69 lines
1.8 KiB
Python
69 lines
1.8 KiB
Python
# -*- coding: utf-8 -*-
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#
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# report_diff.py
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#
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# Date Created: July 11, 2018
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#
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# Author: Michael E. Tryby
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# US EPA - ORD/NRMRL
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#
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# system imports
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import itertools as it
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# third party imports
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import numpy as np
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# project imports
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import nrtest_epanet.output_reader as ordr
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def report_diff(path_test, path_ref, min_cdd):
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for (test, ref) in it.izip(ordr.output_generator(path_test),
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ordr.output_generator(path_ref)):
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if len(test[0]) != len(ref[0]):
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raise ValueError('Inconsistent lengths')
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# Skip over arrays that are equal
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if np.array_equal(test[0], ref[0]):
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continue
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else:
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lre = log_relative_error(test[0], ref[0])
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idx = np.unravel_index(np.argmin(lre), lre.shape)
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if lre[idx] < min_cdd:
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print_diff(idx, lre, test, ref)
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return
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def log_relative_error(q, c):
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'''
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Computes log relative error, a measure of numerical accuracy.
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Single precision machine epsilon is between 2^-24 and 2^-23.
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Reference:
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McCullough, B. D. "Assessing the Reliability of Statistical Software: Part I."
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The American Statistician, vol. 52, no. 4, 1998, pp. 358<35>366.
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JSTOR, JSTOR, www.jstor.org/stable/2685442.
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'''
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diff = np.subtract(q, c)
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tmp_c = np.copy(c)
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# If ref value is small compute absolute error
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tmp_c[np.fabs(tmp_c) <= 1.0e-6] = 1.0
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re = np.fabs(diff)/np.fabs(tmp_c)
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# If re is tiny set lre to number of digits
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re[re < 1.0e-7] = 1.0e-7
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# If re is very large set lre to zero
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re[re > 2.0] = 1.0
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return np.negative(np.log10(re))
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def print_diff(idx, lre, test, ref):
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print("LRE: %f\nIdx: %s\nSut: %f\nRef: %f\n"
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% ((lre[idx]),(idx[0], ref[1]),(test[0][idx[0]]),(ref[0][idx[0]]))) |