Histogram.cc revision 11061:25b53a7195f7
1/* 2 * Copyright (c) 1999-2008 Mark D. Hill and David A. Wood 3 * All rights reserved. 4 * 5 * Redistribution and use in source and binary forms, with or without 6 * modification, are permitted provided that the following conditions are 7 * met: redistributions of source code must retain the above copyright 8 * notice, this list of conditions and the following disclaimer; 9 * redistributions in binary form must reproduce the above copyright 10 * notice, this list of conditions and the following disclaimer in the 11 * documentation and/or other materials provided with the distribution; 12 * neither the name of the copyright holders nor the names of its 13 * contributors may be used to endorse or promote products derived from 14 * this software without specific prior written permission. 15 * 16 * THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS 17 * "AS IS" AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT 18 * LIMITED TO, THE IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR 19 * A PARTICULAR PURPOSE ARE DISCLAIMED. IN NO EVENT SHALL THE COPYRIGHT 20 * OWNER OR CONTRIBUTORS BE LIABLE FOR ANY DIRECT, INDIRECT, INCIDENTAL, 21 * SPECIAL, EXEMPLARY, OR CONSEQUENTIAL DAMAGES (INCLUDING, BUT NOT 22 * LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS OR SERVICES; LOSS OF USE, 23 * DATA, OR PROFITS; OR BUSINESS INTERRUPTION) HOWEVER CAUSED AND ON ANY 24 * THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT LIABILITY, OR TORT 25 * (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY OUT OF THE USE 26 * OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF SUCH DAMAGE. 27 */ 28 29#include <cmath> 30#include <iomanip> 31 32#include "base/intmath.hh" 33#include "mem/ruby/common/Histogram.hh" 34 35using namespace std; 36 37Histogram::Histogram(int binsize, uint32_t bins) 38{ 39 m_binsize = binsize; 40 clear(bins); 41} 42 43Histogram::~Histogram() 44{ 45} 46 47void 48Histogram::clear(int binsize, uint32_t bins) 49{ 50 m_binsize = binsize; 51 clear(bins); 52} 53 54void 55Histogram::clear(uint32_t bins) 56{ 57 m_largest_bin = 0; 58 m_max = 0; 59 m_data.resize(bins); 60 for (uint32_t i = 0; i < bins; i++) { 61 m_data[i] = 0; 62 } 63 64 m_count = 0; 65 m_max = 0; 66 m_sumSamples = 0; 67 m_sumSquaredSamples = 0; 68} 69 70void 71Histogram::doubleBinSize() 72{ 73 assert(m_binsize != -1); 74 uint32_t t_bins = m_data.size(); 75 76 for (uint32_t i = 0; i < t_bins/2; i++) { 77 m_data[i] = m_data[i*2] + m_data[i*2 + 1]; 78 } 79 for (uint32_t i = t_bins/2; i < t_bins; i++) { 80 m_data[i] = 0; 81 } 82 83 m_binsize *= 2; 84} 85 86void 87Histogram::add(int64_t value) 88{ 89 assert(value >= 0); 90 m_max = max(m_max, value); 91 m_count++; 92 93 m_sumSamples += value; 94 m_sumSquaredSamples += (value*value); 95 96 uint32_t index; 97 98 if (m_binsize == -1) { 99 // This is a log base 2 histogram 100 if (value == 0) { 101 index = 0; 102 } else { 103 index = floorLog2(value) + 1; 104 if (index >= m_data.size()) { 105 index = m_data.size() - 1; 106 } 107 } 108 } else { 109 // This is a linear histogram 110 uint32_t t_bins = m_data.size(); 111 112 while (m_max >= (t_bins * m_binsize)) doubleBinSize(); 113 index = value/m_binsize; 114 } 115 116 assert(index < m_data.size()); 117 m_data[index]++; 118 m_largest_bin = max(m_largest_bin, index); 119} 120 121void 122Histogram::add(Histogram& hist) 123{ 124 uint32_t t_bins = m_data.size(); 125 126 if (hist.getBins() != t_bins) { 127 if (m_count == 0) { 128 m_data.resize(hist.getBins()); 129 } else { 130 fatal("Histograms with different number of bins " 131 "cannot be combined!"); 132 } 133 } 134 135 m_max = max(m_max, hist.getMax()); 136 m_count += hist.size(); 137 m_sumSamples += hist.getTotal(); 138 m_sumSquaredSamples += hist.getSquaredTotal(); 139 140 // Both histograms are log base 2. 141 if (hist.getBinSize() == -1 && m_binsize == -1) { 142 for (int j = 0; j < hist.getData(0); j++) { 143 add(0); 144 } 145 146 for (uint32_t i = 1; i < t_bins; i++) { 147 for (int j = 0; j < hist.getData(i); j++) { 148 add(1<<(i-1)); // account for the + 1 index 149 } 150 } 151 } else if (hist.getBinSize() >= 1 && m_binsize >= 1) { 152 // Both the histogram are linear. 153 // We are assuming that the two histograms have the same 154 // minimum value that they can store. 155 156 while (m_binsize > hist.getBinSize()) hist.doubleBinSize(); 157 while (hist.getBinSize() > m_binsize) doubleBinSize(); 158 159 assert(m_binsize == hist.getBinSize()); 160 161 for (uint32_t i = 0; i < t_bins; i++) { 162 m_data[i] += hist.getData(i); 163 164 if (m_data[i] > 0) m_largest_bin = i; 165 } 166 } else { 167 fatal("Don't know how to combine log and linear histograms!"); 168 } 169} 170 171// Computation of standard deviation of samples a1, a2, ... aN 172// variance = [SUM {ai^2} - (SUM {ai})^2/N]/(N-1) 173// std deviation equals square root of variance 174double 175Histogram::getStandardDeviation() const 176{ 177 if (m_count <= 1) 178 return 0.0; 179 180 double variance = 181 (double)(m_sumSquaredSamples - m_sumSamples * m_sumSamples / m_count) 182 / (m_count - 1); 183 return sqrt(variance); 184} 185 186void 187Histogram::print(ostream& out) const 188{ 189 printWithMultiplier(out, 1.0); 190} 191 192void 193Histogram::printPercent(ostream& out) const 194{ 195 if (m_count == 0) { 196 printWithMultiplier(out, 0.0); 197 } else { 198 printWithMultiplier(out, 100.0 / double(m_count)); 199 } 200} 201 202void 203Histogram::printWithMultiplier(ostream& out, double multiplier) const 204{ 205 if (m_binsize == -1) { 206 out << "[binsize: log2 "; 207 } else { 208 out << "[binsize: " << m_binsize << " "; 209 } 210 out << "max: " << m_max << " "; 211 out << "count: " << m_count << " "; 212 // out << "total: " << m_sumSamples << " "; 213 if (m_count == 0) { 214 out << "average: NaN |"; 215 out << "standard deviation: NaN |"; 216 } else { 217 out << "average: " << setw(5) << ((double) m_sumSamples)/m_count 218 << " | "; 219 out << "standard deviation: " << getStandardDeviation() << " |"; 220 } 221 222 for (uint32_t i = 0; i <= m_largest_bin; i++) { 223 if (multiplier == 1.0) { 224 out << " " << m_data[i]; 225 } else { 226 out << " " << double(m_data[i]) * multiplier; 227 } 228 } 229 out << " ]"; 230} 231 232bool 233node_less_then_eq(const Histogram* n1, const Histogram* n2) 234{ 235 return (n1->size() > n2->size()); 236} 237