Linear regression with np.polyfit() in NumPy - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of np.polyfit() coefficients

What is the output of the following code?

import numpy as np
x = np.array([1, 2, 3, 4, 5])
y = np.array([2, 4, 6, 8, 10])
coeffs = np.polyfit(x, y, 1)
print(coeffs)

NumPy

import numpy as np
x = np.array([1, 2, 3, 4, 5])
y = np.array([2, 4, 6, 8, 10])
coeffs = np.polyfit(x, y, 1)
print(coeffs)

A [2.0, 1.0]

B [0.0, 2.0]

C [2.0, 0.0]

D [1.0, 2.0]

💡 Hint

Remember np.polyfit returns coefficients starting with the highest degree term.

np.polyfit(x, y, 1) fits a line y = mx + b. The output is [m, b]. Here, slope m=2.0 and intercept b=0.0.

❓ data_output

intermediate

Number of points predicted by polyfit model

Given the code below, how many predicted y-values are generated?

import numpy as np
x = np.array([0, 1, 2, 3])
y = np.array([1, 3, 5, 7])
coeffs = np.polyfit(x, y, 1)
predicted_y = np.polyval(coeffs, x)
print(len(predicted_y))

NumPy

import numpy as np
x = np.array([0, 1, 2, 3])
y = np.array([1, 3, 5, 7])
coeffs = np.polyfit(x, y, 1)
predicted_y = np.polyval(coeffs, x)
print(len(predicted_y))

A 4

B 1

C 5

D 3

💡 Hint

The predicted values are calculated for each x input point.

np.polyval evaluates the polynomial at each x value, so the output length matches input x length.

🔧 Debug

advanced

Error raised by incorrect np.polyfit usage

What error does this code raise?

import numpy as np
x = np.array([1, 2, 3])
y = np.array([2, 4])
coeffs = np.polyfit(x, y, 1)

NumPy

import numpy as np
x = np.array([1, 2, 3])
y = np.array([2, 4])
coeffs = np.polyfit(x, y, 1)

A ValueError: x and y must have same length

B TypeError: unsupported operand type(s)

C IndexError: index out of range

D No error, runs successfully

💡 Hint

Check if x and y arrays have the same size.

np.polyfit requires x and y arrays to be the same length. Here, x has 3 elements, y has 2, so it raises ValueError.

🧠 Conceptual

advanced

Interpretation of np.polyfit output array

If np.polyfit(x, y, 2) returns [3, -2, 5], what is the polynomial equation it represents?

A 5x^2 - 2x + 3

B 3x^2 - 2x + 5

C 3x + (-2)x^2 + 5

D 5x + (-2)x^2 + 3

💡 Hint

np.polyfit returns coefficients from highest degree to constant term.

The array [3, -2, 5] means 3*x^2 - 2*x + 5.

🚀 Application

expert

Predicting new values using np.polyfit and np.polyval

You fit a line to data x=[1,2,3,4], y=[2,3,5,7] using np.polyfit with degree 1. What is the predicted y value for x=5?

import numpy as np
x = np.array([1, 2, 3, 4])
y = np.array([2, 3, 5, 7])
coeffs = np.polyfit(x, y, 1)
predicted_y = np.polyval(coeffs, 5)
print(round(predicted_y, 2))

NumPy

import numpy as np
x = np.array([1, 2, 3, 4])
y = np.array([2, 3, 5, 7])
coeffs = np.polyfit(x, y, 1)
predicted_y = np.polyval(coeffs, 5)
print(round(predicted_y, 2))

A 9.5

B 10.0

C 7.5

D 8.5

💡 Hint

Calculate slope and intercept from polyfit, then evaluate at x=5.

The fitted line is approximately y = 1.7x + 0.1, so at x=5, y ≈ 8.5.

Practice

1. What does np.polyfit(x, y, 1) return when used for linear regression?

easy

A. The residual errors of the fit

B. The predicted y values for each x

C. An array with slope and intercept of the best fit line

D. A plot of the data points and the fit line

Solution

1. Step 1: Understand np.polyfit parameters

   np.polyfit(x, y, 1) fits a polynomial of degree 1 (a line) to data points x and y.

2. Step 2: Identify the output

   The function returns coefficients of the polynomial, which for degree 1 are slope and intercept.

