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
Related
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/
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/
I have two timestamp strings. I want to find the difference between them in seconds.
I've tried:
from time import gmtime, strptime
a = "Mon 11 Dec 2017 13:54:36 -0700"
b = "Mon 11 Dec 2017 13:54:36 -0000"
time1 = strptime(a, "%a %d %b %Y %H:%M:%S %z")
time2 = strptime(b, "%a %d %b %Y %H:%M:%S %z")
time1-time2
Getting an error: TypeError: unsupported operand type(s) for -: 'time.struct_time' and 'time.struct_time'
So, how do I calculate the difference using package time?
I was successful using package datetime - in the code below, but I think I read that datetime ignores seconds in leap years, or something to that effect. Thus, I am trying to use 'time':
from datetime import datetime
time1 = datetime.strptime(a, "%a %d %b %Y %H:%M:%S %z")
time2 = datetime.strptime(b, "%a %d %b %Y %H:%M:%S %z")
dif = time1 - time2
print(int(dif.total_seconds()))
Thank you very much!
First of all, you're using time.strptime, which returns <class 'time.struct_time'>, and it doesn't support the substract operator, one possible way to achieve what you want would be converting to datetime:
from datetime import datetime
from time import mktime
from time import gmtime, strptime
a = "Mon 11 Dec 2017 13:54:36 -0700"
b = "Mon 11 Dec 2017 13:54:36 -0000"
time1 = strptime(a, "%a %d %b %Y %H:%M:%S %z")
time2 = strptime(b, "%a %d %b %Y %H:%M:%S %z")
print(datetime.fromtimestamp(mktime(time1))-datetime.fromtimestamp(mktime(time2)))
Or even better, use datetime.datetime.strptime so you don't need intermediate conversions.
For a more detailed description of the supported operations of datetime please refer to the section supported operations in the docs here. Especially the section where it says:
If both are aware and have different tzinfo attributes, a-b acts as if
a and b were first converted to naive UTC datetimes first. The result
is (a.replace(tzinfo=None) - a.utcoffset()) - (b.replace(tzinfo=None)
- b.utcoffset()) except that the implementation never overflows.
In any case, maybe your best chance is considering an alternative method like the one proposed in this answer
I have a string like 23 July 1914 and want to convert it to 23/07/1914 date format.
But my code gives error.
from datetime import datetime
datetime_object = datetime.strptime('1 June 2005','%d %m %Y')
print datetime_object
Your error is in the format you are using to strip your string. You use %m as the format specifier for month, but this expects a 0 padded integer representing the month of the year (e.g. 06 for your example). What you want to use is %B, which expects an month of the year written out fully (e.g. June in your example).
For a full explanation of the datetime format specifiers please consult the documentation, and if you have any other issues please check there first.
Here is what you should be doing:
from datetime import datetime
datetime_object = datetime.strptime('1 June 2005','%d %B %Y')
s = datetime_object.strftime("%d/%m/%y")
print(s)
Output:
>>> 01/06/05
You see your strptime requires two parameters.
strptime(string[, format])
And the string will be converted to a datetime object according to a format you specify.
There are various formats
%a - abbreviated weekday name
%A - full weekday name
%b - abbreviated month name
%B - full month name
%c - preferred date and time representation
%C - century number (the year divided by 100, range 00 to 99)
%d - day of the month (01 to 31)
%D - same as %m/%d/%y
%e - day of the month (1 to 31)
%g - like %G, but without the century
%G - 4-digit year corresponding to the ISO week number (see %V).
%h - same as %b
%H - hour, using a 24-hour clock (00 to 23)
The above are some examples. Take a look here for formats
Take a goood look at these two!
%b - abbreviated month name
%B - full month name
It should be in a similar pattern to the string you provide. Confusing take a look at these examples.
>>> datetime.strptime('1 jul 2009','%d %b %Y')
datetime.datetime(2009, 7, 1, 0, 0)
>>> datetime.strptime('1 Jul 2009','%d %b %Y')
datetime.datetime(2009, 7, 1, 0, 0)
>>> datetime.strptime('jul 21 1996','%b %d %Y')
datetime.datetime(1996, 7, 21, 0, 0)
As you can see based on the format the string is turned into a datetime object. Now take a look!
