How to Convert a String Date into Datetime Format in Python

how to convert a string date into datetime format in python?

The particular format for strptime:

datetime.datetime.strptime(string_date, "%Y-%m-%d %H:%M:%S.%f")
#>>> datetime.datetime(2013, 9, 28, 20, 30, 55, 782000)

How to convert string date to datetime format for graphing in python

You can use datetime.datetime.strptime -

import datetime
string = '30.06.2019 07:00:00.000,1.13760,1.13760,1.13760,1.13760,0'.split(',')[0]
d = datetime.datetime.strptime(string, '%d.%m.%Y %H:%M:%S.%f')

or more to the point if you already have the comma-separated fields,

times.append(datetime.datetime.strptime(row[0], '%d.%m.%Y %H:%M:%S.%f')

I would recommend not using time as the name of a variable since that will conflict with the time module which you may have imported. That kind of thing can really be hard to chase down.

Convert a date (string) into datetime format

Datetime documentation has a table of possible formats for strptime and strftime, and %m is labeled as 'Month as a zero-padded decimal number' (i.e. 01,02,...), which is clearly not what you need.

What you do need it %B 'Month as locale’s full name' (i.e. January, February, …, December (en_US); Januar, Februar, …, Dezember (de_DE))

So, your format should be:

from datetime import datetime
datetime.strptime('April 29, 2021', '%B %d, %Y')

How to convert string to datetime format in pandas python?

Use to_datetime. There is no need for a format string since the parser is able to handle it:

In [51]:
pd.to_datetime(df['I_DATE'])

Out[51]:
0   2012-03-28 14:15:00
1   2012-03-28 14:17:28
2   2012-03-28 14:50:50
Name: I_DATE, dtype: datetime64[ns]

To access the date/day/time component use the dt accessor:

In [54]:
df['I_DATE'].dt.date

Out[54]:
0    2012-03-28
1    2012-03-28
2    2012-03-28
dtype: object

In [56]:    
df['I_DATE'].dt.time

Out[56]:
0    14:15:00
1    14:17:28
2    14:50:50
dtype: object

You can use strings to filter as an example:

In [59]:
df = pd.DataFrame({'date':pd.date_range(start = dt.datetime(2015,1,1), end = dt.datetime.now())})
df[(df['date'] > '2015-02-04') & (df['date'] < '2015-02-10')]

Out[59]:
         date
35 2015-02-05
36 2015-02-06
37 2015-02-07
38 2015-02-08
39 2015-02-09

How to convert a date string to different format

I assume I have import datetime before running each of the lines of code below

datetime.datetime.strptime("2013-1-25", '%Y-%m-%d').strftime('%m/%d/%y')

prints "01/25/13".

If you can't live with the leading zero, try this:

dt = datetime.datetime.strptime("2013-1-25", '%Y-%m-%d')
print '{0}/{1}/{2:02}'.format(dt.month, dt.day, dt.year % 100)

This prints "1/25/13".

EDIT: This may not work on every platform:

datetime.datetime.strptime("2013-1-25", '%Y-%m-%d').strftime('%m/%d/%y')

Parse date string and change format

datetime module could help you with that:

datetime.datetime.strptime(date_string, format1).strftime(format2)

For the specific example you could do

>>> import datetime
>>> datetime.datetime.strptime('Mon Feb 15 2010', '%a %b %d %Y').strftime('%d/%m/%Y')
'15/02/2010'
>>>

How to convert string to datetime?

Here to parse a string to date time, then you can:

def convert(s):
    return datetime.strptime(s, '%Y-%m-%dT%H:%M:%SZ')

someobject = convert('2017-08-15T13:34:35Z')
print(someobject.isoformat())

Convert string Jun 1 2005 1:33PM into datetime

datetime.strptime parses an input string in the user-specified format into a timezone-naive datetime object:

>>> from datetime import datetime
>>> datetime.strptime('Jun 1 2005  1:33PM', '%b %d %Y %I:%M%p')
datetime.datetime(2005, 6, 1, 13, 33)

To obtain a date object using an existing datetime object, convert it using .date():

>>> datetime.strptime('Jun 1 2005', '%b %d %Y').date()
date(2005, 6, 1)

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Links:

• strptime docs: Python 2, Python 3

• strptime/strftime format string docs: Python 2, Python 3

• strftime.org format string cheatsheet

Notes:

• strptime = "string parse time"
• strftime = "string format time"

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