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 = 5
df.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):]