(strDate.replace(strDate.split(" ")[4] + " ", ""), '%a %b %d %H:%M:%S %Y')
print(objDate)
-> 2012-01-22 21:32:58
By the way,
1. The example of code that you post is not the one for the problem that you are asking about! (it does not include the +0000 in the string).
2. The error that you share (false format of the date, which has been answered already and described f.ex. here) is for another code, not the one you present above! (Error throws '%d %m %H %M %S %Y' instead of '%a %b %H %M %S %Y').

In your string u are missing the %d to catch the Day of month 01-31
'%a %b %d %H:%M:%S %Y'
If u want also to catch the +0000 u have to use the %z notation
Use this as a refrence to build your string correctly:
http://strftime.org/

Convert string to DateTime for mysql in python

I want to convert my date into DateTime object for MySQL.
My string format is: Mon Aug 27 04:47:45 +0000 2018
Expected Output: 'YYYY-M-D H:mm:ss'

from datetime import datetime
t = datetime.strptime('Mon Aug 27 04:47:45 +0000 2008', '%a %b %d %H:%M:%S % z %Y')
t.strftime('%Y-%m-%d %H:%M:%S')
Refer section 8.1.8
here

If you are using python 3, this solution would work -
from datetime import datetime
x = 'Mon Aug 27 04:47:45 +0000 2018'
x = datetime.strftime(datetime.strptime(x, '%a %b %d %I:%M:%S %z %Y'), '%Y-%m-%d %H:%M:%S')
# OP '2018-08-27 04:47:45'
But for python 2, you might get a ValueError: 'z' is a bad directive.... In that case, you'll either have to use something like pytz or dateutil. The table that you need to look for all these conversions can be found here
Edit: You can't have Expected Output: 'YYYY-M-D H:mm:ss' if you convert your datetime string to datetime object. Datetime object has it's own format. Above gives you a string of the format that you want

from datetime import datetime
date_as_dt_object = datetime.strptime(dt, '%a %b %d %H:%M:%S %z %Y')
You can use date_as_dt_object in a raw query or an ORM. If used in a raw query pass it as a string like:
query = "select * from table where date >" + str(date_as_dt_object)
Check out this list for Python's strftime directives.
http://strftime.org/

Convert date string in excel to date object in python

I have a date in excel which is given in: dd mmm yy format i.e.,
29 Jun 18
How do I convert this string into a date object?
I get the error:
time data '13 Jul 18' does not match format '%d %m %Y'
when I try
datetime.strptime(input, '%d %m %Y')
What should the correct date format be?

Since the year in your excell is only two digits (i.e., 18 and not 2018) you need to use %y instead of %Y in your format string:
datetime.strptime(input, '%d %b %y')
For example:
datetime.strptime( '13 Jul 18', '%d %b %y')
Results with:
datetime.datetime(2018, 7, 13, 0, 0)
See this page for more information about date/time string format.

You can use python datetime module or you can use dateutil parser to parse the string date to valid datetime object. I'd go with dateutil parser as I don't have to define string format. Here is an example
from dateutil.parser import parse
dt = parse("Thu Sep 25 10:36:28 BRST 2003")
Remember to install dateutil by pip install python-dateutil

You would have to import datetime and import xlrd
Use xlrd to open the excel workbook as
book = xlrd.open_workbook("Excel.xlsx")
sheet = book.sheet_by_name("Worksheet")
Use this to convert
obj = datetime.datetime(*xlrd.xldate_as_tuple(sheet.cell(row,column).value, book.datemode))

from datetime import datetime
datetime.strptime("29 Jun 18", "%d %b %y").date()
Here you get a datetime.date object, I don't know if that's good enough for you. I recommend you to visit the documentation on the module

You can make use of strptime which follows the pattern:
datetime.strptime(date_string, format)
example:
from datetime import datetime
dt = datetime.strptime('19 Jul 2017', '%d %b %y')
Hope this helps :)

Your format is incorrect: %b is Locale's short month and %y is two digit year
import time
time.strptime('13 Jul 18', '%d %b %y')
time.struct_time(tm_year=2018, tm_mon=7, tm_mday=13, tm_hour=0, tm_min=0, tm_sec=0, tm_wday=4, tm_yday=194, tm_isdst=-1)