'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()])
Solution 1:[1]
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
Solution 2:[2]
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
Solution 3:[3]
df.groupby(pd.Grouper(key='Date_Time', axis=0, freq='M')).sum()
M for month Y for year D for day
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
Solution | Source |
---|---|
Solution 1 | cs95 |
Solution 2 | |
Solution 3 | Espoir Murhabazi |