KeyError: dictionary lookup failure

A KeyError: dictionary lookup failure occurs when you try to access a key in a dictionary that does not exist. This can happen due to incorrect key names, case sensitivity, or missing data.

---

1. Common Causes and Fixes

Cause 1: Accessing a Non-Existent Key

If you try to access a key that isn’t 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 avoid the error.

print(data.get("city", "Key not found"))  # Output: Key not found

---

Cause 2: Case Sensitivity in Dictionary Keys

Dictionary keys in Python are case-sensitive, meaning "Name" and "name" are different.

Incorrect Code:

data = {"Name": "Alice"}
print(data["name"])  # Error: 'name' key does not match 'Name'

Solution: Convert keys to lowercase for consistency.

print(data.get("name".lower(), "Key not found"))  # Output: Key not found

---

Cause 3: Accessing Keys in a Nested Dictionary Without Proper Checking

If a key exists at a deeper level, directly trying to access it can fail.

Incorrect Code:

data = {"user": {"name": "Alice"}}
print(data["user"]["age"])  # Error: 'age' key does not exist

Solution: Use .get() for nested dictionary access.

print(data["user"].get("age", "Key not found"))  # Output: Key not found

---

Cause 4: Accessing a Key After Deleting It

If a key is removed from the dictionary and accessed later, it causes a KeyError.

Incorrect Code:

data = {"name": "Alice", "age": 25}
del data["age"]
print(data["age"])  # Error: 'age' key was deleted

Solution: Check if the key exists before accessing it.

if "age" in data:
    print(data["age"])
else:
    print("Key not found")  # Output: Key not found

---

Cause 5: Iterating Over a Dictionary and Accessing Missing Keys

When iterating, if you try to access a missing key, it raises a KeyError.

Incorrect Code:

data = {"a": 1, "b": 2}
for key in ["a", "b", "c"]:
    print(data[key])  # Error: 'c' does not exist

Solution: Use .get() to safely access keys.

for key in ["a", "b", "c"]:
    print(data.get(key, "Key not found"))