The error message:
KeyError: 'missing_key'
occurs when you attempt to access a key in a dictionary that does not exist.
1. Common Causes and Fixes
Cause 1: Accessing a Non-Existent Key
If you try to retrieve a value using a key that is not present in the dictionary, Python raises a KeyError
.
Incorrect Code:
data = {"name": "Alice", "age": 25}
print(data["city"]) # Error: 'city' key does not exist
Solution: Use .get()
to return a default value instead of raising an error.
print(data.get("city", "Key not found")) # Output: "Key not found"
Cause 2: Using a Key with a Typo or Case Difference
Dictionaries in Python are case-sensitive, meaning "Name"
and "name"
are different keys.
Incorrect Code:
data = {"Name": "Alice"}
print(data["name"]) # Error: 'name' is different from 'Name'
Solution: Normalize keys using .lower().strip()
.
key = " name ".strip().lower()
data = {"name": "Alice"} # Convert keys to lowercase
print(data.get(key, "Key not found")) # Output: Alice
Cause 3: Using pop()
on a Missing Key
If you call .pop()
on a key that doesn’t exist, Python raises a KeyError
.
Incorrect Code:
data = {"name": "Alice"}
data.pop("age") # Error: 'age' key does not exist
Solution: Provide a default value to .pop()
.
data.pop("age", "Key not found") # Output: "Key not found"
Cause 4: Iterating Over a List of Dictionaries Without Checking Keys
If you’re looping through a list of dictionaries and assume a key always exists, you may run into a KeyError
.
Incorrect Code:
data = [{"name": "Alice"}, {"age": 25}]
for item in data:
print(item["name"]) # Error: The second dictionary has no 'name' key
Solution: Use .get()
to prevent errors.
for item in data:
print(item.get("name", "Key not found"))