>>> datetime.strptime('1 July 2009','%d %b %Y')
Traceback (most recent call last):
File "<pyshell#12>", line 1, in <module>
datetime.strptime('1 July 2009','%d %b %Y')
File "/usr/lib/python3.5/_strptime.py", line 510, in _strptime_datetime
tt, fraction = _strptime(data_string, format)
File "/usr/lib/python3.5/_strptime.py", line 343, in _strptime
(data_string, format))
ValueError: time data '1 July 2009' does not match format '%d %b %Y'
Why error because jun or Jun (short form) stands for %b. When you supply a June it gets confused. Now what to do? Changed the format.
>>> datetime.strptime('1 July 2009','%d %B %Y')
datetime.datetime(2009, 7, 1, 0, 0)
Simple now converting the datetime object is simple enough.
>>> s = datetime.strptime('1 July 2009','%d %B %Y')
>>> s.strftime('%d/%m/%Y')
'01/07/2009
Again the %m is the format for displaying it in months (numbers) read more about them.
The placeholder for "Month as locale’s full name." would be %B not %m:
>>> from datetime import datetime
>>> datetime_object = datetime.strptime('1 June 2005','%d %B %Y')
>>> print(datetime_object)
2005-06-01 00:00:00
>>> print(datetime_object.strftime("%d/%m/%Y"))
01/06/2005
This should work:
from datetime import datetime
print(datetime.strptime('1 June 2005', '%d %B %Y').strftime('%d/%m/%Y'))
print(datetime.strptime('23 July 1914', '%d %B %Y').strftime('%d/%m/%Y'))
For more info you can read about strftime-strptime-behavior
%d means "Day of the month as a zero-padded decimal number."
%m means "Month as a zero-padded decimal number."
Neither day or month are supplied what you tell it to expect. What you need it %B for month (only if your locale is en_US), and %-d for day.
What is the correct format to convert this string Tue Jan 10 2017 13:00:13 GMT 0800 (Taipei Standard Time) to a python date type object using strptime?
I tried the answer from this question and it is not working for me:
date1 = datetime.strptime(strDate1, '%b %d %Y %I:%M%p')
ValueError: time data 'Tue Jan 10 2017 13:00:13 GMT 0800 (Taipei
Standard Time)' does not match format '%b %d %Y %I:%M%p'
You can format the date without timezone and add it later,
import pytz
date1=datetime.strptime('Tue Jan 10 2017 13:00:13', '%a %b %d %Y %H:%M:%S')
tz=pytz.timezone('Asia/Taipei')
tz.localize(date1)
Nominally you would want to be able to use the %z (lowercase z) to convert the TZ offset, however support for this is sketchy. So you can DIY it:
import datetime as dt
import re
PARSE_TIMESTAMP = re.compile('(.*) ([+-]?\d+) \(.*\)$')
def my_datetime_parse(timestamp):
''' return a naive datetime relative to UTC '''
# find the standard time stamp, and the TZ offset and remove extra end
matches = PARSE_TIMESTAMP.match(timestamp).groups()
# convert the timestamp element
timestamp = dt.datetime.strptime(matches[0], '%a %b %d %Y %H:%M:%S %Z')
# calculate the timezone offset
tz_offset = matches[1]
sign = '+'
if tz_offset[0] in '-+':
sign = tz_offset[0]
tz_offset = tz_offset[1:]
tz_offset += '0' * (4 - len(tz_offset))
minutes = int(tz_offset[0:2]) * 60 + int(tz_offset[2:])
if sign == '-':
minutes = -minutes
# add the timezone offset to our time
timestamp += dt.timedelta(minutes=minutes)
return timestamp
date_string = 'Tue Jan 10 2017 13:00:13 GMT +0800 (Taipei Standard Time)'
print(my_datetime_parse(date_string))
This code produces:
2017-01-10 21:00:13
The code removes the (Taipei Standard Time) since it is redundant with the +0800.