Keep percentage of rows of dataframe based on column value

Question

Let's suppose that I have a dataframe like that:

import pandas as pd
df = pd.DataFrame({'id':['A','A', 'A', 'B','B'], 'value':[2, 4, 6, 3, 4]})

I want to filter this only for id = A and keep an x percentage of the rows having id = A .

For example if x=60% then the dataframe should look like that:

  col1  col2
0    A     2
1    A     4
2    B     3
2    B     4

How can I do this efficiently in pandas ?

Just to clarify that it is not necessary that all the id =A rows are the one after each other.

Answer 1

One way is using iloc[] with pd.concat

x = 0.6
cond = df['id'].eq('A')
out = pd.concat((df[cond].iloc[:int(round(df['id'].eq('A').sum() * x))],
                 df[~cond]),sort=False).sort_index()

---

  id  value
0  A      2
1  A      4
3  B      3
4  B      4

Answer 2

You can use df.sample to achieve that easily

ids = ['A']
frac = 0.6
df.groupby('id', group_keys=False).apply(lambda x: x.sample(frac=frac) 
                                                   if x.name in ids else x)

Out:

    id  value
1   A   4
0   A   2
3   B   3
4   B   4