Pandas: Difference between largest and smallest value within group

Question

Given a data frame that looks like this

GROUP VALUE
  1     5
  2     2
  1     10
  2     20
  1     7

I would like to compute the difference between the largest and smallest value within each group. That is, the result should be

GROUP   DIFF
  1      5
  2      18

What is an easy way to do this in Pandas?

What is a fast way to do this in Pandas for a data frame with about 2 million rows and 1 million groups?

Answer 1

Using @unutbu 's df

per timing
unutbu's solution is best over large data sets

import pandas as pd
import numpy as np

df = pd.DataFrame({'GROUP': [1, 2, 1, 2, 1], 'VALUE': [5, 2, 10, 20, 7]})

df.groupby('GROUP')['VALUE'].agg(np.ptp)

GROUP
1     5
2    18
Name: VALUE, dtype: int64

---

np.ptp docs returns the range of an array

---

timing
small df

large df
df = pd.DataFrame(dict(GROUP=np.arange(1000000) % 100, VALUE=np.random.rand(1000000)))

large df
many groups
df = pd.DataFrame(dict(GROUP=np.arange(1000000) % 10000, VALUE=np.random.rand(1000000)))

Answer 2

groupby/agg generally performs best when you take advantage of the built-in aggregators such as 'max' and 'min'. So to obtain the difference, first compute the max and min and then subtract:

import pandas as pd
df = pd.DataFrame({'GROUP': [1, 2, 1, 2, 1], 'VALUE': [5, 2, 10, 20, 7]})
result = df.groupby('GROUP')['VALUE'].agg(['max','min'])
result['diff'] = result['max']-result['min']
print(result[['diff']])

yields

       diff
GROUP      
1         5
2        18

Answer 3

Note: this will get the job done, but @piRSquared's answer has faster methods.

You can use groupby(), min(), and max():

df.groupby('GROUP')['VALUE'].apply(lambda g: g.max() - g.min())