np.where() for conditional selection in NumPy - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of np.where() with simple condition

What is the output of this code snippet?

NumPy

import numpy as np
arr = np.array([10, 15, 20, 25, 30])
result = np.where(arr > 20, arr, 0)
print(result)

A [ 0 0 0 25 30]

B [10 15 20 25 30]

C [10 15 20 0 0]

D [ 0 15 20 25 30]

💡 Hint

np.where(condition, x, y) returns elements from x where condition is True, else from y.

The condition arr > 20 is True for elements 25 and 30 only. So, np.where returns those elements, and 0 for others.

❓ data_output

intermediate

Resulting array shape after np.where() with condition

Given the code below, what is the shape of the resulting array?

NumPy

import numpy as np
arr = np.array([[1, 2], [3, 4], [5, 6]])
result = np.where(arr % 2 == 0, arr, -1)
print(result.shape)

A (2, 3)

B (6,)

C (3, 2)

D (3,)

💡 Hint

np.where returns an array with the same shape as the input when used with three arguments.

The input array arr has shape (3, 2). Using np.where with condition and two arrays returns an array of the same shape.

🔧 Debug

advanced

Identify the error in np.where usage

What error does this code raise?

NumPy

import numpy as np
arr = np.array([1, 2, 3, 4])
result = np.where(arr > 2)
print(result)

A No error, prints indices where condition is True

B SyntaxError: invalid syntax

C ValueError: operands could not be broadcast together

D TypeError: where() missing required argument 'y' (pos 3)

💡 Hint

np.where can be used with one or three arguments. With one argument, it returns indices.

When np.where is called with only the condition, it returns the indices where the condition is True. No error occurs.

🚀 Application

advanced

Using np.where() to replace negative values

Given the array below, which option correctly replaces all negative values with zero using np.where()?

NumPy

import numpy as np
arr = np.array([3, -1, 5, -7, 2])

A np.where(arr > 0, arr, 0)

B np.where(arr < 0, 0, arr)

C np.where(arr <= 0, 0, arr)

D np.where(arr == 0, 0, arr)

💡 Hint

Replace values less than zero with zero, keep others unchanged.

Option B replaces all elements less than zero with 0, keeping others as is. Others either replace wrong values or miss negatives.

🧠 Conceptual

expert

Understanding np.where() output with multiple conditions

Consider this code snippet. What is the output of result?

NumPy

import numpy as np
arr = np.array([1, 2, 3, 4, 5])
result = np.where(arr % 2 == 0, 'even', np.where(arr % 3 == 0, 'div3', 'odd'))
print(result)

A ['even' 'even' 'div3' 'even' 'odd']

B ['odd' 'even' 'odd' 'even' 'odd']

C ['odd' 'even' 'div3' 'odd' 'odd']

D ['odd' 'even' 'div3' 'even' 'odd']

💡 Hint

np.where can be nested to check multiple conditions in order.

For each element: if divisible by 2, label 'even'; else if divisible by 3, label 'div3'; else 'odd'.

Practice

1. What does the np.where() function do in NumPy?

easy

A. Selects elements from two arrays based on a condition

B. Sorts an array in ascending order

C. Calculates the mean of an array

D. Creates a new array filled with zeros

Solution

1. Step 1: Understand the purpose of np.where()

   np.where() is used to choose elements from two arrays depending on a condition.

2. Step 2: Compare with other options

   Sorting, calculating mean, or creating zeros are different functions like np.sort(), np.mean(), and np.zeros().

3. Final Answer:

   Selects elements from two arrays based on a condition -> Option A

4. Quick Check:

   np.where() = conditional selection [OK]

Hint: np.where picks from two arrays based on a condition [OK]

Common Mistakes:

• Confusing np.where() with sorting or aggregation functions
• Thinking np.where() creates new arrays without conditions
• Assuming np.where() only finds indices, not selects values

2. Which of the following is the correct syntax to use np.where() to select elements from arrays a and b based on condition cond?

easy

A. np.where(a, cond, b)

B. np.where(cond, a, b)

C. np.where(a, b, cond)

D. np.where(cond, b, a)

Solution

1. Step 1: Recall np.where() parameter order

   The correct order is np.where(condition, x, y) where elements from x are chosen if condition is True, else from y.

2. Step 2: Match parameters with options

   np.where(cond, a, b) matches this order: np.where(cond, a, b). Others mix the order incorrectly.

3. Final Answer:

   np.where(cond, a, b) -> Option B

4. Quick Check:

   np.where(condition, x, y) correct syntax [OK]

Hint: Remember: condition first, then true array, then false array [OK]

Common Mistakes:

• Swapping the order of arrays and condition
• Putting arrays before the condition
• Using np.where() without parentheses

3. What is the output of the following code?

import numpy as np
arr = np.array([10, 20, 30, 40])
result = np.where(arr > 25, arr, 0)
print(result)

medium

A. [ 0 0 30 40]

B. [10 20 30 40]

C. [10 20 0 0]

D. [30 40 0 0]

Solution

1. Step 1: Evaluate the condition arr > 25

   Elements greater than 25 are 30 and 40 at indices 2 and 3.

2. Step 2: Apply np.where() selection

   For elements > 25, keep original value; else replace with 0. So result is [0, 0, 30, 40].

3. Final Answer:

   [ 0 0 30 40] -> Option A

4. Quick Check:

   Condition true keeps value, false replaces with 0 [OK]

Hint: Check condition element-wise, replace false with second array [OK]

Common Mistakes:

• Confusing which elements satisfy the condition
• Mixing up true and false array values
• Forgetting np.where returns a new array

4. Identify the error in this code snippet:

import numpy as np
x = np.array([1, 2, 3, 4])
y = np.array([10, 20, 30, 40])
result = np.where(x > 2, y)
print(result)

medium

A. Condition x > 2 is invalid syntax

B. Arrays x and y have different lengths

C. Missing third argument for false condition in np.where()

D. np.where() cannot be used with arrays

Solution

1. Step 1: Check np.where() arguments

   np.where() requires three arguments for conditional selection: condition, array if True, array if False.

2. Step 2: Identify missing argument

   The code only provides condition and one array, missing the third argument for false case.

3. Final Answer:

   Missing third argument for false condition in np.where() -> Option C

4. Quick Check:

   np.where needs condition, true array, false array [OK]

Hint: np.where needs three arguments for selection [OK]

Common Mistakes:

• Using np.where with only two arguments for selection
• Assuming np.where can infer missing arrays
• Confusing np.where with np.nonzero

5. Given arrays a = np.array([5, 15, 25, 35]) and b = np.array([100, 200, 300, 400]), how can you create a new array that takes elements from b when a is greater than 20, otherwise from a, using np.where()?

hard

A. np.where(a > 20, a, b)

B. np.where(b > 20, a, b)

C. np.where(a < 20, b, a)

D. np.where(a > 20, b, a)

Solution

1. Step 1: Understand the condition and selection

   We want elements from b where a > 20, else from a.

2. Step 2: Apply np.where() with correct order

   Use np.where(condition, x, y) where x is chosen if condition True, else y. So np.where(a > 20, b, a).

3. Final Answer:

   np.where(a > 20, b, a) -> Option D

4. Quick Check:

   Condition true picks b, false picks a [OK]

Hint: Condition first, then true array, then false array [OK]

Common Mistakes:

• Swapping arrays in np.where() parameters
• Using wrong comparison operator
• Confusing which array to pick on true condition