Python中多列出生日期（DOB）比对并返回重复值的简化实现方案咨询

Great question! Your current np.select approach works perfectly for 3 DOB columns, but it quickly becomes unwieldy if you ever need to add more columns. Here's a flexible, scalable method that works no matter how many DOB columns you have:

Step 1: Define your target columns #

First, we'll identify all the DOB columns in your DataFrame (adjust the matching logic if your column names follow a different pattern):

import pandas as pd

# Identify all columns starting with "DOB_"
dob_cols = [col for col in df.columns if col.startswith('DOB_')]

Step 2: Create a row-wise processing function #

We'll write a function that checks each row to find values that appear at least twice, ignoring invalid entries like NaN or "Did not report":

def get_valid_dob(row):
    # Filter out nulls and non-report entries
    valid_dobs = [val for val in row[dob_cols] if pd.notna(val) and val != "Did not report"]
    # Count how often each valid DOB appears
    dob_counts = pd.Series(valid_dobs).value_counts()
    # Grab the first value that occurs 2+ times (if any)
    frequent_dobs = dob_counts[dob_counts >= 2]
    return frequent_dobs.index[0] if not frequent_dobs.empty else "To check manually"

Step 3: Apply the function to your DataFrame #

Finally, we'll apply this function to every row to create the DOB_Check column:

df['DOB_Check'] = df.apply(get_valid_dob, axis=1)

Why this works better #

• Scalability: Add as many DOB columns as you want—no need to update conditional logic.
• Robustness: Automatically handles NaN and "Did not report" entries without extra conditionals.
• Readability: The logic is straightforward, making it easier to modify later (e.g., if you need to adjust what counts as an invalid entry).

Verification #

Running this code on your sample dataset will produce exactly the output you're expecting:

内容的提问来源于stack exchange，提问作者Melon123