NumPy: Average of every consecutive triplet of elements of a given array
Average every consecutive triplet in an array.
Write a NumPy program to create an array which is the average of every consecutive triplet of elements in a given array.
Sample Solution:
Python Code:
# Importing the NumPy library and aliasing it as 'np'
import numpy as np
# Creating a NumPy array
arr1 = np.array([1, 2, 3, 2, 4, 6, 1, 2, 12, 0, -12, 6])
# Displaying the original array
print("Original array:")
print(arr1)
# Reshaping the array into rows with three elements each and computing the mean along the rows (axis=1)
result = np.mean(arr1.reshape(-1, 3), axis=1)
# Displaying the average of every consecutive triplet of elements of the array
print("Average of every consecutive triplet of elements of the said array:")
print(result)
Sample Output:
Original array: [ 1 2 3 2 4 6 1 2 12 0 -12 6] Average of every consecutive triplet of elements of the said array: [ 2. 4. 5. -2.]
Explanation:
In the above code -
- arr1 = np.array([1,2,3, 2,4,6, 1,2,12, 0,-12,6]): Create a NumPy array 'arr1' with the given values.
- arr1.reshape(-1, 3): Reshape 'arr1' into a 2D array with 3 columns and an automatically inferred number of rows using -1. In this case, it will result in a 4x3 array.
- result = np.mean(arr1.reshape(-1, 3), axis=1): Calculate the mean of the reshaped array along axis 1 (row-wise mean).
- print(result): Finally print() function prints the resulting array containing the row-wise mean values.
Pictorial Presentation:
Python-Numpy Code Editor: