23 июн. 2013 г. · I'm trying to figure out how to group the data by key1 and sum only the data1 values where key2 equals 'one'. Here's what I've tried def f(d,a,b): ...

14 нояб. 2021 г. · We can use Series.where to replace the values that don't match the condition with NaN , then just groupby transform 'sum' since NaN values ...

29 сент. 2022 г. · The exact expected output is unclear, but assuming you want a new column summing the Sell-Buy per Symbol: # ensure we have numeric values ...

20 июл. 2022 г. · I just realized the simple answer: df["Result_"] = np.where(np.isnan(df["units"]), np.nan, df.groupby(["article", ...

15 мая 2022 г. · This is because df[['column1', 'column2']] returns a dataframe which is a two dimensional object. How to fix the error? Hard way: Pass each of ...

13 июн. 2017 г. · I want to select those rows from df that contain an 'ID' for which 'x' is 1 at least twice. For every 'ID' I manage to count the number of times ...

7 июл. 2020 г. · Idea is replace non matched values to 0 in Series.where and then is used GroupBy.transform with sum : mask = df['Num'].isin([17,12]) ...

16 февр. 2023 г. · My actual dataset is large (e.g., about 20,000 IDs and each ID has 240 time periods), and the above code takes long to run. I would appreciate ...

2 мая 2022 г. · One option is a double merge and a groupby: date = pd.Series(df.Date1.unique(), name='Date') step1 = df.merge(date, left_on = 'Date2', ...

24 дек. 2018 г. · If you want to create a Function: def my_agg(x): names = { 'Total_Count': x['Type'].count(), 'Total_Number': x['Number'].sum(), ...