Interpolation for smoothing data in SciPy - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to import the interpolation function from scipy.

SciPy

from scipy.interpolate import [1]

Drag options to blanks, or click blank then click option'

Ainterp1d

Binterpolate

Csmooth

Dinterpolator

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'interpolate' which is the module, not the function.

Trying to import 'smooth' which does not exist in scipy.interpolate.

Using 'interpolator' which is not a valid function.

The interp1d function from scipy.interpolate is used for 1-dimensional interpolation, which helps smooth data.

2fill in blank

medium

Complete the code to create an interpolation function for given x and y data.

SciPy

f = interp1d(x, y, kind=[1])

Drag options to blanks, or click blank then click option'

A'nearest'

B'linear'

C'average'

D'smooth'

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'smooth' which is not a valid kind.

Using 'average' which is not a valid kind.

Confusing 'nearest' with smoothing, but it creates step-like interpolation.

The kind='linear' argument creates a linear interpolation function, which is a common choice for smoothing data.

3fill in blank

hard

Fix the error in the code to evaluate the interpolation function at new points.

SciPy

y_new = f([1])

Drag options to blanks, or click blank then click option'

Ax_new

By_new

Cx

Df

Attempts:

3 left

💡 Hint

Common Mistakes

Passing y_new which is undefined or wrong.

Passing x which is original data, not new points.

Passing the function f itself instead of input values.

To get interpolated values, you call the function f with new x-values, here x_new.

4fill in blank

hard

Fill both blanks to create a smooth interpolation using cubic method and evaluate at new points.

SciPy

f = interp1d(x, y, kind=[1])
y_smooth = f([2])

Drag options to blanks, or click blank then click option'

A'cubic'

Bx_new

C'linear'

Dy_new

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'linear' instead of 'cubic' for smooth interpolation.

Passing y_new instead of x_new to the function.

Confusing the order of blanks.

Using kind='cubic' creates a smooth cubic interpolation. Then, calling f with x_new evaluates the smooth values.

5fill in blank

hard

Fill all three blanks to create a dictionary of smoothed values for words longer than 3 letters.

SciPy

smoothed = [1]: f([2]) for [3] in words if len([3]) > 3}

Drag options to blanks, or click blank then click option'

Aword

Bx_new

Dx

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names for key and loop variable.

Passing x or word to the function instead of x_new.

Not filtering words by length correctly.

This dictionary comprehension creates keys from words longer than 3 letters and values by applying the interpolation function f to x_new. The variable word is used to iterate over words.

Practice

1. What is the main purpose of using interpolation in data smoothing with SciPy?

easy

A. To convert data into categorical variables

B. To remove all data points from a dataset

C. To sort data points in ascending order

D. To estimate values between known data points

Solution

1. Step 1: Understand interpolation concept

   Interpolation is used to estimate or predict values between existing data points to create a smooth curve.

2. Step 2: Identify the correct purpose

   Among the options, only estimating values between known points matches interpolation's goal.

3. Final Answer:

   To estimate values between known data points -> Option D

4. Quick Check:

   Interpolation = estimate between points [OK]

Hint: Interpolation fills gaps between points [OK]

Common Mistakes:

• Confusing interpolation with data sorting
• Thinking interpolation removes data points
• Mixing interpolation with data categorization

2. Which of the following is the correct way to create a linear interpolation function using SciPy's interp1d?

easy

A. interp1d(x, y, type='linear')

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

C. interp1d(x, y, method='linear')

D. interp1d(x, y, mode='linear')

Solution

1. Step 1: Recall interp1d syntax

   The correct parameter to specify interpolation type is kind, not method, type, or mode.

2. Step 2: Match correct parameter usage

   Using kind='linear' is the proper way to create a linear interpolation function.

3. Final Answer:

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

4. Quick Check:

   Parameter for interpolation type = kind [OK]

Hint: Use 'kind' to set interpolation type [OK]

Common Mistakes:

• Using 'method' instead of 'kind'
• Using 'type' or 'mode' parameters incorrectly
• Forgetting to specify interpolation kind

3. Given the code below, what will be the output of y_new?

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='quadratic')
y_new = f(1.5)

medium

A. 2.25

B. 3.5

C. 2.0

D. Error due to insufficient points

Solution

1. Step 1: Understand quadratic interpolation

   Quadratic interpolation fits a parabola through points. At x=1.5, it estimates y between points (1,1) and (2,4).

2. Step 2: Calculate expected value

   Since y = x^2, at 1.5, y = 1.5^2 = 2.25, matching quadratic interpolation output.

3. Final Answer:

   2.25 -> Option A

4. Quick Check:

   Quadratic interp at 1.5 = 2.25 [OK]

Hint: Quadratic interp matches square values here [OK]

Common Mistakes:

• Choosing linear interpolation value
• Assuming error due to points count
• Miscomputing interpolation value

4. Identify the error in the following code snippet for smoothing data using cubic interpolation:

import numpy as np
from scipy.interpolate import interp1d
x = np.array([0, 1, 2])
y = np.array([1, 3, 7])
f = interp1d(x, y, kind='cubic')
y_new = f(1.5)

medium

A. Not enough data points for cubic interpolation

B. Incorrect import statement for interp1d

C. Wrong parameter name for interpolation kind

D. x and y arrays have different lengths

Solution

1. Step 1: Check data points for cubic interpolation

   Cubic interpolation requires at least 4 data points; here only 3 points are provided.

2. Step 2: Identify the error cause

   Because of insufficient points, interp1d will raise an error for cubic kind.

3. Final Answer:

   Not enough data points for cubic interpolation -> Option A

4. Quick Check:

   Cubic interp needs ≥4 points [OK]

Hint: Cubic interpolation needs at least 4 points [OK]

Common Mistakes:

• Assuming import error
• Confusing parameter names
• Ignoring array length mismatch

5. You have noisy temperature data recorded hourly for 10 hours: x = np.arange(10) and y contains noisy values. You want a smooth curve to estimate temperatures every 0.1 hour using cubic interpolation. Which code snippet correctly achieves this smoothing?

hard

A. f = interp1d(x, y, kind='cubic'); x_new = np.arange(0, 9.1, 0.1); y_smooth = f(x_new)

B. f = interp1d(x, y, kind='linear'); x_new = np.linspace(0, 9, 91); y_smooth = f(x_new)

C. f = interp1d(x, y, kind='cubic'); x_new = np.linspace(0, 9, 91); y_smooth = f(x_new)

D. f = interp1d(x, y, kind='quadratic'); x_new = np.arange(0, 10, 0.1); y_smooth = f(x_new)

Solution

1. Step 1: Choose correct interpolation kind and new x values

   Cubic interpolation is requested, so kind='cubic' is correct. For smooth sampling every 0.1 hour over 0 to 9, use np.linspace(0, 9, 91) for evenly spaced points.

2. Step 2: Verify code correctness

   f = interp1d(x, y, kind='cubic'); x_new = np.linspace(0, 9, 91); y_smooth = f(x_new) uses cubic kind and linspace for x_new, then applies f(x_new) to get smooth y values, matching requirements.

3. Final Answer:

   f = interp1d(x, y, kind='cubic'); x_new = np.linspace(0, 9, 91); y_smooth = f(x_new) -> Option C

4. Quick Check:

   Cubic interp + linspace for smooth curve [OK]

Hint: Use cubic kind and linspace for smooth interpolation [OK]

Common Mistakes:

• Using linear or quadratic instead of cubic
• Using arange with wrong stop value
• Not matching x_new range with original data