how to change datetimeindex to just contain date in python

Question

I have a column called date with this values

> DatetimeIndex(['2014-02-19'], dtype='datetime64[ns]', freq=None)
> DatetimeIndex(['2013-02-29'], dtype='datetime64[ns]', freq=None)
> DatetimeIndex(['2018-04-15'], dtype='datetime64[ns]', freq=None)

how do i modify the column to just extract the date values and get rid of words like DatetimeIndex and brackets etc?

> 2014-02-19
> 2013-02-19
> 2018-04-15

The code I wrote I think is pretty incorrect but still attaching it here:

def fundate(x):
    return x[0][0]

df['date'] = df.apply(lambda row : fundate(row['date']), axis = 1)

could someone please help me?

This should work https://stackoverflow.com/questions/52278464/convert-datetimeindex-to-datetime-date-in-pandas — SunilG, May 28 '21 at 07:15

score 0 · Answer 1 · answered May 28 '21 at 07:03

0

Do you mean something like this (not tested)?

def fundate(x):
    return x.strftime("%Y-%m-%d")

From https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.Series.dt.strftime.html

answered May 28 '21 at 07:03

umbe1987

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score 0 · Answer 2 · answered May 28 '21 at 07:05

0

I think this is what you want. It selects the first item from the index and changes it to a string.

df['new_column'] = df['date'].apply(lambda x: x[0].strftime("%Y-%m-%d"))

answered May 28 '21 at 07:05

Rutger

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jezrael · Accepted Answer · 2021-05-28T07:11:30.437

0

If there is object DatetimeIndex use:

df['date'] = df['date'].str[0]

If there is multiple values per row:

df = df.explode('date')

If there are strings and only one value per row:

df["date"] = df["date"].str.extract(r"(\d{4}-\d{2}-\d{2})", expand=False)

If possible multipel values per row:

df["date"] = df["date"].str.findall(r"(\d{4}-\d{2}-\d{2})")
df = df.explode('date')

edited May 28 '21 at 07:11

answered May 28 '21 at 07:05

jezrael

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Utsav · Answer 4 · 2021-05-28T07:28:36.660

We are using regex to fetch date element between square brackets of input string.

If you need output in object/str

df.date = df["date"].str.extract('\[(.*?)]', expand=True).replace("'","", regex=True)

Output

    date
0   2014-02-19
1   2013-02-28
2   2018-04-15

Output Type

df.info()

<class 'pandas.core.frame.DataFrame'>
RangeIndex: 3 entries, 0 to 2
Data columns (total 1 columns):
 #   Column  Non-Null Count  Dtype 
---  ------  --------------  ----- 
 0   date    3 non-null      object
dtypes: object(1)
memory usage: 152.0+ bytes

If output is required as datetime type

df.date = df["date"].str.extract('\[(.*?)]', expand=True).replace("'","", regex=True)
df.date = pd.to_datetime(df.date)
df

Output

    date
0   2014-02-19
1   2013-02-28
2   2018-04-15

Output Type

df.info()
<class 'pandas.core.frame.DataFrame'>
RangeIndex: 3 entries, 0 to 2
Data columns (total 1 columns):
 #   Column  Non-Null Count  Dtype         
---  ------  --------------  -----         
 0   date    3 non-null      datetime64[ns]
dtypes: datetime64[ns](1)
memory usage: 152.0 bytes

Input DF

df = pd.DataFrame({
    'date':["DatetimeIndex(['2014-02-19'], dtype='datetime64[ns]', freq=None)",
"DatetimeIndex(['2013-02-28'], dtype='datetime64[ns]', freq=None)",
"DatetimeIndex(['2018-04-15'], dtype='datetime64[ns]', freq=None)"]
})

df

score 0 · Answer 5 · answered May 28 '21 at 07:16

You can do:

import pandas as pd
import datetime as dt
rng = pd.date_range('2015-02-24', periods=5, freq='T')
df = pd.DataFrame({ 'date': rng })
    
df['date'] = df['date'].dt.strftime("%Y-%m-%d")

to convert a pd.Series with dtype datetime64ns to a string

If you like to convert that pd.Series to an list you can do:

datelist = df['date'].tolist()

how to change datetimeindex to just contain date in python

5 Answers5