1

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.

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])
Kins
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Wiz123
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2 Answers2

1

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.

Computation for different array sizes for the two solutions.

Kins
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1

How about a applying a simple boolean mask:

import numpy as np

np.random.seed(42)
arr = np.round(np.random.random(10), 3)
tol = 0.1

print(arr)
# [0.375 0.951 0.732 0.599 0.156 0.156 0.058 0.866 0.601 0.708]

# set to 0 values below `tol`
arr[np.abs(arr) < tol] = 0.0

print(arr)
# [0.375 0.951 0.732 0.599 0.156 0.156 0.    0.866 0.601 0.708]
norok2
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