Histogram.cc revision 9497:2759161b9d7f
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 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 fatal("Histograms with different number of bins cannot be combined!"); 128 } 129 130 m_max = max(m_max, hist.getMax()); 131 m_count += hist.size(); 132 m_sumSamples += hist.getTotal(); 133 m_sumSquaredSamples += hist.getSquaredTotal(); 134 135 // Both histograms are log base 2. 136 if (hist.getBinSize() == -1 && m_binsize == -1) { 137 for (int j = 0; j < hist.getData(0); j++) { 138 add(0); 139 } 140 141 for (uint32_t i = 1; i < t_bins; i++) { 142 for (int j = 0; j < hist.getData(i); j++) { 143 add(1<<(i-1)); // account for the + 1 index 144 } 145 } 146 } else if (hist.getBinSize() >= 1 && m_binsize >= 1) { 147 // Both the histogram are linear. 148 // We are assuming that the two histograms have the same 149 // minimum value that they can store. 150 151 while (m_binsize > hist.getBinSize()) hist.doubleBinSize(); 152 while (hist.getBinSize() > m_binsize) doubleBinSize(); 153 154 assert(m_binsize == hist.getBinSize()); 155 156 for (uint32_t i = 0; i < t_bins; i++) { 157 m_data[i] += hist.getData(i); 158 159 if (m_data[i] > 0) m_largest_bin = i; 160 } 161 } else { 162 fatal("Don't know how to combine log and linear histograms!"); 163 } 164} 165 166// Computation of standard deviation of samples a1, a2, ... aN 167// variance = [SUM {ai^2} - (SUM {ai})^2/N]/(N-1) 168// std deviation equals square root of variance 169double 170Histogram::getStandardDeviation() const 171{ 172 if (m_count <= 1) 173 return 0.0; 174 175 double variance = 176 (double)(m_sumSquaredSamples - m_sumSamples * m_sumSamples / m_count) 177 / (m_count - 1); 178 return sqrt(variance); 179} 180 181void 182Histogram::print(ostream& out) const 183{ 184 printWithMultiplier(out, 1.0); 185} 186 187void 188Histogram::printPercent(ostream& out) const 189{ 190 if (m_count == 0) { 191 printWithMultiplier(out, 0.0); 192 } else { 193 printWithMultiplier(out, 100.0 / double(m_count)); 194 } 195} 196 197void 198Histogram::printWithMultiplier(ostream& out, double multiplier) const 199{ 200 if (m_binsize == -1) { 201 out << "[binsize: log2 "; 202 } else { 203 out << "[binsize: " << m_binsize << " "; 204 } 205 out << "max: " << m_max << " "; 206 out << "count: " << m_count << " "; 207 // out << "total: " << m_sumSamples << " "; 208 if (m_count == 0) { 209 out << "average: NaN |"; 210 out << "standard deviation: NaN |"; 211 } else { 212 out << "average: " << setw(5) << ((double) m_sumSamples)/m_count 213 << " | "; 214 out << "standard deviation: " << getStandardDeviation() << " |"; 215 } 216 217 for (uint32_t i = 0; i <= m_largest_bin; i++) { 218 if (multiplier == 1.0) { 219 out << " " << m_data[i]; 220 } else { 221 out << " " << double(m_data[i]) * multiplier; 222 } 223 } 224 out << " ]"; 225} 226 227bool 228node_less_then_eq(const Histogram* n1, const Histogram* n2) 229{ 230 return (n1->size() > n2->size()); 231} 232