How to Group and Map Data in Pandas: 5 Practical Methods

1. Introduction

A question was posted in a Python community about how to process the following data, where a list of numeric identifiers needs to be mapped to a list of string codes.

num = [1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 2.0, 2.0, 2.0, 2.0, 2.0, 2.0, 2.0, 2.0, 3.0, 3.0, 3.0, 3.0, 3.0, 3.0]

data = ['201825301001', '201825301002', '201825301004', '201825301005', '201825301006', '201825301007', '201825301008', '201825301009', '201825301010', '201825301011', '201825301012', '201825301013', '201825301014', '201825301015', '201825301016', '201825301017', '201825301018', '201825301019', '201825305001', '201825305002']

2. Implementation

Method 1

A suggestion was given to read the data from an Excel file, group by the numeric column, and convert the groups to a dictionary.

import pandas as pd

df = pd.read_excel('1.xlsx', names=['num', 'date'])
df = df.groupby('num').agg(list)
res = df.to_dict()['date']
print(res)

The output matches the expected mapping.

Method 2

Another approach uses a simple loop to build the dictionary.

num=[1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 2.0, 2.0, 2.0, 2.0, 2.0, 2.0, 2.0, 2.0, 3.0, 3.0, 3.0, 3.0, 3.0, 3.0]
num=[int(i) for i in num]
data=['201825301001', '201825301002', '201825301004', '201825301005', '201825301006', '201825301007', '201825301008', '201825301009', '201825301010', '201825301011', '201825301012', '201825301013', '201825301014', '201825301015', '201825301016', '201825301017', '201825301018', '201825301019', '201825305001', '201825305002']
result={}
for k,v in zip(num,data):
    if k in result.keys():
        result.get(k).append(v)
    else:
        result[k]=[v]
print(result)

Method 3

Using itertools.groupby to achieve the same result in a more functional style.

from itertools import groupby
num=[1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 2.0, 2.0, 2.0, 2.0, 2.0, 2.0, 2.0, 2.0, 3.0, 3.0, 3.0, 3.0, 3.0, 3.0]
data=['201825301001', '201825301002', '201825301004', '201825301005', '201825301006', '201825301007', '201825301008', '201825301009', '201825301010', '201825301011', '201825301012', '201825301013', '201825301014', '201825301015', '201825301016', '201825301017', '201825301018', '201825301019', '201825305001', '201825305002']
result = {k: [i[1] for i in v] for k, v in groupby(zip(num, data), key=lambda x: int(x[0]))}
print(result)

Method 4

A variant that builds the dictionary with a single pass.

from itertools import groupby
num=[1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 2.0, 2.0, 2.0, 2.0, 2.0, 2.0, 2.0, 2.0, 3.0, 3.0, 3.0, 3.0, 3.0, 3.0]
data=['201825301001', '201825301002', '201825301004', '201825301005', '201825301006', '201825301007', '201825301008', '201825301009', '201825301010', '201825301011', '201825301012', '201825301013', '201825301014', '201825301015', '201825301016', '201825301017', '201825301018', '201825301019', '201825305001', '201825305002']
result={}
for k,v in zip(num,data):
    result[k]=result.get(k,[])+[v]
result={int(k):result.get(k,[])+[v] for k,v in zip(num,data)}
print(result)

Method 5

Leveraging Pandas DataFrame grouping directly.

num=[1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 2.0, 2.0, 2.0, 2.0, 2.0, 2.0, 2.0, 2.0, 3.0, 3.0, 3.0, 3.0, 3.0, 3.0]
data=['201825301001', '201825301002', '201825301004', '201825301005', '201825301006', '201825301007', '201825301008', '201825301009', '201825301010', '201825301011', '201825301012', '201825301013', '201825301014', '201825301015', '201825301016', '201825301017', '201825301018', '201825301019', '201825305001', '201825305002']
import pandas as pd
df = pd.DataFrame({'num': num, 'data': data})
df = df.groupby('num').agg(list)
res = df.to_dict()['data']
print(res)

3. Summary

The article presents five concrete Pandas‑based solutions for converting parallel lists of numeric keys and string values into a dictionary that groups strings by their numeric key, demonstrating multiple coding styles—from simple loops to groupby and DataFrame aggregation—so readers can choose the approach that best fits their workflow.