Python dictionaries are case-sensitive, meaning "Name" and "name" are treated as different keys. If you try to access a key using the wrong case, you’ll get a KeyError.
1. Causes and Fixes
Cause 1: Mismatched Case When Accessing a Dictionary Key
If the key in the dictionary is capitalized differently than how you’re trying to access it, Python raises a KeyError.
Incorrect Code:
data = {"Name": "Alice"}
print(data["name"]) # Error: 'name' is different from 'Name'
Solution: Use .get() to avoid errors.
print(data.get("name", "Key not found")) # Output: "Key not found"
Solution: Convert keys to lowercase for consistency.
data = {"name": "Alice"} # Store keys in lowercase
print(data.get("name", "Key not found")) # Output: Alice
Cause 2: Inconsistent Key Cases in a Loop
If dictionary keys are mixed case, trying to access them without standardizing their format can cause errors.
Incorrect Code:
data = {"Name": "Alice", "Age": 25}
for key in ["name", "age"]:
print(data[key]) # Error: 'name' key does not match 'Name'
Solution: Convert dictionary keys to lowercase before accessing them.
data = {k.lower(): v for k, v in data.items()}
for key in ["name", "age"]:
print(data.get(key, "Key not found"))
Cause 3: Using .pop() or .setdefault() with Case Issues
If you use .pop() or .setdefault() with the wrong case, you might get a KeyError.
Incorrect Code:
data = {"Name": "Alice"}
data.pop("name") # Error: 'name' does not match 'Name'
Solution: Convert keys to lowercase before using them.
data = {"Name": "Alice"}
data_lower = {k.lower(): v for k, v in data.items()}
print(data_lower.pop("name", "Key not found")) # Output: Alice