Create Pandas Column based on Index Values

Question

I'm sure this has been asked before but I can't find the correct way to go about doing this. I have a dataframe with a datetime index

df =              col2
     test      
     | 04-05-2018 | 56 |
     | 04-06-2018 | 23 |
     | 04-07-2018 | 87 |
     | 04-08-2018 | 11 |

What I would like to do is to access (and consequently create a new column from) the month property of the datetime index. I have tried to just access the property using print(df['test'].datetime.month) but I receive the expected KeyError (I know that the title of the dataframe is 'test' and not the index itself, I just don't know what to do).

Any help is greatly appreciated, and thanks in advance.

You can access the index as a column after calling df.reset_index() or the index directly with df.index — G. Anderson
– G. Anderson, Commented Oct 24, 2019 at 19:39
thanks both, didn't know i could do it that way! @QuangHoang first solution is what i'm looking for as I would like to keep it as the index — poorpractice
– poorpractice, Commented Oct 24, 2019 at 19:45

Gorlomi · Accepted Answer · 2019-10-24 19:58:01Z

1

Hope this is what you're looking for:

import pandas as pd

data  = [['04-05-2018',56] ,
         ['04-06-2018',23],
         ['04-07-2018',87],
         ['04-08-2018',11]]

df = pd.DataFrame(data, columns =['test', 'col2'])

df.set_index('test', inplace=True)

df['MONTH'] = pd.to_datetime(df.index).month

df


           col2 Month
test        
04-05-2018  56  4
04-06-2018  23  4
04-07-2018  87  4
04-08-2018  11  4

answered Oct 24, 2019 at 19:58

Gorlomi

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Comments

Walid Chiko · Accepted Answer · 2019-10-24 20:20:16Z

1

import pandas as pd

data  = [['04-05-2018',56] ,
         ['04-06-2018',23],
         ['04-07-2018',87],
         ['04-08-2018',11]]

df = pd.DataFrame(data, columns =['test', 'col2'])
df.set_index('test', inplace=True)
df

            col2
test            
04-05-2018    56
04-06-2018    23
04-07-2018    87
04-08-2018    11

df=dfT.reset_index()
df['Months'] = list(dfT['test'].astype(str).str[3:5])
df

            col2 Months
test                   
04-05-2018    56     05
04-06-2018    23     06
04-07-2018    87     07
04-08-2018    11     08

Done ✅✅

edited Oct 24, 2019 at 20:20

answered Oct 24, 2019 at 19:58

Walid Chiko

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Comments

Alexey · Accepted Answer · 2019-10-24 20:10:54Z

0

Like this?

import pandas as pd

ind = ['04-05-2018','04-06-2018', '04-07-2018', '04-08-2018']    
col2 = [56,23,87,11]

df = pd.DataFrame(col2, index = ind, columns = ['col2'])

Out[1]:
            col2
04-05-2018  56
04-06-2018  23
04-07-2018  87
04-08-2018  11


df['month'] = pd.to_datetime(df.index).month
df

Out[2]:

           col2 month
04-05-2018  56  4
04-06-2018  23  4
04-07-2018  87  4
04-08-2018  11  4

upd for zeros in first char

df['month_s'] = [str(mnth).zfill(2) for mnth in pd.to_datetime(df.index).month]
df

Out[3]:


           col2 month   month_s
04-05-2018  56     4    04
04-06-2018  23     4    04
11-07-2018  87     4    04
04-08-2018  11     4    04

edited Oct 24, 2019 at 20:10

answered Oct 24, 2019 at 19:44

Alexey

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1 Comment

poorpractice Over a year ago

Basically yes! I was just trying to keep the leading zero for the month - is that possible?

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