基于pandas python中分组的第n个十分位数的虚拟

Question

I have a pandas dataframe like this:

import pandas as pd
df = {'Person' : ['A', 'A', 'A', 'B', 'B', 'B', 'B', 'C', 'C', 'C', 'C', 'C', 'C', 'D', 'D', 'D', 'E', 'E', 'E', 'E', 'F', 'F', 'F', 'F', 'G', 'G', 'G', 'G', 'G', 'H', 'H', 'H', 'H', 'I', 'I', 'I', 'I', 'I', 'J', 'J', 'J', 'J', 'J', 'J', 'K', 'K', 'K', 'K', 'K', 'L', 'L','L'],
      'Score' : [18, 17, 15, 10, 11, 12, 15, 15, 16, 16, 16, 15, 18, 10, 12, 12, 8, 7, 10, 9, 5, 4, 2, 4, 10, 12, 11, 12, 10, 3, 1, 5, 6, 18, 19, 20, 16, 19, 10, 12, 11, 13, 10, 12, 20, 20, 20, 19, 19, 7, 12, 15], 
      'Group' : [1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 1, 1, 1, 1, 3, 3, 3, 3, 3, 3, 3, 3, 3, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 1, 1, 1]}
df = pd.DataFrame(df, columns = ['Person', 'Group', 'Score', 'Dummy'])
df

Therefore, I would like to create a dummy that takes the value of 1 when an individual score is higher than or equal to the 8th decile of the group, and zero otherwise. For instance, I can calculate the decile per group using:

df.groupby("Group")["Score"].quantile(0.8)

Group
1    15.0
2    19.2
3    12.0
Name: Score, dtype: float64

I want to create a new dummy variable that takes the value of 1 when the score of group 1 is higher than or equal to 15.0, the score of group 2 is higher than or equal to 19.2, and the score of group 3 is higher than or equal to 12.0, and zero otherwise.

The outcome variable would therefore look like this:

df = {'Person' : ['A', 'A', 'A', 'B', 'B', 'B', 'B', 'C', 'C', 'C', 'C', 'C', 'C', 'D', 'D', 'D', 'E', 'E', 'E', 'E', 'F', 'F', 'F', 'F', 'G', 'G', 'G', 'G', 'G', 'H', 'H', 'H', 'H', 'I', 'I', 'I', 'I', 'I', 'J', 'J', 'J', 'J', 'J', 'J', 'K', 'K', 'K', 'K', 'K', 'L', 'L','L'],
      'Score' : [18, 17, 15, 10, 11, 12, 15, 15, 16, 16, 16, 15, 18, 10, 12, 12, 8, 7, 10, 9, 5, 4, 2, 4, 10, 12, 11, 12, 10, 3, 1, 5, 6, 18, 19, 20, 16, 19, 10, 12, 11, 13, 10, 12, 20, 20, 20, 19, 19, 7, 12, 15], 
      'Group' : [1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 1, 1, 1, 1, 3, 3, 3, 3, 3, 3, 3, 3, 3, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 1, 1, 1], 
      'Dummy' : [1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1]}
df = pd.DataFrame(df, columns = ['Person', 'Group', 'Score', 'Dummy'])
df

What would be the most direct way to do this?

Answer 1

This is just a map:

quantiles = df.groupby("Group")["Score"].quantile(0.8)

df['Dummy'] = (df['Score'] >= df['Group'].map(quantiles)).astype(int)

Output (head):

   Person  Group  Score  Dummy
0       A      1     18      1
1       A      1     17      1
2       A      1     15      1
3       B      2     10      0
4       B      2     11      0
5       B      2     12      0
6       B      2     15      0
7       C      2     15      0
8       C      2     16      0
9       C      2     16      0
10      C      2     16      0

Answer 2

We can use transform here, to cast the quantile to each row:

q = df.groupby("Group")["Score"].transform('quantile', q=0.8)
df['Dummy'] = df['Score'].ge(q).astype(int)

print(df.head(10))

  Person  Group  Score  Dummy
0      A      1     18      1
1      A      1     17      1
2      A      1     15      1
3      B      2     10      0
4      B      2     11      0
5      B      2     12      0
6      B      2     15      0
7      C      2     15      0
8      C      2     16      0
9      C      2     16      0