I have seen a variant of this question asked that keeps the top n rows of each group in a pandas dataframe and the solutions use n as an absolute number rather than a percentage here Pandas get topmost n records within each group. However, in my dataframe, each group has different numbers of rows in it and I want to keep the top n% rows of each group. How would I approach this problem?
-
By "top n%" I assume you mean "first n%". Is that correct? – jpp Nov 17 '18 at 22:42
-
@jpp Yes. Though I do need to get the next n% for the next dataframe and keep doing this 100/n times. – camelCase Nov 18 '18 at 03:13
2 Answers
You can construct a Boolean series of flags and filter before you groupby
. First let's create an example dataframe and look at the number of row for each unique value in the first series:
np.random.seed(0)
df = pd.DataFrame(np.random.randint(0, 2, (10, 3)))
print(df[0].value_counts())
0 6
1 4
Name: 0, dtype: int64
Then define a fraction, e.g. 50% below, and construct a Boolean series for filtering:
n = 0.5
g = df.groupby(0)
flags = (g.cumcount() + 1) <= g[1].transform('size') * n
Then apply the condition, set the index as the first series and (if required) sort the index:
df = df.loc[flags].set_index(0).sort_index()
print(df)
1 2
0
0 1 1
0 1 1
0 1 0
1 1 1
1 1 0
As you can see, the resultant dataframe only has 3 0
indices and 2 1
indices, in each case half the number in the original dataframe.

- 159,742
- 34
- 281
- 339
Here is another option which builds on some of the answers in the post you mentioned
First of all here is a quick function to either round up or round down. If we want the top 30% of rows of a dataframe 8 rows long then we would try to take 2.4 rows. So we will need to either round up or down.
My preferred option is to round up. This is because, for eaxample, if we were to take 50% of the rows, but had one group which only had one row, we would still keep that one row. I kept this separate so that you can change the rounding as you wish
def round_func(x, up=True):
'''Function to round up or round down a float'''
if up:
return int(x+1)
else:
return int(x)
Next I make a dataframe to work with and set a parameter p
to be the fraction of the rows from each group that we should keep. Everything follows and I have commented it so that hopefully you can follow.
import pandas as pd
df = pd.DataFrame({'id':[1,1,1,2,2,2,2,3,4],'value':[1,2,3,1,2,3,4,1,1]})
p = 0.30 # top fraction to keep. Currently set to 80%
df_top = df.groupby('id').apply( # group by the ids
lambda x: x.reset_index()['value'].nlargest( # in each group take the top rows by column 'value'
round_func(x.count().max()*p))) # calculate how many to keep from each group
df_top = df_top.reset_index().drop('level_1', axis=1) # make the dataframe nice again
df looked like this
id value
0 1 1
1 1 2
2 1 3
3 2 1
4 2 2
5 2 3
6 2 4
7 3 1
8 4 1
df_top looks like this
id value
0 1 3
1 2 4
2 2 3
3 3 1
4 4 1

- 322
- 2
- 8
-
1Here I have assumed you wanted the 100*p% largest values. If you just wanted the first (going down the dataframe) then jpp's answer is better. – James Fulton Nov 17 '18 at 23:25
-
thanks for the great answer! however when the dataframe is more than 2 columns, how to use your solution to display all other columns (from the original df)? – Mark K Mar 06 '23 at 15:10