How to change the datetime format in Pandas

Compared to the first answer, I will recommend to use dt.strftime() first, and then pd.to_datetime(). In this way, it will still result in the datetime data type.

For example,

import pandas as pd

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

df['DOB1'] = df['DOB'].dt.strftime('%m/%d/%Y')
print(df.dtypes)

df['DOB1'] = pd.to_datetime(df['DOB1'])
print(df.dtypes)

Changing the format but not changing the type:

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

The below code worked for me instead of the previous one:

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

The below code changes to the ‘datetime’ type and also formats in the given format string.

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

You can try this. It’ll convert the date format to DD-MM-YYYY:

df['DOB'] = pd.to_datetime(df['DOB'], dayfirst = True)

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 this answer.

I will suppose that your column DOB already has the datetime64 type (you have 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

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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()

Below is the code that worked for me. And we need to be very careful for format. The below link will be definitely useful for knowing your exiting format and changing into the desired format (follow the strftime() and strptime() format codes in strftime() and strptime() Behavior):

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