GroupImputer

Fill missing values using the average of similar rows, instead of one flat value for the entire column.

Standard pandas.fillna(mean) fills every missing value with the same number — even if a "Sales" salary and a "Tech" salary have nothing in common. GroupImputer fills each missing value using the average for its own group (e.g. its department), with a safe fallback when a group has no data at all.

Installation

pip install imputekit

Quick Start

import pandas as pd
from imputekit import GroupImputer

df = pd.DataFrame({
    "Department": ["Sales", "Sales", "Sales", "Tech", "Tech", "Tech"],
    "Salary":     [60000,   45000,   None,    50000,  90000,  None],
})

filler = GroupImputer(group_by="Department", target="Salary")
df_filled = filler.fit_transform(df)

print(df_filled)
print(filler.report())

Output:

Imputation Report
-----------------
Target column          : Salary
Grouped by              : Department
Statistic used          : mean
Missing values filled   : 2
  filled by group avg   : 2
  filled by fallback    : 0
Already present values  : 4

How It Works

1. fit() groups the data by the column(s) you specify and computes the mean, median, or mode of the target column for each group.
2. transform() fills each missing value using its group's statistic.
3. If a group has no valid data at all, it falls back to the overall column statistic instead of leaving the value empty.
4. report() tells you exactly how many values were filled, and by which method — so nothing happens silently.

API

GroupImputer(group_by, target, strategy="mean")

• group_by: a column name, or list of column names, to group by
• target: the column containing missing values to fill
• strategy: "mean", "median", or "mode"

Methods: .fit(df), .transform(df), .fit_transform(df), .report()

Why not just use df.groupby(...).transform(...)?

You can! For a single quick fix, plain pandas is fine. GroupImputer is useful when you want the fallback safety net, multiple group columns, and a clear report of what was filled and how — without rewriting that logic every time.

License

MIT