Python2迁移至Python3遇字典取值错误：'int'对象不可下标访问

Hey there! Let's break down why this error is popping up and fix it for you.

What's Causing the Error? #

The line laP = float(rd[0]) is throwing a TypeError because rd is the integer 0 at that point—not the list you expect. This happens when myDict.get(key.strip(), 0) can't find the key in your dictionary, so it returns the default value 0.

The root issue is a Python 2 vs Python 3 string/bytes difference:

• In your dictionary myDict, all keys are regular Unicode strings (like 'Aabenraa').
• In Python 3, data.kommune.encode("utf-8") converts the string to a bytes object (e.g., b'Aabenraa').
• Python 3 treats str and bytes as completely separate types—so even if the content matches, b'Aabenraa' won't match the 'Aabenraa' key in your dictionary. Hence, get() returns 0.

Step-by-Step Fix #

1. Remove the encode("utf-8") call
   Since Python 3 uses Unicode strings by default, your dictionary keys and data.kommune should already be compatible. Change the key line to:

   key = data.kommune.strip()
   

2. Add a safety check for missing keys
   To avoid similar errors if a kommune isn't in your dictionary, update the code to handle missing keys gracefully:

   myDict = littleUglyDataCollectionInTheSourceCode()
   key = data.kommune.strip()
   rd = myDict.get(key)  # No default value here—returns None if key is missing
   
   if rd is None:
       print(f"Warning: Kommune '{key}' not found in the data!")
       # Handle this case (e.g., return a default value, log an error, etc.)
       return  # Or adjust based on what your function needs to do
   
   laP = float(rd[0])  # average precipitation
   midV = float(rd[1]) # Middelværdi Klimagrid
   

Why This Worked in Python 2 #

In Python 2, strings (str) were essentially byte strings, and Unicode strings were unicode objects. When you called encode("utf-8") on a Python 2 unicode string, you got a str object that matched the keys in your dictionary. Python 3's strict separation of str (Unicode) and bytes breaks this compatibility.

Give these changes a try—your code should now correctly fetch the list from the dictionary and avoid the TypeError!

内容的提问来源于stack exchange，提问作者Rasmus Kierudsen