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#!/usr/bin/python
# Copyright (C) 2015-2016 Free Software Foundation, Inc.
# This file is part of the GNU C Library.
#
# The GNU C Library is free software; you can redistribute it and/or
# modify it under the terms of the GNU Lesser General Public
# License as published by the Free Software Foundation; either
# version 2.1 of the License, or (at your option) any later version.
#
# The GNU C Library is distributed in the hope that it will be useful,
# but WITHOUT ANY WARRANTY; without even the implied warranty of
# MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
# Lesser General Public License for more details.
#
# You should have received a copy of the GNU Lesser General Public
# License along with the GNU C Library; if not, see
# <http://www.gnu.org/licenses/>.
"""Functions to import benchmark data and process it"""
import json
try:
import jsonschema as validator
except ImportError:
print('Could not find jsonschema module.')
raise
def mean(lst):
"""Compute and return mean of numbers in a list
The numpy average function has horrible performance, so implement our
own mean function.
Args:
lst: The list of numbers to average.
Return:
The mean of members in the list.
"""
return sum(lst) / len(lst)
def split_list(bench, func, var):
""" Split the list into a smaller set of more distinct points
Group together points such that the difference between the smallest
point and the mean is less than 1/3rd of the mean. This means that
the mean is at most 1.5x the smallest member of that group.
mean - xmin < mean / 3
i.e. 2 * mean / 3 < xmin
i.e. mean < 3 * xmin / 2
For an evenly distributed group, the largest member will be less than
twice the smallest member of the group.
Derivation:
An evenly distributed series would be xmin, xmin + d, xmin + 2d...
mean = (2 * n * xmin + n * (n - 1) * d) / 2 * n
and max element is xmin + (n - 1) * d
Now, mean < 3 * xmin / 2
3 * xmin > 2 * mean
3 * xmin > (2 * n * xmin + n * (n - 1) * d) / n
3 * n * xmin > 2 * n * xmin + n * (n - 1) * d
n * xmin > n * (n - 1) * d
xmin > (n - 1) * d
2 * xmin > xmin + (n-1) * d
2 * xmin > xmax
Hence, proved.
Similarly, it is trivial to prove that for a similar aggregation by using
the maximum element, the maximum element in the group must be at most 4/3
times the mean.
Args:
bench: The benchmark object
func: The function name
var: The function variant name
"""
means = []
lst = bench['functions'][func][var]['timings']
last = len(lst) - 1
while lst:
for i in range(last + 1):
avg = mean(lst[i:])
if avg > 0.75 * lst[last]:
means.insert(0, avg)
lst = lst[:i]
last = i - 1
break
bench['functions'][func][var]['timings'] = means
def do_for_all_timings(bench, callback):
"""Call a function for all timing objects for each function and its
variants.
Args:
bench: The benchmark object
callback: The callback function
"""
for func in bench['functions'].keys():
for k in bench['functions'][func].keys():
if 'timings' not in bench['functions'][func][k].keys():
continue
callback(bench, func, k)
def compress_timings(points):
"""Club points with close enough values into a single mean value
See split_list for details on how the clubbing is done.
Args:
points: The set of points.
"""
do_for_all_timings(points, split_list)
def parse_bench(filename, schema_filename):
"""Parse the input file
Parse and validate the json file containing the benchmark outputs. Return
the resulting object.
Args:
filename: Name of the benchmark output file.
Return:
The bench dictionary.
"""
with open(schema_filename, 'r') as schemafile:
schema = json.load(schemafile)
with open(filename, 'r') as benchfile:
bench = json.load(benchfile)
validator.validate(bench, schema)
do_for_all_timings(bench, lambda b, f, v:
b['functions'][f][v]['timings'].sort())
return bench
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