I am quite new to programming so please excuse if I don't use the correct terms.
I want to convert a list of dates into another format. Until now it was easy because the dates that I had to convert were in this format: %d-%b-%Y (example 13-Jan-2023). So I could just "tell" python that it was in this format and that it should convert it to : %d/%m/%Y (example 13/01/2023).
However, the initial date format has changed to 13-janv.-2023 (per example), so the months are now in french and abbreviated (janv. ; févr. ; mars ; avr. ; mai ; juin ; juil. ; sept. ; oct. ; nov. ; déc.
I can imagine that there must be a way to define a dictionary such as "Jan : janv. ", and then define it as a new format but I have not succeeded yet. Your help will be appreciated.
Thanks
I am sorry, I have not found a similar question...
localeis your friend for region-specific topics.
from datetime import datetime as dt
import locale
locale.setlocale(locale.LC_TIME, "fr_FR")
date_fr = '13-janv.-2023'
dt.strptime(date_fr, '%d-%b-%Y').strftime('%d/%m/%Y')
Returns your desired result. But be aware! After setting your locale, you can't parse your initial ('13-Jan-2023') format anyore until your change the locale back.
Explanation:
strptime is to parse your input, hence the p in the function name
strftime is to format your date/time, hence the f.
Related
I have a datatime string (which comes from django/python) which looks like so:
datatime_str='2020-08-18 16:48:13.722422+00:00'
I then do, in Matlab 2018a:
fmt_dt='yyyy-MM-dd HH:mm:ss.SSSSSS+HH:mm';
datetime(datatime_str,'TimeZone','local','Format',fmt_dt);
and I get:
2020-08-18 00:00:13.722422+00:00
I am not sure what it is that I am doing wrong, but the result is obviously wrong :(
Any help would be great
Yes the +00:00 should be formatted as timezone, not hours, minute. However, you can set the display format as you would like, and this could be different from the input. The default Matlab datetime display format discards the fractional seconds (and also the timezone I think). For example:
fmt_dt_input='yyyy-MM-dd HH:mm:ss.SSSSSSxxxxx';
fmt_dt_show='yyyy-MM-dd HH:mm:ss.SSSSSS xxxxx';
datatime_str='2020-08-18 16:48:13.722422+00:00';
t = datetime(datatime_str,'InputFormat',fmt_dt_input,'TimeZone','local','Format',fmt_dt_show)
Has as output: 2020-08-18 16:48:13.722422 +00:00
EDIT: btw this info on datetime can be found here
your input string contains a UTC offset at the end, +00:00 which you parse as hours and minutes - that is why they are set to 00:00 in the result. Use e.g.
datetime('2020-08-18 16:48:13.722422+00:00', 'InputFormat', 'yyyy-MM-dd HH:mm:ss.SSSSSSZ', 'TimeZone', 'UTC')
instead (change the TimeZone parameter to whatever you need).
Here's an array of datetime values:
array = np.array(['2016-05-01T00:00:59.3+10:00', '2016-05-01T00:02:59.4+10:00',
'2016-05-01T00:03:59.4+10:00', '2016-05-01T00:13:00.1+10:00',
'2016-05-01T00:22:00.5+10:00', '2016-05-01T00:31:01.1+10:00'],
dtype=object)
pd.to_datetime is very good at inferring datetime formats.
array = pd.to_datetime(array)
print(array)
DatetimeIndex(['2016-04-30 14:00:59.300000', '2016-04-30 14:02:59.400000',
'2016-04-30 14:03:59.400000', '2016-04-30 14:13:00.100000',
'2016-04-30 14:22:00.500000', '2016-04-30 14:31:01.100000'],
dtype='datetime64[ns]', freq=None)
How can I dynamically figure out what datetime format pd.to_datetime inferred? Something like: %Y-%m-%dT... (sorry, my datetime foo is really bad).
I don't think it's possible to do this in full generality in pandas.
As mentioned in other comments and answers, the internal function _guess_datetime_format is close to being what you ask for, but it has strict criteria for what constitutes a guessable format and so it will only work for a restricted class of datetime strings.
These criteria are set out in the _guess_datetime_format function on these lines and you can also see some examples of good and bad formats in the test_parsing script.
Some of the main points are:
year, month and day must each be present and identifiable
the year must have four digits
exactly six digits must be used if using microseconds
you can't specify a timezone
This means that it will fail to guess the format for datetime strings in the question despite them being a valid ISO 8601 format:
>>> from pandas.core.tools.datetimes import _guess_datetime_format_for_array
>>> array = np.array(['2016-05-01T00:00:59.3+10:00'])
>>> _guess_datetime_format_for_array(array)
# returns None
In this case, dropping the timezone and padding the microseconds to six digits is enough to make pandas to recognise the format:
>>> array = np.array(['2016-05-01T00:00:59.300000']) # six digits, no tz
>>> _guess_datetime_format_for_array(array)
'%Y-%m-%dT%H:%M:%S.%f'
This is probably as good as it gets.
