Python中inplace参数使用及DataFrame数据清洗报错问题咨询

Hey Jason, let's work through these two Pandas issues you're facing—they're super common pitfalls, so don't sweat it!

The issue here is that when you run df.iloc[:, 0:8].replace('???', np.nan), Pandas returns a copy of that sliced DataFrame with the changes applied, but it doesn't modify your original df unless you explicitly tell it to. Here are two solid fixes:

• Option 1: Use the inplace=True parameter
  Add inplace=True directly to the replace call to make changes to the sliced subset in place, which updates the original DataFrame:

  df.iloc[:, 0:8].replace('???', np.nan, inplace=True)
  

  A quick heads-up: Sometimes chained indexing + inplace can trigger warnings about setting on a copy. If you run into that, the second option is more reliable.

• Option 2: Assign the modified subset back to the original DataFrame
  This is the clearer, safer approach for most cases—you explicitly overwrite the target columns with the cleaned data:

  df.iloc[:, 0:8] = df.iloc[:, 0:8].replace('???', np.nan)
  

  Now when you check df, you'll see the NaNs instead of "???".

Let's break down that error: TypeError: ('replace() argument 2 must be str, not float', 'occurred at index Transport')

The problem is that once some elements get converted to np.nan (which is a float), your cleaning function tries to call replace() on those float values—and floats don't have a replace() method. Here's how to fix this:

Quick fix for your function #

Update your cleaning function to only run replace() on string elements. Non-string elements (like existing NaNs) get returned as-is:

import numpy as np

def cleaning(x):
    if isinstance(x, str):
        return x.replace("???", np.nan)
    return x

# Apply the function and assign back to the target columns
df.iloc[:, 0:8] = df.iloc[:, 0:8].applymap(cleaning)

Even better: Ditch applymap entirely #

applymap is slow for large datasets because it iterates over every single element. Since you're just replacing a specific string, using Pandas' built-in replace() is way more efficient (and avoids the type error altogether):

df.iloc[:, 0:8] = df.iloc[:, 0:8].replace('???', np.nan)

This handles all string elements matching "???" automatically, and leaves non-string values (like NaNs) untouched.

内容的提问来源于stack exchange，提问作者Jason