[FIXED] Regenerating an array based on tolerance value in Python

Issue

I have an array Sum with some elements being very small. I want to print this array based on a tolerance value, tol such that if any element is less than this, it becomes zero. I present the expected output.

import numpy as np
tol = 1e-12

Sum = np.array([ 7.37551766e-08, -4.99999992e-17,  0.00000000e+00,  5.00000028e-16,
       -4.99999984e-17,  0.00000000e+00, -4.83360759e-07])

for i in range(0,len(Sum)):
    if (abs(Sum[i]) < tol):
        Sum[i]==0
print(Sum)

The expected output is

[array([ 7.37551766e-08, 0,  0.00000000e+00,  0,
       0,  0.00000000e+00, -4.83360759e-07])]

Solution

Two easy solutions:

Solution 1

You can apply a simple mask:

Sum[np.abs(Sum) < tol] = 0.0

This is a simple one-liner, ideal if you only want to zero-out some elements.

Solution 2

You could map over the array with a function that returns 0 when input is under threshold:

def f(x, threshold):
   return x if x>threshold else 0.0

tol = 1e-12
Sum = np.vectorize(lambda x: f(x, tol))(Sum)

The advantage of this solution is that you can easily do more complex operations on your elements by modifying the function f(x, threshold). The drawback is that is a little bit longer.

Time efficiency

Using a simple boolean mask is way more efficient, see the figure and note the logarithmic axes.

Answered By - Kins