Python Pandas Group by date using datetime data

Question:

I have a datetime column Date_Time that I wish to groupby without creating a new column. Is this possible? I tried the following and it does not work.

df = pd.groupby(df,by=[df['Date_Time'].date()])
Asked By: GoBlue_MathMan

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Source

Answers:

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
Answered By: jezrael

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
Answered By: piRSquared
df.groupby(pd.Grouper(key='Date_Time', axis=0, freq='M')).sum()
  • M for month
  • Y for year
  • D for day
Answered By: Avijit Das