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'.