How to split a date column into separate day , month ,year column in pandas

Question:

I have a dataset df:

               Dewptm   Fog  Humidity   Pressurem     Tempm      Wspdm  Rainfall
datetime_utc                            
1996-11-01    11.666667 0.0  52.916667  -2659.666667  22.333333  2.466667   0
1996-11-02    10.458333 0.0  48.625000  1009.833333   22.916667  8.028571   0
1996-11-03    12.041667 0.0  55.958333  1010.500000   21.791667  4.804545   0
1996-11-04    10.222222 0.0  48.055556  1011.333333   22.722222  1.964706   0
...

Here is df.columns:

Index(['Dewptm', 'Fog', 'Humidity', 'Pressurem', 'Rain', 'Tempm', 'Wspdm',
       'Rainfall'],
      dtype='object')

How could I split datetime_utc column into the year, month and day column?

I tried:

df["day"] = df['datetime_utc'].map(lambda x: x.day)
df["month"] = df['datetime_utc'].map(lambda x: x.month)
df["year"] = df['datetime_utc'].map(lambda x: x.year)

Error:

KeyError: 'datetime_utc'

Also

pd.concat([df.drop('datetime_utc', axis = 1), 
          (df.datetime_utc.str.split("-).str[:3].apply(pd.Series)
          .rename(columns={0:'year', 1:'month', 2:'day'}))], axis = 1)

I am getting error:

KeyError: "['datetime_utc'] not found in axis"

The problem I am facing is the column datetime_utc is the default index column in my dataset.

Asked By: Ratnesh

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Answers:

The problem is that datetime_utc is in your index instead a column, so you have to access your index to be able to make your new columns:

df['day'] = df.index.day
df['month'] = df.index.month
df['year'] = df.index.year

print(df)
                 Dewptm  Fog   Humidity    Pressurem      Tempm     Wspdm  
datetime_utc                                                                
1996-11-01    11.666667  0.0  52.916667 -2659.666667  22.333333  2.466667   
1996-11-02    10.458333  0.0  48.625000  1009.833333  22.916667  8.028571   
1996-11-03    12.041667  0.0  55.958333  1010.500000  21.791667  4.804545   
1996-11-04    10.222222  0.0  48.055556  1011.333333  22.722222  1.964706   

              Rainfall  day  month  year  
datetime_utc                              
1996-11-01           0    1     11  1996  
1996-11-02           0    2     11  1996  
1996-11-03           0    3     11  1996  
1996-11-04           0    4     11  1996  

If you want datetime_utc as a column you have to reset your index and then you can access the datetime methods with dt.month, dt.year and dt.day like following:

# Reset our index so datetime_utc becomes a column
df.reset_index(inplace=True)

# Create new columns
df['day'] = df['datetime_utc'].dt.day
df['month'] = df['datetime_utc'].dt.month
df['year'] = df['datetime_utc'].dt.year

print(df)
  datetime_utc     Dewptm  Fog   Humidity    Pressurem      Tempm     Wspdm  
0   1996-11-01  11.666667  0.0  52.916667 -2659.666667  22.333333  2.466667   
1   1996-11-02  10.458333  0.0  48.625000  1009.833333  22.916667  8.028571   
2   1996-11-03  12.041667  0.0  55.958333  1010.500000  21.791667  4.804545   
3   1996-11-04  10.222222  0.0  48.055556  1011.333333  22.722222  1.964706   

   Rainfall  day  month  year  
0         0    1     11  1996  
1         0    2     11  1996  
2         0    3     11  1996  
3         0    4     11  1996  

Note if your index is not in datetime type yet, use the following before you try to extract year, month and day:

df.index = pd.to_datetime(df.index)
Answered By: Erfan

A one-liner version is to call timetuple() on each Timestamps, which returns a tuple similar to datetime.datetime.timetuple. Since only year, month, day are needed, slice the first 3 elements.

# if datetime_utc is index
df.index = pd.to_datetime(df.index)           # <-- omit if index is already datetime64[ns]
df[['year', 'month', 'day']] = [x.timetuple()[:3] for x in df.index.tolist()]


# if datetime_utc is a column
df['datetime_utc'] = pd.to_datetime(df['datetime_utc'])  # <-- omit if datetime_utc is already datetime64[ns]
df[['year', 'month', 'day']] = df['datetime_utc'].apply(lambda x: x.timetuple()[:3]).tolist()
Answered By: cottontail