Parametric interpolation in SciPy - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to import the parametric spline function from scipy.

SciPy

from scipy.interpolate import [1]

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

AUnivariateSpline

Bsplprep

Cgriddata

Dinterp1d

Attempts:

3 left

💡 Hint

Common Mistakes

Choosing interpolation functions meant for 1D data like interp1d.

Confusing griddata with parametric spline functions.

The splprep function is used for parametric spline interpolation in scipy.

2fill in blank

medium

Complete the code to create a parametric spline representation of points x and y.

SciPy

tck, u = splprep([x, y], s=[1])

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

A0

B1

CNone

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Using None or negative values for smoothing which are invalid here.

Using smoothing > 0 which allows approximation, not exact fit.

Setting s=0 forces the spline to pass through all points exactly.

3fill in blank

hard

Fix the error in the code to evaluate the spline at new parameter values u_new.

SciPy

x_new, y_new = splev([1], tck)

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

Au_new

Btck

Cx

Dy

Attempts:

3 left

💡 Hint

Common Mistakes

Passing the spline representation tck as the first argument instead of the parameter values.

Passing original data points instead of parameter values.

The splev function evaluates the spline at the new parameter values u_new.

4fill in blank

hard

Fill both blanks to create a smooth parametric curve and evaluate it at 100 points.

SciPy

tck, u = splprep([x, y], s=[1])
u_new = np.linspace(0, 1, [2])

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

A0

B50

C100

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using smoothing > 0 which approximates instead of fitting exactly.

Choosing too few points for evaluation leading to a rough curve.

Use s=0 for exact fit and 100 points for smooth evaluation.

5fill in blank

hard

Fill all three blanks to create a parametric spline, evaluate it, and plot the smooth curve.

SciPy

tck, u = splprep([x, y], s=[1])
u_new = np.linspace(0, 1, [2])
x_new, y_new = splev([3], tck)
plt.plot(x_new, y_new)
plt.show()

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

A0

B50

Cu_new

D100

Attempts:

3 left

💡 Hint

Common Mistakes

Using the wrong variable for evaluation in splev.

Choosing smoothing > 0 which changes the curve shape.

Set smoothing to zero for exact fit, evaluate at 100 points, and use u_new for evaluation.

Practice

1. What is the main purpose of parametric interpolation in data science?

easy

A. To create smooth curves where both x and y depend on a parameter

B. To sort data points in ascending order

C. To calculate the average of a dataset

D. To find the maximum value in a list

Solution

1. Step 1: Understand parametric interpolation concept

   Parametric interpolation creates smooth curves by defining both x and y as functions of a parameter, usually denoted t.

2. Step 2: Compare with other options

   Sorting, averaging, and finding maximum values are unrelated to creating smooth curves from points.

3. Final Answer:

   To create smooth curves where both x and y depend on a parameter -> Option A

4. Quick Check:

   Parametric interpolation = smooth curves with parameter [OK]

Hint: Parametric means both x and y depend on a parameter [OK]

Common Mistakes:

• Confusing interpolation with sorting or averaging
• Thinking only y depends on the parameter
• Assuming parametric interpolation finds max or min values

2. Which of the following is the correct way to import the functions needed for parametric interpolation in SciPy?

easy

A. from scipy.interpolate import splprep, splev

B. import scipy.interpolate as spint

C. from scipy import interpolate

D. import splprep, splev from scipy.interpolate

Solution

1. Step 1: Recall correct import syntax

   The correct syntax to import specific functions is 'from module import function1, function2'.

2. Step 2: Check each option

   from scipy.interpolate import splprep, splev uses correct syntax. import scipy.interpolate as spint imports the module but not functions directly. from scipy import interpolate imports the module but not functions directly. import splprep, splev from scipy.interpolate has incorrect syntax.

3. Final Answer:

   from scipy.interpolate import splprep, splev -> Option A

4. Quick Check:

   Correct import uses 'from ... import ...' [OK]

Hint: Use 'from scipy.interpolate import splprep, splev' [OK]

Common Mistakes:

• Using incorrect import syntax
• Importing the whole module without functions
• Trying to import functions with wrong order

3. What will be the output of the following code snippet?

import numpy as np
from scipy.interpolate import splprep, splev

x = np.array([0, 1, 2, 3])
y = np.array([0, 1, 0, 1])
tck, u = splprep([x, y], s=0)
new_points = splev(np.linspace(0, 1, 3), tck)
print(np.round(new_points, 2))

medium

A. Error due to incorrect input arrays

B. [[0. 1. 2. ]
[0. 1. 0. ]]

C. [[0. 1.5 3. ]
[0. 0.5 1. ]]

D. [[0. 1.25 2.5 ]
[0. 0.75 0.5 ]]

Solution

1. Step 1: Understand splprep and splev usage

   splprep creates a parametric spline for points (x, y). splev evaluates the spline at given parameter values.

2. Step 2: Evaluate splev at 3 points between 0 and 1

   Using np.linspace(0,1,3) gives [0, 0.5, 1]. Evaluating spline at these points returns smooth interpolated coordinates close to original points.

3. Final Answer:

   [[0. 1.25 2.5 ] [0. 0.75 0.5 ]] -> Option D

4. Quick Check:

   Parametric spline evaluated at 3 points = [[0. 1.25 2.5 ] [0. 0.75 0.5 ]] [OK]

Hint: Remember splev outputs arrays for x and y at given parameters [OK]

Common Mistakes:

• Confusing input arrays with output
• Expecting original points instead of interpolated
• Misunderstanding parameter range for splev

4. Identify the error in the following code for parametric interpolation:

import numpy as np
from scipy.interpolate import splprep, splev

x = [0, 1, 2]
y = [0, 1]
tck, u = splprep([x, y], s=0)
new_points = splev(np.linspace(0, 1, 5), tck)
print(new_points)

medium

A. Missing import statement for numpy

B. x and y arrays have different lengths causing an error

C. Incorrect use of splev function parameters

D. No error, code runs fine

Solution

1. Step 1: Check input arrays length

   x has length 3, y has length 2. splprep requires x and y to have the same length.

2. Step 2: Understand consequence of length mismatch

   Different lengths cause splprep to raise an error because points must be paired.

3. Final Answer:

   x and y arrays have different lengths causing an error -> Option B

4. Quick Check:

   Input arrays must be same length [OK]

Hint: Check that x and y arrays have equal length before splprep [OK]

Common Mistakes:

• Ignoring array length mismatch
• Assuming splev parameters are wrong instead
• Forgetting to import numpy (though it is imported here)

5. You have scattered points representing a closed shape. How can you use SciPy's parametric interpolation to create a smooth closed curve through these points?

hard

A. Use splprep with s>0 to ignore the closed shape

B. Use splev with s=0 to smooth the curve

C. Use splprep with parameter per=1 to indicate a periodic (closed) curve

D. Use numpy.linspace to close the curve manually

Solution

1. Step 1: Understand how to create closed curves with splprep

   Setting per=1 in splprep tells it to create a periodic spline, which closes the curve smoothly.

2. Step 2: Evaluate options

   splev evaluates the spline but does not control closure. s>0 smooths but does not close. numpy.linspace generates parameter values but does not close the curve.

3. Final Answer:

   Use splprep with parameter per=1 to indicate a periodic (closed) curve -> Option C

4. Quick Check:

   per=1 in splprep closes the curve [OK]

Hint: Set per=1 in splprep for closed curves [OK]

Common Mistakes:

• Using splev or s parameter to close curve incorrectly
• Not setting per=1 for closed shapes
• Trying to close curve manually with linspace