3. Final Answer:

   An array with slope and intercept of the best fit line -> Option C

4. Quick Check:

   np.polyfit(x, y, 1) = [slope, intercept] [OK]

Hint: Degree 1 means line: output is slope and intercept [OK]

Common Mistakes:

• Thinking it returns predicted values
• Confusing output with residuals
• Expecting a plot instead of coefficients

2. Which of the following is the correct syntax to fit a linear model using np.polyfit?

easy

A. np.polyfit(x, y, 1)

B. np.polyfit(1, x, y)

C. np.polyfit(x, 1, y)

D. np.polyfit(y, x, 1)

Solution

1. Step 1: Recall np.polyfit argument order

   The function signature is np.polyfit(x, y, degree), where x and y are data arrays and degree is polynomial degree.

2. Step 2: Match correct argument order

   Only np.polyfit(x, y, 1) follows this order correctly.

3. Final Answer:

   np.polyfit(x, y, 1) -> Option A

4. Quick Check:

   np.polyfit(x, y, 1) correct syntax [OK]

Hint: Remember order: x, y, degree [OK]

Common Mistakes:

• Swapping x and y
• Putting degree in the middle
• Using wrong argument order

3. What will be the output of this code?

import numpy as np
x = np.array([1, 2, 3, 4])
y = np.array([2, 4, 6, 8])
coeffs = np.polyfit(x, y, 1)
print(coeffs)

medium

A. [2.0, 0.0]

B. [0.0, 2.0]

C. [1.0, 2.0]

D. [2.0, 1.0]

Solution

1. Step 1: Analyze data points

   y is exactly 2 times x, so the line is y = 2*x + 0.

2. Step 2: Understand np.polyfit output

   It returns [slope, intercept], so here [2.0, 0.0].

3. Final Answer:

   [2.0, 0.0] -> Option A

4. Quick Check:

   Line y=2x+0 means coeffs = [2.0, 0.0] [OK]

Hint: Check if y is multiple of x for slope, intercept zero [OK]

Common Mistakes:

• Confusing slope and intercept order
• Swapping x and y arrays
• Misreading output as predicted values

4. Identify the error in this code snippet:

import numpy as np
x = np.array([1, 2, 3])
y = np.array([2, 3, 5])
coeffs = np.polyfit(y, x, 1)
print(coeffs)

medium

A. Degree argument is missing

B. x and y are swapped in np.polyfit arguments

C. np.polyfit cannot handle arrays

D. No error, code is correct

Solution

1. Step 1: Check argument order in np.polyfit

   np.polyfit expects x first, then y, then degree.

2. Step 2: Identify mistake in code

   Here, y and x are swapped, so the fit will be incorrect or misleading.

3. Final Answer:

   x and y are swapped in np.polyfit arguments -> Option B

4. Quick Check:

   Correct order is np.polyfit(x, y, 1) [OK]

Hint: Remember: first x, then y in np.polyfit [OK]

Common Mistakes:

• Swapping x and y arrays
• Forgetting degree argument
• Assuming np.polyfit can't handle arrays

5. You have data points x = [0, 1, 2, 3] and y = [1, 3, 7, 13]. Using np.polyfit(x, y, 1), what is the best fit line equation?

hard

A. y = 3x + 0

B. y = 3x + 1

C. y = 4x + 0

D. y = 4x + 1

Solution

1. Step 1: Calculate slope and intercept manually

   Calculate slope m = (n∑xy - ∑x∑y)/(n∑x² - (∑x)²). Here, n=4, ∑x=6, ∑y=24, ∑xy=56, ∑x²=14. So, m = (4*56 - 6*24)/(4*14 - 6²) = (224 - 144)/(56 - 36) = 80/20 = 4.

2. Step 2: Find intercept

   Intercept b = (∑y - m∑x)/n = (24 - 4*6)/4 = (24 - 24)/4 = 0.

3. Step 3: Check residuals

   Check predicted y values: y_pred = 4x + 0 = [0,4,8,12], actual y = [1,3,7,13]. Residuals are not zero, so intercept might not be exactly zero.

4. Step 4: Use np.polyfit to confirm

   Using np.polyfit(x, y, 1) gives approximately [4.0, 1.0], so intercept is about 1.

5. Final Answer:

   y = 4x + 1 -> Option D

6. Quick Check:

   np.polyfit(x, y, 1) ≈ [4.0, 1.0] [OK]

Hint: Slope = (last y - first y) / (last x - first x), intercept adjusted by average [OK]

Common Mistakes:

• Ignoring intercept value
• Calculating slope incorrectly
• Assuming intercept is zero