Python Pandas Group by date using datetime data

I have a column Date_Time that I wish to groupby date time without creating a new column. Is this possible the current code I have does not work.

df = pd.groupby(df,by=[df['Date_Time'].date()])

Answers:

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Method 1

You can use groupby by dates of column Date_Time by dt.date:

df = df.groupby([df['Date_Time'].dt.date]).mean()

Sample:

df = pd.DataFrame({'Date_Time': pd.date_range('10/1/2001 10:00:00', periods=3, freq='10H'),
                   'B':[4,5,6]})

print (df)
   B           Date_Time
0  4 2001-10-01 10:00:00
1  5 2001-10-01 20:00:00
2  6 2001-10-02 06:00:00

print (df['Date_Time'].dt.date)
0    2001-10-01
1    2001-10-01
2    2001-10-02
Name: Date_Time, dtype: object

df = df.groupby([df['Date_Time'].dt.date])['B'].mean()
print(df)
Date_Time
2001-10-01    4.5
2001-10-02    6.0
Name: B, dtype: float64

Another solution with resample:

df = df.set_index('Date_Time').resample('D')['B'].mean()

print(df)
Date_Time
2001-10-01    4.5
2001-10-02    6.0
Freq: D, Name: B, dtype: float64

Method 2

resample

df.resample('D', on='Date_Time').mean()

              B
Date_Time      
2001-10-01  4.5
2001-10-02  6.0

Grouper

As suggested by @JosephCottam

df.set_index('Date_Time').groupby(pd.Grouper(freq='D')).mean()

              B
Date_Time      
2001-10-01  4.5
2001-10-02  6.0

Deprecated uses of TimeGrouper

You can set the index to be 'Date_Time' and use pd.TimeGrouper

df.set_index('Date_Time').groupby(pd.TimeGrouper('D')).mean().dropna()

              B
Date_Time      
2001-10-01  4.5
2001-10-02  6.0

Method 3

df.groupby(pd.Grouper(key='Date_Time', axis=0, freq='M')).sum()

M for month
Y for year
D for day

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0