Searching and counting elements in Python - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to count how many times the number 5 appears in the list.

Python

numbers = [1, 5, 3, 5, 7, 5]
count_fives = numbers.[1](5)
print(count_fives)

Drag options to blanks, or click blank then click option'

Aindex

Bcount

Cfind

Dsearch

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'index' instead of 'count' which only finds the first position.

The count method counts how many times a value appears in a list.

2fill in blank

medium

Complete the code to check if the word 'apple' is in the list of fruits.

Python

fruits = ['banana', 'orange', 'apple', 'grape']
if 'apple' [1] fruits:
    print('Found apple!')

Drag options to blanks, or click blank then click option'

Ain

Bhas

Cnot

Dis

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'is' which checks identity, not membership.

The keyword in checks if an item exists in a list.

3fill in blank

hard

Fix the error in the code to find the index of the first occurrence of 7 in the list.

Python

values = [4, 7, 2, 7, 9]
position = values.[1](7)
print(position)

Drag options to blanks, or click blank then click option'

Alocate

Bfind

Csearch

Dindex

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'find' which is not a list method and causes an error.

The index method returns the position of the first matching element in a list.

4fill in blank

hard

Fill both blanks to create a dictionary with words as keys and their lengths as values, but only for words longer than 4 letters.

Python

word_lengths = {word: [1] for word in words if len(word) [2] 4}

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Alen(word)

B<=

C>

Dword

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<=' instead of '>' which selects shorter words.

Use len(word) to get the length, and > to filter words longer than 4 letters.

5fill in blank

hard

Fill all three blanks to create a dictionary with uppercase words as keys and their counts as values, but only for words that appear more than once.

Python

word_counts = { [1] : words.count([2]) for [3] in words if words.count([3]) > 1 }

Drag options to blanks, or click blank then click option'

Aword.upper()

Bword

Ditem

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names inconsistently.

Use word.upper() for keys, count the word itself, and iterate with word.

Practice

1. Which Python operator checks if an element exists in a list?

easy

A. in

B. count()

C. find()

D. exists()

Solution

1. Step 1: Understand the purpose of in

   The in operator checks if an element is present in a list or other collection.

2. Step 2: Compare with other options

   count() counts occurrences, find() and exists() are not valid list operators in Python.

3. Final Answer:

   in -> Option A

4. Quick Check:

   Use in to check membership [OK]

Hint: Use in to check presence quickly [OK]

Common Mistakes:

• Confusing count() with membership check
• Using non-existent methods like find()
• Trying to use exists() which is invalid

2. Which of the following is the correct syntax to count how many times the number 5 appears in list nums?

easy

A. count(nums, 5)

B. nums.count(5)

C. nums.count = 5

D. nums.count[5]

Solution

1. Step 1: Identify the correct method call

   To count occurrences, use the list method count() with the element as argument: nums.count(5).

2. Step 2: Check syntax of other options

   count(nums, 5) is invalid syntax, nums.count = 5 assigns a value incorrectly, and nums.count[5] is invalid indexing.

3. Final Answer:

   nums.count(5) -> Option B

4. Quick Check:

   Use list.count(element) to count [OK]

Hint: Use list.count(value) to count occurrences [OK]

Common Mistakes:

• Using function call syntax incorrectly
• Assigning instead of calling method
• Using square brackets instead of parentheses

3. What is the output of this code?

fruits = ['apple', 'banana', 'apple', 'cherry']
print(fruits.count('apple'))

medium

A. Error

B. 1

C. 3

D. 2

Solution

1. Step 1: Understand the list contents

   The list fruits contains 'apple' twice, 'banana' once, and 'cherry' once.

2. Step 2: Apply count() method

   fruits.count('apple') counts how many times 'apple' appears, which is 2.

3. Final Answer:

   2 -> Option D

4. Quick Check:

   Counting 'apple' in list = 2 [OK]

Hint: Count returns how many times item appears [OK]

Common Mistakes:

• Counting unique items instead of occurrences
• Expecting index instead of count
• Confusing count with length

4. Find the error in this code that tries to count how many times 10 appears in numbers:

numbers = [10, 20, 10, 30]
count = numbers.count[10]
print(count)

medium

A. Using square brackets instead of parentheses for count method

B. Variable name 'count' is reserved and cannot be used

C. List 'numbers' is not defined

D. Missing import for count function

Solution

1. Step 1: Identify method call syntax

   Methods in Python are called with parentheses, not square brackets. numbers.count[10] is invalid syntax.

2. Step 2: Correct the syntax

   It should be numbers.count(10) to count occurrences of 10.

3. Final Answer:

   Using square brackets instead of parentheses for count method -> Option A

4. Quick Check:

   Method calls need parentheses, not brackets [OK]

Hint: Use parentheses () to call methods, not brackets [] [OK]

Common Mistakes:

• Using [] instead of () for method calls
• Thinking count is a function needing import
• Assuming variable names are reserved

5. Given a list data = [0, 1, 2, 0, 3, 0, 4], which code snippet counts how many zeros are in the list and prints a message only if zeros exist?

hard

A. if 0 not in data:
print(f\"Zeros found: {data.count(0)}\")

B. print(f\"Zeros found: {data.count(0)}\")

C. if data.count(0) > 0:
print(f\"Zeros found: {data.count(0)}\")

D. if data.contains(0):
print(f\"Zeros found: {data.count(0)}\")

Solution

1. Step 1: Understand the goal

   We want to count zeros and print only if there is at least one zero.

2. Step 2: Analyze each option

   if 0 not in data:
       print(f\"Zeros found: {data.count(0)}\")

   checks if 0 is NOT in data, printing only when NO zeros -- incorrect.

   print(f\"Zeros found: {data.count(0)}\")

   always prints, even if count is zero.

   if data.count(0) > 0:
       print(f\"Zeros found: {data.count(0)}\")

   checks if count > 0 then prints the count. Correct and efficient.

   if data.contains(0):
       print(f\"Zeros found: {data.count(0)}\")

   uses invalid contains().

3. Step 3: Choose best option

   if data.count(0) > 0:
       print(f\"Zeros found: {data.count(0)}\")

   is correct, checking count once and printing only if zeros exist.

4. Final Answer:

   if data.count(0) > 0: print(f\"Zeros found: {data.count(0)}\") -> Option C

5. Quick Check:

   Check count > 0 before printing [OK]

Hint: Check count > 0 to confirm presence before printing [OK]

Common Mistakes:

• Using invalid method contains()
• Printing count without checking if zero exists
• Using not in which prints when zeros are absent