Let's break down what's going wrong here and fix it step by step. Your line of code:

ID=int(os.path.split(imagePath[-1]).split('.')[1])

throws this error because os.path.split() returns a tuple, not a string. When you pass imagePath[-1] to os.path.split(), it spits out a pair like ('/path/to/your/images', 'face.123.jpg') — a tuple containing the directory path and the filename. Trying to call .split('.') on this tuple will fail, since tuples don't have that method.

Here are a few straightforward fixes depending on what imagePath[-1] actually contains:

1. If imagePath[-1] is just the filename (no directory path) #

Skip os.path.split() entirely and split the filename string directly:

# Example: imagePath[-1] = "user.456.png"
ID = int(imagePath[-1].split('.')[1])

2. If imagePath[-1] is a full path (directory + filename) #

First extract just the filename using os.path.basename(), then split it:

# Example: imagePath[-1] = "/home/datasets/face.789.jpg"
filename = os.path.basename(imagePath[-1])
ID = int(filename.split('.')[1])

3. A more robust way (handles extensions cleanly) #

If you want to avoid accidentally splitting on dots in the filename itself (unlikely for face dataset IDs, but good practice), use os.path.splitext() to separate the filename from its extension first:

filename = os.path.basename(imagePath[-1])  # Get just the filename if needed
name_without_ext = os.path.splitext(filename)[0]  # e.g., "face.123" from "face.123.jpg"
ID = int(name_without_ext.split('.')[1])

All these approaches ensure you're calling .split('.') on a string, not a tuple, so the error should disappear.

内容的提问来源于stack exchange，提问作者Vinayak Pattar