Top N rows vs rows with Top N unique values using pandas-CodePudding

I have a pandas dataframe like as shown below

import pandas as pd

data ={'Count':[1,1,2,3,4,2,1,1,2,1,3,1,3,6,1,1,9,3,3,6,1,5,2,2,0,2,2,4,0,1,3,2,5,0,3,3,1,2,2,1,6,2,3,4,1,1,3,3,4,3,1,1,4,2,3,0,2,2,3,1,3,6,1,8,4,5,4,2,1,4,1,1,1,2,3,4,1,1,1,3,2,0,6,2,3,2,9,10,2,1,2,3,1,2,2,3,2,1,8,4,0,3,3,5,12,1,5,13,6,13,7,3,5,2,3,3,1,1,5,15,7,9,1,1,1,2,2,2,4,3,3,2,4,1,2,9,3,1,3,0,0,4,0,1,0,1,0]}

df = pd.DataFrame(data)

I would like to do the below

a) Top 5 rows (this will return only 5 rows)

b) Rows with Top 5 unique values (this can return N > 5 rows if the top 5 values are repeating). See my example screenshot below where we have 8 rows for selecting top 5 unique values

While am able to get Top 5 rows by using the below

df.nlargest(5,['Count'])

However, when I try the below for b), I don't get the expected output

df.nlargest(5,['Count'],keep='all')

I expect my output to be like as below

CodePudding user response：

Are you after top 5 unique values or largest top five values?

df =(df.assign(top5rows=np.where(df.index.isin(df.head(5).index),'Y','N'),
              top5unique=np.where(df.index.isin(df.drop_duplicates(keep='first').head(5).index), 'Y','N')))

or did you need

df =(df.assign(top5rows=np.where(df.index.isin(df.head(5).index),'Y','N'),
              top5unique=np.where(df['Count'].isin(list(df['Count'].unique()[:5])),'Y','N')))

    Count top5rows top5unique
0       1        Y          Y
1       1        Y          Y
2       2        Y          Y
3       3        Y          Y
4       4        Y          Y
5       2        N          Y
6       1        N          Y
7       1        N          Y
8       2        N          Y
9       1        N          Y
10      3        N          Y
11      1        N          Y
12      3        N          Y
13      6        N          Y
14      1        N          Y