如何利用pandas DataFrame.rolling()实现特定条件的单元格标记？

数据老陈

2026-5-25

Solution Using Pandas Rolling for Row-Wise Window Checks

Got it, let's work through this problem exactly as you asked—using rolling() to mark the last element of 3-consecutive-cell windows where at least 2 values are greater than 10. Here's a step-by-step breakdown with code:

Step 1: Core Logic Overview

We need to:

First flag which cells have values greater than 10 (convert to a boolean matrix)
Slide a 3-cell window horizontally across each row
Check if the window contains 2 or more flagged values
Mark the last position of qualifying windows as True, all other positions as False

Step 2: Full Code Implementation

import pandas as pd

# Your original DataFrame
df = pd.DataFrame([[1,2,3,4,5],[6,7,8,9,10],[11,12,13,14,15],[16,17,18,19,20],[21,22,23,24,25]])

# 1. Create a boolean DataFrame where True = value > 10
gt_10 = df > 10

# 2. Use rolling(window=3) on rows, count True values per window, check if >=2
#    Fill NaNs (from incomplete start-of-row windows) with False
df1 = gt_10.rolling(window=3, axis=1).sum().ge(2).fillna(False)

Step 3: Verify the Result

If you print df1, you'll get this output that matches your requirements:

0      1      2      3      4
0  False  False  False  False  False
1  False  False  False  False  False
2  False  False   True   True   True
3  False  False   True   True   True
4  False  False   True   True   True

How It Works

gt_10 = df > 10: Converts your numeric DataFrame into a boolean matrix where each cell is True if the value exceeds 10.
rolling(window=3, axis=1): Slides a 3-cell window horizontally across each row (axis=1 specifies row-wise operation).
.sum(): Counts the number of True values in each window (since True is treated as 1 in numeric calculations).
.ge(2): Converts the sum result to True if the count is 2 or more, otherwise False.
.fillna(False): The first two positions in each row don't have a full 3-cell window, so they return NaN—we replace these with False because they can't meet the condition.