interp1d for 1D interpolation in SciPy - Step-by-Step Execution

Concept Flow - interp1d for 1D interpolation

Start with known points (x, y)

↓

Create interp1d function

↓

Input new x values

↓

interp1d calculates interpolated y

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Output interpolated y values

We start with known data points, create an interpolation function, input new x values, and get interpolated y values.

Execution Sample

SciPy

from scipy.interpolate import interp1d
x = [0, 1, 2]
y = [0, 1, 4]
f = interp1d(x, y)
new_x = [0.5, 1.5]
result = f(new_x)

This code creates an interpolation function from points (0,0), (1,1), (2,4) and finds y values at 0.5 and 1.5.

Execution Table

Step	Action	Input	Process	Output
1	Define known points	x=[0,1,2], y=[0,1,4]	Store points for interpolation	Points stored
2	Create interp1d function	x, y	Build linear interpolation function	Function f created
3	Input new x values	new_x=[0.5,1.5]	Pass new_x to f	Calculate interpolated y
4	Calculate y at 0.5	0.5	Linear interpolate between (0,0) and (1,1)	0.5
5	Calculate y at 1.5	1.5	Linear interpolate between (1,1) and (2,4)	2.5
6	Return result	new_x	Collect interpolated values	[0.5, 2.5]

💡 All new_x values processed, interpolation complete

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 6
x	undefined	[0,1,2]	[0,1,2]	[0,1,2]	[0,1,2]
y	undefined	[0,1,4]	[0,1,4]	[0,1,4]	[0,1,4]
f	undefined	undefined	interp1d function	interp1d function	interp1d function
new_x	undefined	undefined	undefined	[0.5,1.5]	[0.5,1.5]
result	undefined	undefined	undefined	undefined	[0.5, 2.5]

Key Moments - 2 Insights

Why does interp1d only work between the known x points?

interp1d by default does linear interpolation only within the range of x. Outside this range, it cannot estimate values unless specified. See execution_table steps 4 and 5 where interpolation happens between known points.

What happens if new_x contains a value exactly equal to a known x?

interp1d returns the exact y value for that x without interpolation. This is because the function matches the known point directly, as shown in execution_table step 4 for x=0.5 (between points) and would be direct if x=1.

Visual Quiz - 3 Questions

Look at the execution_table, what is the interpolated y value at new_x=1.5?

A 2.5

B 1.5

C 3.0

D 4.0

💡 Hint

Check step 5 in execution_table where y at 1.5 is calculated.

According to variable_tracker, what is the value of 'result' after step 6?

A [0, 1]

B [1.0, 4.0]

C [0.5, 2.5]

D undefined

💡 Hint

Look at the 'result' row under 'After Step 6' in variable_tracker.

If we add new_x=3 to the input, what will happen by default?

A interp1d will extrapolate and give a value

B interp1d will raise an error

C interp1d will return NaN

D interp1d will ignore the value

💡 Hint

interp1d by default does not extrapolate beyond known x; see key_moments about interpolation range.

Concept Snapshot

interp1d creates a function for 1D interpolation
Input: known x and y points
Output: interpolated y for new x values
Default: linear interpolation within x range
Outside range: raises error unless allowed
Use f(new_x) to get interpolated results

Full Transcript

We start with known data points x and y. Then we create an interpolation function f using interp1d. When we input new x values, the function calculates y values by linear interpolation between the known points. For example, at x=0.5, y is 0.5, and at x=1.5, y is 2.5. The function only works within the range of known x values by default. If a new x is outside this range, an error occurs unless extrapolation is enabled. The result is an array of interpolated y values corresponding to the new x inputs.

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