Pandas · groupby

groupby follows the split-apply-combine pattern: split the rows into groups by key, apply a function to each group, and combine the results.

• df.groupby("branch")["cgpa"].mean(): one number per group.
• Named aggregation: .agg(avg=("cgpa", "mean"), n=("cgpa", "size")) gives clean column names.
• Several keys: groupby(["branch", "year"]) produces a MultiIndex.
• transform returns a result the same length as the input, e.g. each row’s deviation from its group mean.
• filter keeps whole groups that satisfy a condition.

import pandas as pd

df = pd.DataFrame({
    "branch": ["ICT", "ICT", "CSE", "CSE", "ME", "ICT", "ME"],
    "year":   [1, 2, 1, 2, 1, 1, 2],
    "cgpa":   [8.9, 7.1, 8.4, 7.9, 6.8, 9.2, 7.5],
})
print(df.groupby("branch")["cgpa"].mean().round(2))
print(df.groupby("branch").agg(avg=("cgpa", "mean"), best=("cgpa", "max"), n=("cgpa", "size")).round(2))
print(df.groupby(["branch", "year"])["cgpa"].mean())

df["vs_branch_avg"] = (df["cgpa"] - df.groupby("branch")["cgpa"].transform("mean")).round(2)
print(df)
print(df.groupby("branch").filter(lambda g: len(g) >= 3)["branch"].unique())

Output:

branch
CSE    8.15
ICT    8.40
ME     7.15
Name: cgpa, dtype: float64
         avg  best  n
branch               
CSE     8.15   8.4  2
ICT     8.40   9.2  3
ME      7.15   7.5  2
branch  year
CSE     1       8.40
        2       7.90
ICT     1       9.05
        2       7.10
ME      1       6.80
        2       7.50
Name: cgpa, dtype: float64
  branch  year  cgpa  vs_branch_avg
0    ICT     1   8.9           0.50
1    ICT     2   7.1          -1.30
2    CSE     1   8.4           0.25
3    CSE     2   7.9          -0.25
4     ME     1   6.8          -0.35
5    ICT     1   9.2           0.80
6     ME     2   7.5           0.35
<StringArray>
['ICT']
Length: 1, dtype: str