Pandas - get first n-rows based on percentage
I want to pop first 5% of record
There is no built-in method but you can do this:
You can multiply the total number of rows to your percent and use the result as parameter for head method.
n = 5df.head(int(len(df)*(n/100)))So if your dataframe contains 1000 rows and n = 5% you will get the first 50 rows.
I've extended Mihai's answer for my usage and it may be useful to people out there.The purpose is automated top-n records selection for time series sampling, so you're sure you're taking old records for training and recent records for testing.
# having # import pandas as pd # df = pd.DataFrame... def sample_first_prows(data, perc=0.7): import pandas as pd return data.head(int(len(data)*(perc)))train = sample_first_prows(df)test = df.iloc[max(train.index):]
may be this will help:
tt = tmp.groupby('id').apply(lambda x: x.head(int(len(x)*0.05))).reset_index(drop=True)