DataFrame手机号列空格替换求助：将第6位空格替换为数字8

Let's break down what went wrong and how to fix your mobile number issue:

Why Your Original Code Failed #

Your code df["Mobile No"] = df["Mobile No"].str.replace(' ', ' ') does two unhelpful things:

• It replaces spaces with identical spaces—so no actual change happens to your data.
• If your "Mobile No" column wasn't stored as a string type (e.g., it was an integer), using the .str accessor would convert all values to NaN, since numeric types don't support string operations directly.

Step-by-Step Solution #

First, ensure your column is treated as string data to avoid NaN issues:

# Convert the column to string type
df["Mobile No"] = df["Mobile No"].astype(str)

Then, use one of these methods to replace the 6th-position space with an 8:

Method 1: Regular Expression (Clean & Concise)

This regex targets exactly the pattern you described: 5 characters, followed by a space, then 4 characters. It replaces the space with an 8:

df["Mobile No"] = df["Mobile No"].str.replace(r'^(.{5}) (.{4})$', r'\18\2', regex=True)

• ^(.{5}): Captures the first 5 characters
• : Matches the space at the 6th position
• (.{4})$: Captures the last 4 characters
• \18\2: Replaces the match with the first captured group, an 8, then the second captured group

Method 2: Custom Function (More Explicit)

If you prefer clearer control over the logic, use apply() with a function that checks for the space and fixes it:

def repair_mobile_number(phone):
    # Check if the number has a space at the 6th position (index 5, 0-based)
    if len(phone) == 10 and phone[5] == ' ':
        return phone[:5] + '8' + phone[6:]
    # Return unchanged if no space found
    return phone

df["Mobile No"] = df["Mobile No"].apply(repair_mobile_number)

Example Test #

For your sample number '88888 8888', both methods will convert it to '8888888888' as expected.

内容的提问来源于stack exchange，提问作者Optimizer