Python math library | isclose() method
Last Updated : 19 Sep, 2023
In the Python math module, math.isclose() the method is used to determine whether two floating point numbers are close in value. To use this function in Python you must import the math module.
Syntax: isclose(a, b, rel_tol = 1e-09, abs_tol 0.0) Parameters: rel_tol: maximum difference for being considered “close”, relative to the magnitude of the input values abs_tol: maximum difference for being considered “close”, regardless of the magnitude of the input values -> rel_tol and abs_tol can be changed by using keyword argument, or by simply providing directly as according to their positions in the parameter list. Return Value : Return True if a is close in value to b, and False otherwise.
Without Using any Tolerance
For the values to be considered close, the difference between them must be smaller than at least one of the tolerances. If both torelances are provided, the absolute tolerance is checked first, and if exceeded the the relative tolerance is checked.
Python3
import math
print(math.isclose(2.005, 2.005))
print(math.isclose(2.005, 2.004))
print(math.isclose(2.006, 2.005))
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Output:
True False False
Using Absolute Tolerance
The absolute difference between the two values is checked against specified absolute tolerance (abs_tol), if the abs_tol is exceed mathisclose() returns False and True if otherwise.
Python3
import math
print(math.isclose(2.005, 2.125, abs_tol = 0.25))
print(math.isclose(2.547, 2.0048, abs_tol = 0.5))
print(math.isclose(2.0214, 2.00214, abs_tol = 0.02))
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Output:
True False True
You can change absolute tolerance, as in above case absolute tolerance is different in all three cases.
Using Relative Tolerance
By default, math.isclose() provides a default rel_tol = 1e-09 = 0.0000000010. If no tolerance is specified then this is used.
However, rel_tol can be specified to meet requirements.
Python3
import math
a = 2.005
b = 2.004
calc_rel_diff = abs(a - b) / ((a + b) / 2)
print("Calculated relative difference = {:.6f}".format(calc_rel_diff))
print("Default relative difference used in math.isclose() = {:.10f}".format(1e-09))
print("Returns: {} --> calc_rel_diff = {} > default rel_tol = {:.10f}".format(math.isclose(a, b), calc_rel_diff, 1e-09))
rel_tol = 5e-04
print("Returns: {} --> calc_rel_diff = {} < specified rel_tol = {:.4f}".format(math.isclose(a, b, rel_tol=rel_tol),calc_rel_diff, rel_tol))
|
Output:
Calculated relative difference = 0.000499 Default relative difference used in math.isclose() = 0.0000000010 Returns: False --> calc_rel_diff = 0.0004988775255674181 > default rel_tol = 0.0000000010 Returns: True --> calc_rel_diff = 0.0004988775255674181 < specified rel_tol = 0.0005
Combining both tolerances
Using both abs_tol (absolute tolerance) and rel_tol (relative tolerance) together in the math.isclose() function allows you to define a more precise and flexible criterion for determining if two numbers are “close enough” based on both absolute and relative considerations.
abs_tol is checked first. If the absolute difference between the numbers is within abs_tol, the function returns True.
If the absolute difference is greater than abs_tol, the function then checks the relative difference based on rel_tol. If the relative difference is within rel_tol, the function returns True.
If neither the absolute tolerance nor the relative tolerance conditions are met, the function returns False.
The example below illustrate a case where the absolute tolerance is exceeded while the relative tolerance is not exceeded giving a positive comparison.
Python3
import math
a = 2.005
b = 2.004
calc_rel_diff = abs(a - b) / ((a + b) / 2)
calc_abs_diff = abs(a-b)
print("Calculated relative difference = {:.6f}".format(calc_rel_diff))
print("Calculated absolute difference = {:.6f}".format(calc_abs_diff))
abs_rol = 4e-04
rel_tol = 5e-04
print("Returns: {} --> Though the calculated absolute difference = {:.6f} > specified abs_tol = {:.6f}, the calc_rel_diff = {:.6f} < specified rel_tol = {:.6f} which is checked last if the absolute difference is exceeded.".format(math.isclose(a, b, rel_tol=rel_tol, abs_tol=abs_rol),calc_abs_diff, abs_rol, calc_rel_diff, rel_tol))
|
Output:
Calculated relative difference = 0.000499 Calculated absolute difference = 0.001000 Returns: True --> Though the calculated absolute difference = 0.001000 > specified abs_tol = 0.000400, the calc_rel_diff = 0.000499 < specified rel_tol = 0.000500 which is checked last if the absolute difference is exceeded.
If both tolerances were exceeded then math.isclose() would return False.
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