np.random.default_rng() modern approach in NumPy - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of random integers with default_rng

What is the output of this code snippet using np.random.default_rng()?

NumPy

import numpy as np
rng = np.random.default_rng(42)
result = rng.integers(1, 5, size=4)
print(result.tolist())

A [1, 3, 2, 2]

B [3, 1, 2, 4]

C [4, 1, 1, 4]

D [2, 4, 3, 1]

💡 Hint

Remember that integers(low, high) generates numbers from low (inclusive) to high (exclusive).

The default_rng with seed 42 produces a fixed sequence. The call rng.integers(1, 5, size=4) generates four integers between 1 and 4 inclusive. The exact output list is [4, 1, 1, 4].

❓ data_output

intermediate

Shape of array from random normal distribution

What is the shape of the array produced by this code?

NumPy

import numpy as np
rng = np.random.default_rng()
arr = rng.normal(loc=0, scale=1, size=(3, 2))
print(arr.shape)

A (3, 2)

B (2, 3)

C (6,)

D (3,)

💡 Hint

The size parameter defines the shape of the output array.

The rng.normal function generates samples from a normal distribution. The size=(3, 2) means the output array has 3 rows and 2 columns, so shape is (3, 2).

🔧 Debug

advanced

Identify the error in random choice usage

What error does this code raise?

NumPy

import numpy as np
rng = np.random.default_rng()
result = rng.choice([10, 20, 30], size=3, replace=False, p=[0.2, 0.5])

A ValueError: probabilities do not sum to 1

B TypeError: choice() got an unexpected keyword argument 'p'

C No error, runs successfully

D ValueError: probabilities must be same length as array

💡 Hint

Check the length of the probability array compared to the choices array.

The p parameter must have the same length as the array to choose from. Here, the array has length 3 but p has length 2, causing a ValueError.

🧠 Conceptual

advanced

Understanding seed behavior in default_rng

Which statement about np.random.default_rng(seed) is true?

A Using the same seed always produces the same random sequence.

B The seed must be a float between 0 and 1.

C default_rng() ignores the seed parameter completely.

D The seed controls the shape of the output array.

💡 Hint

Think about reproducibility in random number generation.

The seed initializes the random number generator. Using the same seed ensures the same sequence of random numbers, which helps with reproducibility.

🚀 Application

expert

Generate reproducible shuffled DataFrame rows

You have a pandas DataFrame df. Which code snippet correctly shuffles its rows reproducibly using np.random.default_rng()?

NumPy

import pandas as pd
import numpy as np
df = pd.DataFrame({'A': range(5), 'B': list('abcde')})

A

rng = np.random.default_rng(42)
shuffled_df = df.sample(frac=1, random_state=rng.bit_generator)

B

rng = np.random.default_rng(42)
shuffled_df = df.iloc[rng.permutation(len(df))].reset_index(drop=True)

C

rng = np.random.default_rng(42)
shuffled_df = df.sample(frac=1, random_state=42)

D

rng = np.random.default_rng(42)
shuffled_df = df.sample(frac=1, random_state=rng)

💡 Hint

Pandas sample expects an int or np.random.RandomState for random_state, but default_rng returns a Generator.

Pandas sample does not accept np.random.Generator directly as random_state. Using rng.permutation to shuffle indices and then reindexing is the correct reproducible approach.