Removing dictionary entries in Python - Mini Project: Build & Apply

Removing dictionary entries

📖 Scenario: You are managing a small store's inventory. Some products are out of stock and need to be removed from the inventory list.

🎯 Goal: Build a program that removes products with zero stock from the inventory dictionary.

📋 What You'll Learn

Create a dictionary called inventory with product names as keys and stock counts as values.

Create a list called out_of_stock to hold products with zero stock.

Use a for loop to find products with zero stock and add them to out_of_stock.

Remove all products in out_of_stock from inventory.

Print the updated inventory dictionary.

💡 Why This Matters

🌍 Real World

Managing inventory is common in stores and warehouses to keep track of available products and remove items that are not currently sellable.

💼 Career

Knowing how to manipulate dictionaries and remove entries based on conditions is useful for data cleaning and management tasks in many programming jobs.

1

Create the inventory dictionary

Create a dictionary called inventory with these exact entries: 'apples': 10, 'bananas': 0, 'oranges': 5, 'pears': 0, 'grapes': 8.

Python

# Create the inventory dictionary with given products and stock counts
# Your code here

Hint

Use curly braces {} to create the dictionary with the exact keys and values.

2

Create the out_of_stock list

Create an empty list called out_of_stock to store products with zero stock.

Python

inventory = {'apples': 10, 'bananas': 0, 'oranges': 5, 'pears': 0, 'grapes': 8}
# Create an empty list called out_of_stock
# Your code here

Hint

Use square brackets [] to create an empty list.

3

Find and collect out-of-stock products

Use a for loop with variables product and stock to iterate over inventory.items(). Inside the loop, if stock is 0, add product to the out_of_stock list.

Python

inventory = {'apples': 10, 'bananas': 0, 'oranges': 5, 'pears': 0, 'grapes': 8}
out_of_stock = []
# Use a for loop to find products with zero stock and add them to out_of_stock
# Your code here

Hint

Use inventory.items() to get product and stock pairs. Use append() to add to the list.

4

Remove out-of-stock products and print inventory

Use a for loop to iterate over out_of_stock and remove each product from inventory using del. Then print the updated inventory dictionary.

Python

inventory = {'apples': 10, 'bananas': 0, 'oranges': 5, 'pears': 0, 'grapes': 8}
out_of_stock = []
for product, stock in inventory.items():
    if stock == 0:
        out_of_stock.append(product)
# Remove products in out_of_stock from inventory
# Print the updated inventory
# Your code here

Hint

Use del inventory[product] to remove a key from the dictionary.

Practice

1. What does the pop(key) method do when used on a Python dictionary?

easy

A. Removes the key and returns its value

B. Removes the key without returning its value

C. Returns the value without removing the key

D. Deletes the entire dictionary

Solution

1. Step 1: Understand the pop method

   The pop(key) method removes the specified key from the dictionary and returns its value.

2. Step 2: Compare with other methods

   Unlike del dict[key], pop returns the removed value.

3. Final Answer:

   Removes the key and returns its value -> Option A

4. Quick Check:

   pop(key) = remove key + return value [OK]

Hint: pop returns value and removes key from dict [OK]

Common Mistakes:

• Thinking pop only returns value without removal
• Confusing pop with del which doesn't return value
• Assuming pop deletes entire dictionary

2. Which of the following is the correct syntax to remove a key 'age' from dictionary person without returning its value?

easy

A. person.delete('age')

B. person.pop('age')

C. person.remove('age')

D. del person['age']

Solution

1. Step 1: Identify syntax for deletion without return

   To remove a key without returning its value, use del dict[key].

2. Step 2: Check options

   del person['age'] uses del person['age'], which is correct syntax. Options C and D are invalid methods.

3. Final Answer:

   del person['age'] -> Option D

4. Quick Check:

   del dict[key] removes key without return [OK]

Hint: Use del dict[key] to remove key without return [OK]

Common Mistakes:

• Using pop when no return is needed
• Using non-existent methods like remove or delete
• Wrong syntax like person.delete('age')

3. What is the output of this code?

data = {'a': 1, 'b': 2, 'c': 3}
value = data.pop('b')
print(value, data)

medium

A. 2 {'a': 1, 'b': 2, 'c': 3}

B. None {'a': 1, 'c': 3}

C. 2 {'a': 1, 'c': 3}

D. KeyError

Solution

1. Step 1: Use pop to remove key 'b'

   The pop('b') removes key 'b' and returns its value 2.

2. Step 2: Print returned value and updated dictionary

   After removal, dictionary is {'a': 1, 'c': 3}. Printing value and dict shows: 2 {'a': 1, 'c': 3}.

3. Final Answer:

   2 {'a': 1, 'c': 3} -> Option C

4. Quick Check:

   pop returns value and removes key [OK]

Hint: pop returns value and removes key from dict [OK]

Common Mistakes:

• Expecting original dict unchanged
• Thinking pop returns None
• Confusing pop with del which returns nothing

4. What error will this code produce?

info = {'name': 'Alice', 'age': 30}
info.pop('gender')

medium

A. None

B. KeyError

C. Returns 'gender'

D. SyntaxError

Solution

1. Step 1: Understand pop behavior without default

   Calling pop on a missing key without a default raises a KeyError.

2. Step 2: Check code

   Since 'gender' is not in info, info.pop('gender') raises KeyError.

3. Final Answer:

   KeyError -> Option B

4. Quick Check:

   pop missing key without default = KeyError [OK]

Hint: pop missing key without default causes KeyError [OK]

Common Mistakes:

• Expecting None instead of error
• Thinking pop returns the key name
• Confusing with pop(key, default)

5. Given items = {'apple': 5, 'banana': 0, 'cherry': 7}, which code removes all entries with value 0 or less safely without errors?

hard

A. for k in list(items):
if items[k] <= 0:
items.pop(k)

B. for k in items:
if items[k] <= 0:
del items[k]

C. items.pop('banana')

D. del items['banana']

Solution

1. Step 1: Avoid modifying dict while iterating

   Directly deleting keys while iterating causes runtime errors. Using list(items) copies keys safely.

2. Step 2: Use pop to remove keys with value ≤ 0

   Loop over copied keys, check value, and use pop to remove safely.

3. Final Answer:

   for k in list(items): if items[k] <= 0: items.pop(k) -> Option A

4. Quick Check:

   Iterate on list copy + pop keys safely [OK]

Hint: Iterate on list(items) to safely remove keys while looping [OK]

Common Mistakes:

• Deleting keys directly while iterating causes error
• Using pop without copying keys list
• Removing only one key instead of all matching