interp1d for 1D interpolation in SciPy - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of linear interpolation with interp1d

What is the output of the following code snippet using interp1d for linear interpolation?

SciPy

import numpy as np
from scipy.interpolate import interp1d

x = np.array([0, 1, 2, 3])
y = np.array([0, 1, 4, 9])
f = interp1d(x, y)
result = f(1.5)
print(result)

A 2.5

B 2.0

C 3.0

D Error: ValueError

💡 Hint

Linear interpolation finds the value between two points on a straight line.

The points at x=1 and x=2 have y=1 and y=4 respectively. Linear interpolation at x=1.5 is halfway between 1 and 4, so (1+4)/2 = 2.5.

❓ data_output

intermediate

Number of interpolated points returned

Given the code below, how many values does the interpolation function return?

SciPy

import numpy as np
from scipy.interpolate import interp1d

x = np.array([0, 2, 4, 6])
y = np.array([0, 4, 16, 36])
f = interp1d(x, y)
values = f([1, 3, 5])
print(len(values))

A 4

B 3

C 1

D Error: TypeError

💡 Hint

The input to the interpolation function is a list of points.

The function is called with three points [1, 3, 5], so it returns three interpolated values.

🔧 Debug

advanced

Error raised by extrapolation with default interp1d

What error does the following code raise when trying to interpolate a value outside the original x range?

SciPy

import numpy as np
from scipy.interpolate import interp1d

x = np.array([0, 1, 2])
y = np.array([0, 1, 4])
f = interp1d(x, y)
print(f(3))

A ValueError: A value (3.0) in x_new is above the interpolation range's maximum value (2)

B IndexError: index out of bounds

C TypeError: unsupported operand type(s)

D No error, outputs 9

💡 Hint

By default, interp1d does not allow extrapolation outside the x range.

The default interp1d raises a ValueError when asked to interpolate outside the original x range unless 'fill_value' or 'bounds_error' is set.

🧠 Conceptual

advanced

Effect of 'kind' parameter in interp1d

Which statement correctly describes the effect of changing the 'kind' parameter in interp1d?

A 'kind' controls the extrapolation behavior beyond the data range.

B 'kind' sets the data type of the output array, e.g., int or float.

C 'kind' changes the interpolation method, e.g., linear, nearest, cubic, affecting smoothness and shape of the curve.

D 'kind' specifies the axis along which interpolation is performed.

💡 Hint

Think about how interpolation can be done in different ways.

The 'kind' parameter selects the interpolation method, such as 'linear' for straight lines, 'nearest' for stepwise, or 'cubic' for smooth curves.

🚀 Application

expert

Using interp1d to fill missing data in a time series

You have a time series with missing values at some timestamps. Which code snippet correctly uses interp1d to fill missing values by interpolation?

SciPy

import numpy as np
from scipy.interpolate import interp1d

time = np.array([0, 1, 2, 4, 5])
values = np.array([10, np.nan, 14, 20, 22])

# Fill missing values code here

print(filled_values)

A filled_values = values.fillna(method='interpolate')

B

f = interp1d(time, values)
filled_values = f(time)

C filled_values = np.interp(time, time, values)

D

mask = ~np.isnan(values)
f = interp1d(time[mask], values[mask])
filled_values = values.copy()
filled_values[np.isnan(values)] = f(time[np.isnan(values)])

💡 Hint

You must exclude missing values when creating the interpolation function.

Option D correctly creates an interpolation function using only known values and fills missing points by evaluating at missing timestamps.

Practice

1. What is the main purpose of interp1d in scipy?

easy

A. To sort data points in ascending order

B. To estimate values between known data points

C. To calculate the derivative of a function

D. To perform matrix multiplication

Solution

1. Step 1: Understand the function role

   interp1d is used to create a function that estimates values between given points.

2. Step 2: Compare options with purpose

   Only To estimate values between known data points describes estimating values between known points, which matches interp1d's purpose.

3. Final Answer:

   To estimate values between known data points -> Option B

4. Quick Check:

   interp1d = estimate between points [OK]

Hint: interp1d fills gaps between points [OK]

Common Mistakes:

• Confusing interpolation with sorting
• Thinking it calculates derivatives
• Mixing it up with matrix operations

2. Which of the following is the correct way to create a linear interpolation function using interp1d with x and y arrays?

easy

A. f = interp1d(x, y, method='linear')

B. f = interp1d(y, x, kind='linear')

C. f = interp1d(x, y, kind='linear')

D. f = interp1d(x, y, kind='smooth')

Solution

1. Step 1: Check parameter order and names

   interp1d takes x and y arrays in order, and the interpolation type is set by kind.

