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140 lines
4.1 KiB
Python
140 lines
4.1 KiB
Python
#!/usr/bin/python
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# Copyright (C) 2015-2025 Free Software Foundation, Inc.
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# This file is part of the GNU C Library.
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#
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# The GNU C Library is free software; you can redistribute it and/or
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# modify it under the terms of the GNU Lesser General Public
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# License as published by the Free Software Foundation; either
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# version 2.1 of the License, or (at your option) any later version.
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#
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# The GNU C Library is distributed in the hope that it will be useful,
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# but WITHOUT ANY WARRANTY; without even the implied warranty of
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# MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
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# Lesser General Public License for more details.
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#
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# You should have received a copy of the GNU Lesser General Public
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# License along with the GNU C Library; if not, see
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# <https://www.gnu.org/licenses/>.
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"""Functions to import benchmark data and process it"""
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import json
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try:
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import jsonschema as validator
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except ImportError:
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print('Could not find jsonschema module.')
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raise
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def mean(lst):
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"""Compute and return mean of numbers in a list
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The numpy average function has horrible performance, so implement our
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own mean function.
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Args:
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lst: The list of numbers to average.
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Return:
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The mean of members in the list.
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"""
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return sum(lst) / len(lst)
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def split_list(bench, func, var):
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""" Split the list into a smaller set of more distinct points
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Group together points such that the difference between the smallest
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point and the mean is less than 1/3rd of the mean. This means that
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the mean is at most 1.5x the smallest member of that group.
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mean - xmin < mean / 3
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i.e. 2 * mean / 3 < xmin
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i.e. mean < 3 * xmin / 2
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For an evenly distributed group, the largest member will be less than
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twice the smallest member of the group.
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Derivation:
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An evenly distributed series would be xmin, xmin + d, xmin + 2d...
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mean = (2 * n * xmin + n * (n - 1) * d) / 2 * n
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and max element is xmin + (n - 1) * d
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Now, mean < 3 * xmin / 2
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3 * xmin > 2 * mean
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3 * xmin > (2 * n * xmin + n * (n - 1) * d) / n
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3 * n * xmin > 2 * n * xmin + n * (n - 1) * d
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n * xmin > n * (n - 1) * d
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xmin > (n - 1) * d
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2 * xmin > xmin + (n-1) * d
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2 * xmin > xmax
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Hence, proved.
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Similarly, it is trivial to prove that for a similar aggregation by using
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the maximum element, the maximum element in the group must be at most 4/3
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times the mean.
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Args:
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bench: The benchmark object
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func: The function name
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var: The function variant name
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"""
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means = []
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lst = bench['functions'][func][var]['timings']
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last = len(lst) - 1
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while lst:
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for i in range(last + 1):
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avg = mean(lst[i:])
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if avg > 0.75 * lst[last]:
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means.insert(0, avg)
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lst = lst[:i]
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last = i - 1
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break
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bench['functions'][func][var]['timings'] = means
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def do_for_all_timings(bench, callback):
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"""Call a function for all timing objects for each function and its
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variants.
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Args:
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bench: The benchmark object
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callback: The callback function
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"""
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for func in bench['functions'].keys():
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for k in bench['functions'][func].keys():
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if 'timings' not in bench['functions'][func][k].keys():
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continue
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callback(bench, func, k)
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def compress_timings(points):
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"""Club points with close enough values into a single mean value
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See split_list for details on how the clubbing is done.
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Args:
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points: The set of points.
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"""
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do_for_all_timings(points, split_list)
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def parse_bench(filename, schema_filename):
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"""Parse the input file
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Parse and validate the json file containing the benchmark outputs. Return
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the resulting object.
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Args:
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filename: Name of the benchmark output file.
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Return:
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The bench dictionary.
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"""
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with open(schema_filename, 'r') as schemafile:
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schema = json.load(schemafile)
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with open(filename, 'r') as benchfile:
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bench = json.load(benchfile)
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validator.validate(bench, schema)
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return bench
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