Python Pandas DataFrame重塑：合并对应重复列Var3/Var4至Var1/Var2

Alright, let's tackle this DataFrame reshaping task. The goal is to take the paired columns (Var1/Var3, Var2/Var4) and turn them into additional rows under the original Var1/Var2 headers, while keeping the Type associated correctly. Here's a straightforward approach that gets the job done:

Step-by-Step Solution #

First, let's start with your original DataFrame:

import pandas as pd
import numpy as np

df = pd.DataFrame({'Type' : ['A', 'A', 'B'], 'Var1' : [1.0, 2.0, 3.0], 'Var2' : [21.0, 22.0, 23.0], 'Var3' : [np.nan, 4.0, 5.0], 'Var4' : [np.nan, 24.0, 25.0] })

We can break this down into three simple steps:

1. Extract original valid rows: Grab the Type, Var1, and Var2 columns as the first part of our result.
2. Align paired columns to original structure: Take Type, Var3, and Var4, then rename Var3 to Var1 and Var4 to Var2 so they match the first part's format.
3. Combine and clean: Stack the two parts together, drop rows with missing values (these are the invalid entries where both Var3 and Var4 were NaN), and reset the index for a clean output.

Here's the code that puts this all together:

# Get the original rows with Var1/Var2
part1 = df[['Type', 'Var1', 'Var2']]

# Convert Var3/Var4 to match Var1/Var2 column names
part2 = df[['Type', 'Var3', 'Var4']].rename(columns={'Var3': 'Var1', 'Var4': 'Var2'})

# Combine datasets, remove invalid NaN rows, and reset index
result_df = pd.concat([part1, part2]).dropna().reset_index(drop=True)

Output Result #

Running this code will give you exactly the target structure you wanted:

Why This Works #

• pd.concat() stacks the two DataFrames vertically, so we retain all original rows plus the converted Var3/Var4 rows.
• dropna() automatically removes the row where both Var3 and Var4 were NaN—since that row would have missing values in the converted Var1/Var2 columns, which is exactly the entry we don't want to keep.
• reset_index(drop=True) cleans up the index to be sequential, matching your target output's structure.

内容的提问来源于stack exchange，提问作者WZhao