问题
I need to calculate the mean per day of the colums duration and km for the rows with value ==1 and values = 0.
df
Out[20]:
Date duration km value
0 2015-03-28 09:07:00.800001 0 0 0
1 2015-03-28 09:36:01.819998 1 2 1
2 2015-03-30 09:36:06.839997 1 3 1
3 2015-03-30 09:37:27.659997 nan 5 0
4 2015-04-22 09:51:40.440003 3 7 0
5 2015-04-23 10:15:25.080002 0 nan 1
how can I modify this solution in order to have the means duration_value0, duration_value1, km_value0 and km_value1?
df = df.set_index('Date').groupby(pd.Grouper(freq='d')).mean().dropna(how='all')
print (df)
duration km
Date
2015-03-28 0.5 1.0
2015-03-30 1.5 4.0
2015-04-22 3.0 7.0
2015-04-23 0.0 0.0
回答1:
I think you are looking pivot table i.e
df.pivot_table(values=['duration','km'],columns=['value'],index=df['Date'].dt.date,aggfunc='mean')
Output:
duration km
value 0 1 0 1
Date
2015-03-28 0.0 1.0 0.0 2.0
2015-03-30 NaN 1.0 5.0 3.0
2015-04-22 3.0 NaN 7.0 NaN
2015-04-23 NaN 0.0 NaN NaN
In [24]:
If you want the new column names like distance0,distance1 ... You can use list comprehension i.e if you store the pivot table in ndf
ndf.columns = [i[0]+str(i[1]) for i in ndf.columns]
Output:
duration0 duration1 km0 km1
Date
2015-03-28 0.0 1.0 0.0 2.0
2015-03-30 NaN 1.0 5.0 3.0
2015-04-22 3.0 NaN 7.0 NaN
2015-04-23 NaN 0.0 NaN NaN
回答2:
I believe doing a group by Date as well as value should do it.
Call dfGroupBy.mean followed by df.reset_index to get your desired output:
In [713]: df.set_index('Date')\
.groupby([pd.Grouper(freq='d'), 'value'])\
.mean().reset_index(1, drop=True)
Out[713]:
duration km
Date
2015-03-28 0.0 0.0
2015-03-28 1.0 2.0
2015-03-30 NaN 5.0
2015-03-30 1.0 3.0
2015-04-22 3.0 7.0
2015-04-23 0.0 NaN
来源：https://stackoverflow.com/questions/45839539/mean-of-a-grouped-by-pandas-dataframe