2. Step 2: Validate options

   f = interp1d(x, y, kind='linear') uses correct order and kind='linear'. f = interp1d(y, x, kind='linear') swaps x and y, which is wrong. f = interp1d(x, y, method='linear') uses wrong parameter name method. f = interp1d(x, y, kind='smooth') uses invalid kind 'smooth'.

3. Final Answer:

   f = interp1d(x, y, kind='linear') -> Option C

4. Quick Check:

   Correct syntax uses kind='linear' [OK]

Hint: Use kind='linear' and x,y in order [OK]

Common Mistakes:

• Swapping x and y arrays
• Using wrong parameter name 'method'
• Using unsupported kind values

3. Given the code:

import numpy as np
from scipy.interpolate import interp1d
x = np.array([0, 1, 2, 3])
y = np.array([0, 1, 4, 9])
f = interp1d(x, y, kind='linear')
print(f(1.5))

What is the output?

medium

A. 2.5

B. 3.25

C. 4.5

D. Error

Solution

1. Step 1: Understand interpolation at 1.5

   Between x=1 (y=1) and x=2 (y=4), linear interpolation at 1.5 is halfway: (1 + 4)/2 = 2.5.

2. Step 2: Confirm calculation

   Linear interpolation formula: y = y1 + (x - x1)*(y2 - y1)/(x2 - x1) = 1 + (1.5-1)*(4-1)/(2-1) = 1 + 0.5*3 = 2.5.

3. Final Answer:

   2.5 -> Option A

4. Quick Check:

   Linear interp at 1.5 = 2.5 [OK]

Hint: Linear interp averages between points [OK]

Common Mistakes:

• Using wrong points for interpolation
• Calculating midpoint incorrectly
• Expecting quadratic interpolation result

4. What is wrong with this code snippet?

from scipy.interpolate import interp1d
x = [0, 1, 2]
y = [1, 4]
f = interp1d(x, y)
print(f(1.5))

medium

A. x and y arrays have different lengths

B. interp1d requires x to be sorted descending

C. Missing kind parameter causes error

D. Cannot interpolate at 1.5 because it's outside x range

Solution

1. Step 1: Check lengths of x and y

   x has 3 elements, y has 2 elements. They must be the same length for interpolation.

2. Step 2: Understand interp1d requirements

   Different lengths cause a ValueError. Sorting or kind parameter is not mandatory. 1.5 is within x range (0 to 2).

3. Final Answer:

   x and y arrays have different lengths -> Option A

4. Quick Check:

   Equal length arrays needed [OK]

Hint: Check x and y lengths match [OK]

Common Mistakes:

• Assuming kind is mandatory
• Thinking x must be descending
• Confusing interpolation range limits

5. You have temperature data recorded every hour: x = [0, 1, 2, 3, 4] hours and y = [15, 17, 20, 22, 21] degrees Celsius. You want to estimate temperature at half-hour intervals using cubic interpolation. Which code correctly creates the interpolation function and estimates temperature at 2.5 hours?

hard

A. f = interp1d(x, y, kind='linear')\nprint(f(2.5))

B. f = interp1d(x, y, kind='quadratic')\nprint(f(2.5))

C. f = interp1d(y, x, kind='cubic')\nprint(f(2.5))

D. f = interp1d(x, y, kind='cubic')\nprint(f(2.5))

Solution

1. Step 1: Identify correct parameters for cubic interpolation

   Use kind='cubic' with x and y arrays in correct order.

2. Step 2: Check options

   f = interp1d(x, y, kind='cubic')\nprint(f(2.5)) uses correct order and cubic kind. f = interp1d(x, y, kind='linear')\nprint(f(2.5)) uses linear, not cubic. f = interp1d(y, x, kind='cubic')\nprint(f(2.5)) swaps x and y, which is wrong. f = interp1d(x, y, kind='quadratic')\nprint(f(2.5)) uses quadratic kind instead of cubic.

3. Final Answer:

   f = interp1d(x, y, kind='cubic') print(f(2.5)) -> Option D

4. Quick Check:

   Cubic kind with correct x,y order [OK]

Hint: Use kind='cubic' and correct x,y order [OK]

Common Mistakes:

• Swapping x and y arrays
• Using quadratic instead of cubic
• Choosing linear instead of cubic for smoothness