I have a problem in groupby 2 column , I got an error

Question

df['USDT_volume'].groupby(pd.Grouper(freq='60min', key=df['Datetime'])).mean().dropna()

my error: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

Please, provide a minimal reproducible example https://stackoverflow.com/help/minimal-reproducible-example and https://stackoverflow.com/questions/20109391/how-to-make-good-reproducible-pandas-examples — alec_djinn, Aug 29 '22 at 12:32

mozway · Answer 1 · 2022-08-29T12:17:54.667

0

the key parameter of pandas.Grouper should be a string, not a Series:

pd.Grouper(freq='60min', key='Datetime')

full command:

df.groupby(pd.Grouper(freq='60min', key='Datetime'))['USDT_volume'].mean().dropna()

edited Aug 29 '22 at 12:17

answered Aug 29 '22 at 12:03

mozway

but I got this: 'The grouper name Datetime is not found' – Sajede Ks77 Aug 29 '22 at 12:15
I want to calculate mean of df ['USDT_volume'] for an each hour in df['Datetime'], help me please – Sajede Ks77 Aug 29 '22 at 12:18
There was another error in your command, you should slice after grouping (otherwise the grouping column is not available) – mozway Aug 29 '22 at 12:18
error: Only valid with DatetimeIndex, TimedeltaIndex or PeriodIndex, but got an instance of 'Index' – Sajede Ks77 Aug 29 '22 at 12:24
Please provide a fully reproducible example. – mozway Aug 29 '22 at 12:26
I use this : df.groupby(pd.Grouper(freq='60min', key='Datetime'))['USDT_volume'].mean().dropna() – Sajede Ks77 Aug 29 '22 at 12:28
I meant a (**minimal**!) reproducible example of the DataFrame – mozway Aug 29 '22 at 12:28

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