How to change the datetime format in pandas

Question

My dataframe has a DOB column (example format 1/1/2016) which by default gets converted to pandas dtype \'object\': DOB object

Answer 1

You can use dt.strftime if you need to convert datetime to other formats (but note that then dtype of column will be object (string)):

import pandas as pd

df = pd.DataFrame({'DOB': {0: '26/1/2016', 1: '26/1/2016'}})
print (df)
         DOB
0  26/1/2016 
1  26/1/2016

df['DOB'] = pd.to_datetime(df.DOB)
print (df)
         DOB
0 2016-01-26
1 2016-01-26

df['DOB1'] = df['DOB'].dt.strftime('%m/%d/%Y')
print (df)
         DOB        DOB1
0 2016-01-26  01/26/2016
1 2016-01-26  01/26/2016

Answer 2

There is a difference between

• the content of a dataframe cell (a binary value) and
• its presentation (displaying it) for us, humans.

So the question is: How to reach the appropriate presentation of my datas without changing the data / data types themselves?

Here is the answer:

• If you use the Jupyter notebook for displaying your dataframe, or
• if you want to reach a presentation in the form of an HTML file (even with many prepared superfluous id and class attributes for further CSS styling — you may or you may not use them),

use styling. Styling don't change data / data types of columns of your dataframe.

Now I show you how to reach it in the Jupyter notebook — for a presentation in the form of HTML file see the note near the end of the question.

I will suppose that your column DOB already has the type datetime64 (you shown that you know how to reach it). I prepared a simple dataframe (with only one column) to show you some basic styling:

• Not styled:

     df
  

          DOB
0  2019-07-03
1  2019-08-03
2  2019-09-03
3  2019-10-03

• Styling it as mm/dd/yyyy:

     df.style.format({"DOB": lambda t: t.strftime("%m/%d/%Y")})
  

          DOB
0  07/03/2019
1  08/03/2019
2  09/03/2019
3  10/03/2019

• Styling it as dd-mm-yyyy:

     df.style.format({"DOB": lambda t: t.strftime("%d-%m-%Y")}) 
  

          DOB
0  03-07-2019
1  03-08-2019
2  03-09-2019
3  03-10-2019

---

Be careful!
The returning object is NOT a dataframe — it is an object of the class Styler, so don't assign it back to df:

Don´t do this:

df = df.style.format({"DOB": lambda t: t.strftime("%m/%d/%Y")})    # Don´t do this!

(Every dataframe has its Styler object accessible by its .style property, and we changed this df.style object, not the dataframe itself.)

---

Questions and Answers:

• Q: Why your Styler object (or an expression returning it) used as the last command in a Jupyter notebook cell displays your (styled) table, and not the Styler object itself?

• A: Because every Styler object has a callback method ._repr_html_() which returns an HTML code for rendering your dataframe (as a nice HTML table).

  Jupyter Notebook IDE calls this method automatically to render objects which have it.

---

Note:

You don't need the Jupyter notebook for styling (i.e. for nice outputting a dataframe without changing its data / data types).

A Styler object has a method render(), too, if you want to obtain a string with the HTML code (e.g. for publishing your formatted dataframe to the Web, or simply present your table in the HTML format):

df_styler = df.style.format({"DOB": lambda t: t.strftime("%m/%d/%Y")})
HTML_string = df_styler.render()

Answer 3

The below code worked for me instead of the previous one - try it out !

df['DOB']=pd.to_datetime(df['DOB'].astype(str), format='%m/%d/%Y')

Answer 4

Changing the format but not changing the type:

df['date'] = pd.to_datetime(df["date"].dt.strftime('%Y-%m'))

Answer 5

Below code changes to 'datetime' type and also formats in the given format string. Works well!

df['DOB']=pd.to_datetime(df['DOB'].dt.strftime('%m/%d/%Y'))

Answer 6

Below is the code worked for me, And we need to be very careful for format. Below link will be definitely useful for knowing your exiting format and changing into desired format(Follow strftime() and strptime() Format Codes on below link):

https://docs.python.org/3/library/datetime.html#strftime-and-strptime-behavior.

data['date_new_format'] = pd.to_datetime(data['date_to_be_changed'] , format='%b-%y')