If pd.to_datetime is not asked to infer the format of the array, or given a format string to try, it will just try and parse each string separately and hope that it is successful. Crucially, it does not need to infer a format in advance to do this.
First, pandas parses the string assuming it is (approximately) a ISO 8601 format. This begins in a call to _string_to_dts and ultimately hits the low-level parse_iso_8601_datetime function that does the hard work.
You can check if your string is able to be parsed in this way using the _test_parse_iso8601 function. For example:
from pandas._libs.tslib import _test_parse_iso8601
def is_iso8601(string):
try:
_test_parse_iso8601(string)
return True
except ValueError:
return False
The dates in the array you give are recognised as this format:
>>> is_iso8601('2016-05-01T00:00:59.3+10:00')
True
But this doesn't deliver what the question asks for and I don't see any realistic way to recover the exact format that is recognised by the parse_iso_8601_datetime function.
If parsing the string as a ISO 8601 format fails, pandas falls back to using the parse() function from the third-party dateutil library (called by parse_datetime_string). This allows a fantastic level of parsing flexibility but, again, I don't know of any good way to extract the recognised datetime format from this function.
If both of these two parsers fail, pandas either raises an error, ignores the string or defaults to NaT (depending on what the user specifies). No further attempt is made to parse the string or guess the format of the string.
DateInfer (PyDateInfer) library allows to infer dates based on the sequence of available dates:
github.com/wdm0006/dateinfer
Usage from docs:
>>> import dateinfer
>>> dateinfer.infer(['Mon Jan 13 09:52:52 MST 2014', 'Tue Jan 21 15:30:00 EST 2014'])
'%a %b %d %H:%M:%S %Z %Y'
>>>
Disclaimer: I have used and then contributed to this library
You can use _guess_datetime_format from core.tools to get the format. ie
from pandas.core.tools import datetimes as tools
tools._guess_datetime_format(pd.to_datetime(array).format()[0][:10])
Output :
'%Y-%m-%d'
To know more about this method you can see here. Hope it helps.
If I knew the format in which a string represents date-time information, then I can easily use datetime.datetime.strptime(s, fmt). However, without knowing the format of the string beforehand, would it be possible to determine whether a given string contains something that could be parsed as a datetime object with the right format string?
Obviously, generating every possible format string to do an exhaustive search is not a feasible idea. I also don't really want to write one function with many format strings hardcoded into it.
Does anyone have any thoughts on how this can be accomplished (perhaps some sort of regex?)?
What about fuzzyparsers:
Sample inputs:
jan 12, 2003
jan 5
2004-3-5
+34 -- 34 days in the future (relative to todays date)
-4 -- 4 days in the past (relative to todays date)
Example usage:
>>> from fuzzyparsers import parse_date
>>> parse_date('jun 17 2010') # my youngest son's birthday
datetime.date(2010, 6, 17)
Install with:
$ pip install fuzzyparsers
You can use parser from dateutil
Example usage:
from dateutil import parser
dt = parser.parse("Aug 28 1999 12:00AM")
For example I give the date as:
2/12/2015
The result should be:
February/Thursday/2015
I tried to do with if but I'm not getting the result. It would be nice if you could tell me the long way (without using built in functions (like datetime and others) too much). I'm new to python and not much is taught in my school.
You don't have to use datetime too much, simply parse the date and output it in whatever format you want
from datetime import datetime
d = "2/12/2015"
print(datetime.strptime(d,"%m/%d/%Y").strftime("%B/%A/%Y"))
February/Thursday/2015
A = Locale’s full weekday name.
B = Locale’s full month name.
Y = Year with century as a decimal number.
All the format directives are here
You could create a dict mapping but you will find datetime is lot simpler.
I want to know how to convert different format dates to expected format in python .
ex : i want to get this format : 2/29/2012
['2012-02-01 // 2012-02-28', '2/15/2012', '2/13/2012', '2/14/2012', '2/23/2012', '2/18/2012', '2/29/2012']
How to check today date in the range '2012-02-01 // 2012-02-28'
Share your suggestions
Use the datetime library in python. You can just compare two different datetime.datetime objects. And you can separate the year, month, date thing to put it in anyform you want.
Check this link for all the library details.
http://docs.python.org/library/datetime.html
Hope that helped.
The dateutil python library parses a wider variety of date formats than the standard datetime module.