First SciPy computation - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of SciPy integration

What is the output of this code that uses SciPy to integrate a simple function?

SciPy

from scipy.integrate import quad

def f(x):
    return x**2

result, error = quad(f, 0, 1)
print(round(result, 3))

A 0.500

B 1.000

C 0.333

D 0.250

💡 Hint

Think about the integral of x squared from 0 to 1.

The integral of x^2 from 0 to 1 is 1/3, which is approximately 0.333.

❓ data_output

intermediate

Result of SciPy root finding

What is the value of root found by SciPy's root_scalar for the function f(x) = x^3 - 1 near x=1?

SciPy

from scipy.optimize import root_scalar

def f(x):
    return x**3 - 1

sol = root_scalar(f, bracket=[0, 2])
print(round(sol.root, 3))

A 0.000

B 1.000

C -1.000

D 2.000

💡 Hint

The cube root of 1 is?

The root of x^3 - 1 = 0 is x = 1, so the root_scalar finds 1.000.

❓ visualization

advanced

Plotting a Gaussian PDF with SciPy

Which option produces the correct plot of a Gaussian probability density function (PDF) using SciPy?

SciPy

import numpy as np
import matplotlib.pyplot as plt
from scipy.stats import norm

x = np.linspace(-3, 3, 100)

# Plot code here
plt.plot(x, norm.pdf(x))
plt.title('Gaussian PDF')
plt.show()

A Plots a flat horizontal line

B Plots a straight line increasing from left to right

C Plots a curve with two peaks

D Plots a bell-shaped curve centered at 0

💡 Hint

The normal distribution has one peak at the mean.

The norm.pdf function returns the Gaussian PDF values, which form a bell-shaped curve centered at 0.

🧠 Conceptual

advanced

Understanding SciPy's optimize.minimize output

After running SciPy's optimize.minimize on a function, what does the 'success' attribute in the result indicate?

A 'success' is True if the optimizer found a solution meeting the criteria

B 'success' is the minimum value found by the optimizer

C 'success' is the number of iterations performed

D 'success' is the initial guess provided to the optimizer

💡 Hint

Think about what success means in optimization.

The 'success' attribute is a boolean that tells if the optimizer successfully found a solution.

🔧 Debug

expert

Identify the error in this SciPy interpolation code

What error will this code raise when run?

SciPy

import numpy as np
from scipy.interpolate import interp1d

x = np.array([0, 1, 2])
y = np.array([0, 1])

f = interp1d(x, y)
print(f(1.5))

A ValueError: x and y must have same length

B TypeError: unsupported operand type(s)

C IndexError: index out of range

D No error, prints 1.5

💡 Hint

Check the lengths of x and y arrays.

interp1d requires x and y arrays to have the same length. Here y has length 2 but x has length 3.

Practice

1. What is the main purpose of the SciPy library in Python?

easy

A. To manage databases

B. To create web applications

C. To perform scientific and mathematical computations easily

D. To design user interfaces

Solution

1. Step 1: Understand SciPy's role

   SciPy is designed for scientific and mathematical computations. Options for managing databases, creating web applications, and designing user interfaces do not match.

2. Final Answer:

   To perform scientific and mathematical computations easily -> Option C

3. Quick Check:

   SciPy = scientific math library [OK]

Hint: SciPy is for science and math, not web or UI [OK]

Common Mistakes:

• Confusing SciPy with web frameworks
• Thinking SciPy manages databases
• Assuming SciPy is for UI design

2. Which of the following is the correct way to import the SciPy square root function?

easy

A. from scipy import sqrt

B. from scipy.special import sqrt

C. from scipy import math.sqrt

D. from scipy import sqrt as sp_sqrt

Solution

1. Step 1: Recall SciPy import syntax

   sqrt is in scipy.special, so from scipy.special import sqrt works. Direct from scipy import sqrt fails as sqrt is not in top-level scipy (A,D); math.sqrt is separate module (B).

2. Final Answer:

   from scipy.special import sqrt -> Option B

3. Quick Check:

   from scipy.special import sqrt [OK]

Hint: Use scipy.special for sqrt [OK]

Common Mistakes:

• Importing sqrt directly from scipy
• Using incorrect module paths
• Trying to import math.sqrt from scipy

3. What is the output of this code?

import numpy as np
from scipy import special
x = np.array([0, 1, 2])
y = special.expit(x)
print(y)

medium

A. [0.5 0.73105858 0.88079708]

B. [1. 2. 7.3890561]

C. [0. 1. 2.]

D. [0. 0.5 0.75]

Solution

1. Step 1: Understand special.expit function

   special.expit computes sigmoid 1 / (1 + exp(-x)): sigmoid(0)=0.5, sigmoid(1)≈0.73105858, sigmoid(2)≈0.88079708 matching [0.5 0.73105858 0.88079708].

2. Final Answer:

   [0.5 0.73105858 0.88079708] -> Option A

3. Quick Check:

   sigmoid([0,1,2]) = [0.5 0.73105858 0.88079708] [OK]

Hint: expit = sigmoid function, check values at 0,1,2 [OK]

Common Mistakes:

• Confusing expit with exponential
• Expecting raw input array as output
• Mixing up output with other functions

4. Identify the error in this code snippet:

from scipy import sqrt
result = sqrt(16)
print(result)

medium

A. print statement syntax is incorrect

B. sqrt function requires two arguments

C. Missing numpy import for sqrt

D. sqrt is not directly in scipy, import from scipy.special

Solution

1. Step 1: Check import correctness

   sqrt is not in top-level scipy (needs scipy.special); function takes one argument; no numpy required; print is correct.

2. Final Answer:

   sqrt is not directly in scipy, import from scipy.special -> Option D

3. Quick Check:

   Import path error = sqrt is not directly in scipy, import from scipy.special [OK]

Hint: sqrt is in scipy.special, not scipy directly [OK]

Common Mistakes:

• Assuming sqrt is in scipy root module
• Thinking sqrt needs multiple arguments
• Confusing print syntax with Python 2

5. You want to compute the Bessel function of the first kind for values [0, 1, 2] using SciPy. Which code snippet correctly does this?

hard

A. from scipy.special import jv
import numpy as np
x = np.array([0,1,2])
result = jv(0, x)
print(result)

B. import scipy
x = [0,1,2]
result = scipy.bessel(0, x)
print(result)

C. from scipy import bessel
x = [0,1,2]
result = bessel.jv(0, x)
print(result)

D. from scipy.special import jn
x = [0,1,2]
result = jn(x, 0)
print(result)

Solution

1. Step 1: Identify correct function and import

   Bessel function of first kind: jv(order, x) from scipy.special. Code with from scipy.special import jv, np.array x, jv(0, x) works; others have wrong imports/functions/arguments.

2. Final Answer:

   from scipy.special import jv import numpy as np x = np.array([0,1,2]) result = jv(0, x) print(result) -> Option A

3. Quick Check:

   jv(0, x) from scipy.special [OK]

Hint: Use jv(order, x) from scipy.special for Bessel function [OK]

Common Mistakes:

• Using wrong function names like bessel or jn incorrectly
• Swapping arguments order in jn or jv
• Not importing numpy for array input