Random data generation in Python - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of random integer generation

What is the output of this code snippet?

Python

import random
random.seed(10)
print(random.randint(1, 10))

A 5

B 7

C 1

D 10

💡 Hint

The seed fixes the random number sequence.

With seed 10, random.randint(1, 10) returns 7.

❓ Predict Output

intermediate

Output of random choice from list

What is the output of this code snippet?

Python

import random
random.seed(3)
colors = ['red', 'green', 'blue']
print(random.choice(colors))

A red

B blue

C green

D IndexError

💡 Hint

Seed controls the choice result.

With seed 3, random.choice picks 'red' from the list.

❓ Predict Output

advanced

Output of random.sample with repeated elements

What is the output of this code snippet?

Python

import random
random.seed(1)
items = [1, 2, 2, 3, 4]
sample = random.sample(items, 3)
print(sample)

A ValueError

B [2, 1, 4]

C [2, 3, 4]

D [1, 2, 3]

💡 Hint

random.sample picks unique indices, not unique values.

random.sample picks 3 unique elements by index; duplicates in list are allowed.

❓ Predict Output

advanced

Output of random.choices with weights

What is the output of this code snippet?

Python

import random
random.seed(0)
items = ['a', 'b', 'c']
weights = [10, 1, 1]
result = random.choices(items, weights=weights, k=5)
print(result)

A ['a', 'a', 'b', 'a', 'a']

B ['a', 'a', 'a', 'a', 'a']

C ['b', 'a', 'a', 'c', 'a']

D TypeError

💡 Hint

Weights influence probability but do not guarantee all same picks.

With seed 0 and weights, random.choices returns ['a', 'a', 'b', 'a', 'a'].

🧠 Conceptual

expert

Effect of random.seed on random.random() sequence

Which option correctly describes the output of this code snippet?

Python

import random
random.seed(42)
values = [round(random.random(), 3) for _ in range(4)]
print(values)

A [0.639, 0.025, 0.275, 0.736]

B [0.639, 0.025, 0.275, 0.223, 0.736]

C [0.639, 0.025, 0.275, 0.223]

D TypeError

💡 Hint

random.seed fixes the sequence of random.random() calls.

With seed 42, the first four random.random() rounded to 3 decimals are [0.639, 0.025, 0.275, 0.223].

Practice

1. What does the random.randint(a, b) function do in Python?

easy

A. Returns a random float between a and b

B. Returns a random integer N such that a ≤ N ≤ b

C. Returns a random element from a list

D. Shuffles the elements of a list in place

Solution

1. Step 1: Understand the function purpose

   random.randint(a, b) generates a random integer between two given numbers a and b, inclusive.

2. Step 2: Compare options with function behavior

   Returns a random integer N such that a ≤ N ≤ b correctly describes this behavior. Options A, C, and D describe other functions like random.uniform, random.choice, and random.shuffle.

3. Final Answer:

   Returns a random integer N such that a ≤ N ≤ b -> Option B

4. Quick Check:

   random.randint = random integer in range [OK]

Hint: randint returns integers between two numbers inclusive [OK]

Common Mistakes:

• Confusing randint with random float functions
• Thinking randint returns a list element
• Mixing up randint with shuffle

2. Which of the following is the correct way to import the random module and use choice to pick a random element from a list items?

easy

A. import random; random.choice(items)

B. from random import randint; choice(items)

C. import random.choice; choice(items)

D. import random; random.randint(items)

Solution

1. Step 1: Check import syntax

   To use choice, you must import the random module fully or import choice specifically. import random; random.choice(items) imports the module correctly.

2. Step 2: Verify function usage

   import random; random.choice(items) calls random.choice(items), which is correct. from random import randint; choice(items) imports randint but tries to call choice without import. import random.choice; choice(items) has invalid import syntax. import random; random.randint(items) calls randint with a list, which is incorrect.

3. Final Answer:

   import random; random.choice(items) -> Option A

4. Quick Check:

   Correct import and call = import random; random.choice(items) [OK]

Hint: Import random module fully to use choice function [OK]

Common Mistakes:

• Importing wrong functions
• Calling functions without module prefix
• Using randint instead of choice

3. What is the output of this code?

import random
items = ['apple', 'banana', 'cherry']
random.shuffle(items)
print(items)

medium

A. SyntaxError because shuffle returns a new list

B. ['apple', 'banana', 'cherry'] (always same order)

C. A new list with one random item from items

D. A randomly shuffled list of the original items

Solution

1. Step 1: Understand random.shuffle behavior

   random.shuffle rearranges the list elements in place randomly. It does not return a new list.

2. Step 2: Analyze the print output

   After shuffling, printing items shows the same list but with elements in random order. So output is a shuffled list, not the original order or a single item.

3. Final Answer:

   A randomly shuffled list of the original items -> Option D

4. Quick Check:

   shuffle changes list order in place [OK]

Hint: shuffle changes list order in place, no new list returned [OK]

Common Mistakes:

• Expecting shuffle to return a new list
• Thinking shuffle picks one random item
• Assuming list order stays same

4. The following code tries to pick a random element from a list but causes an error. What is the problem?

import random
items = ['red', 'green', 'blue']
print(random.choice(items, 1))

medium

A. random.choice needs the list to be converted to a tuple

B. random.choice requires the list to be sorted first

C. random.choice does not take two arguments

D. random.choice only works with strings, not lists

Solution

1. Step 1: Check random.choice function signature

   random.choice takes exactly one argument: a sequence (like a list). It returns one random element.

2. Step 2: Identify the error cause

   The code passes two arguments (items and 1), which is invalid and causes a TypeError.

3. Final Answer:

   random.choice does not take two arguments -> Option C

4. Quick Check:

   choice takes one argument only [OK]

Hint: choice takes only one argument: the sequence [OK]

Common Mistakes:

• Passing extra arguments to choice
• Thinking choice returns multiple items
• Confusing choice with sample

5. You want to generate a dictionary where keys are numbers from 1 to 5 and values are random integers between 10 and 20. Which code correctly does this?

hard

A. import random
result = {i: random.randint(10, 20) for i in range(1, 6)}

B. import random
result = {random.randint(10, 20): i for i in range(1, 6)}

C. import random
result = {i: random.choice(range(10, 20)) for i in range(1, 6)}

D. import random
result = dict(random.randint(10, 20) for i in range(1, 6))

Solution

1. Step 1: Understand dictionary comprehension syntax

   We want keys as numbers 1 to 5 and values as random integers between 10 and 20. The syntax is {key: value for key in iterable}.

2. Step 2: Check each option

   import random result = {i: random.randint(10, 20) for i in range(1, 6)} correctly uses i as key and random.randint(10, 20) as value for each i in 1 to 5.
   import random result = {random.randint(10, 20): i for i in range(1, 6)} swaps keys and values incorrectly.
   import random result = {i: random.choice(range(10, 20)) for i in range(1, 6)} uses random.choice(range(10, 20)) which produces integers 10-19 excluding 20, unlike randint(10,20).
   import random result = dict(random.randint(10, 20) for i in range(1, 6)) tries to convert a generator of integers to dict, which causes an error.

3. Final Answer:

   import random result = {i: random.randint(10, 20) for i in range(1, 6)} -> Option A

4. Quick Check:

   Correct dict comprehension with randint [OK]

Hint: Use dict comprehension with randint for random values [OK]

Common Mistakes:

• Swapping keys and values
• Using dict() on generator of ints
• Using choice with range(10,20) excludes 20