Group a DataFrame by months and an additional column

Question

I have the next DataFrame:

data={
 'date':['02/01/2023', '03/01/2023', '12/01/2023', '16/01/2023', '23/01/2023', '03/02/2023', '14/02/2023', '17/02/2023', '17/02/2023', '20/02/2023'],
'amount':[-2.6, -230.0, -9.32, -13.99, -12.99, -50.0, -5.84, -6.6, -11.95, -20.4],
'concept':['FOOD', 'REPAIR', 'HEALTH', 'NO CLASSIFIED', 'NO CLASSIFIED', 'REPAIR', 'FOOD', 'NO CLASSIFIED', 'FOOD', 'HEALTH']     
      }
df = pd.DataFrame(data)

I need to group the information first by months and then by the concept of each item. I tried something this:

df.groupby(['date','concept']).sum()

And it works for an individual day, but I need the same but grouped by the entire month.

I tried also converting that df.date to datetime values: df.date = pd.to_datetime(df.date,dayfirst=True), but I don’t know how to indicate that the grouping should be by the each entire month.

The result I need would be something like this:

date	concept	amount
Jan-23	FOOD	-2.6
	HEALTH	-9.32
	NO CLASSIFIED	-26.98
	REPAIR	-230
Feb-23	FOOD	-17.79
	HEALTH	-20.4
	NO CLASSIFIED	-6.6
	REPAIR	-50

Asked By: nekovolta

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Source

Answer 1

You can use datetime module. First convert the date column to a datetime object and then create new year and month columns accessing by ‘dt’. Then groupby and sum. try below code:

import datetime

df["date"] = pd.to_datetime(df["date"],dayfirst=True)
df["month"] = df["date"].dt.month
df["year"] = df["date"].dt.year
df.drop("date",axis=1,inplace = True)

result = df.groupby(["year","month","concept"]).sum()

or you can use :

df["date"] = pd.to_datetime(df["date"],dayfirst=True)
df['month_year'] = df['date'].dt.to_period('M')
df.drop("date",axis=1,inplace = True)
result = df.groupby(["month_year","concept"]).sum()

Answered By: Ugur Yigit

Group a DataFrame by months and an additional column

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