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.