我有N个不同的日志文件来自我们设备上运行的N个不同的服务。我想将N个文件合并到一个文件中,保持按时间顺序排列。文件大小可以从几KB到GB。
N个日志文件具有相同的格式,如:
********** LOGGING SESSION STARTED ************
* Hmsoa Version: 2.4.0.12
* Exe Path: c:\program files (x86)\silicon biosystems\deparray300a_driver\deparray300a_driver.exe
* Exe Version: 1.6.0.154
************************************************
TIME = 2017/02/01 11:12:12,180 ; THID = 4924; CAT = ; LVL = 1000; LOG = API 'Connect'->Enter;
TIME = 2017/02/01 11:12:12,196 ; THID = 4924; CAT = ; LVL = 1000; LOG = API 'Connect'->Exit=0;
TIME = 2017/02/01 11:12:12,196 ; THID = 4924; CAT = ; LVL = 1000; LOG = API 'CCisProxyLocal CONNECT - ok'->Enter;
TIME = 2017/02/01 11:12:12,196 ; THID = 4924; CAT = ; LVL = 1000; LOG = API 'CRecoveryAxesProxyLocal CONNECT - ok'->Enter;
TIME = 2017/02/01 11:12:12,196 ; THID = 4924; CAT = ; LVL = 1000; LOG = API 'CAmplifierProxyLocalV3 CONNECT - ok'->Enter;
TIME = 2017/02/01 11:12:12,196 ; THID = 4924; CAT = ; LVL = 1000; LOG = API 'SYSTEM_DIAGNOSIS_GET'->Enter;
TIME = 2017/02/01 11:12:12,211 ; THID = 4924; CAT = ; LVL = 1000; LOG = API 'SYSTEM_DIAGNOSIS_GET'->Exit=0;
TIME = 2017/02/01 11:12:12,211 ; THID = 4924; CAT = ; LVL = 1000; LOG = API 'LBL_SQUARE_SET'->Enter;
TIME = 2017/02/01 11:12:12,219 ; THID = 4924; CAT = ; LVL = 1000; LOG = API 'LBL_SQUARE_SET'->Exit=0;
由于我已有N个不同的文件,到目前为止我所做的是应用一个外部排序算法,为每个文件读取一行:
#include "stdafx.h"
#include "boost/regex.hpp"
#include "boost/lexical_cast.hpp"
#include "boost\filesystem.hpp"
#include <string>
#include <fstream>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <climits>
#include <ctime>
namespace fs = boost::filesystem;
static const boost::regex expression(R"(^(?:(?:TIME\s=\s\d{4}\/\d{2}\/\d{2}\s)|(?:@))([0-9:.,]+))");
static const boost::regex nameFileEx(R"(^[\d\-\_]+(\w+\s?\w+|\w+))");
static const std::string path("E:\\2017-02-01");
//static const std::string path("E:\\TestLog");
unsigned long time2Milleseconds(const std::string & time)
{
int a, b, c, d;
if (sscanf_s(time.c_str(), "%d:%d:%d,%d", &a, &b, &c, &d) >= 3)
return a * 3600000 + b * 60000 + c * 1000 + d;
}
void readAllFilesUntilLine7(std::vector<std::pair<std::ifstream, std::string>> & vifs)
{
std::string line;
for (int i = 0; i < vifs.size(); ++i)
{
int lineNumber = 0;
while (lineNumber != 7 && std::getline(vifs[i].first, line))
{
++lineNumber;
}
}
}
void checkRegex(std::vector<std::pair<std::ifstream, std::string>> & vifs, std::vector<unsigned long> & logTime, std::vector<std::string> & lines, int index, int & counter)
{
std::string line;
boost::smatch what;
if (std::getline(vifs[index].first, line))
{
if (boost::regex_search(line, what, expression))
{
logTime[index] = time2Milleseconds(what[1]);
}
lines[index] = line;
}
else
{
--counter;
logTime[index] = ULONG_MAX;
}
}
void mergeFiles(std::vector<std::pair<std::ifstream, std::string>> & vifs, std::vector<unsigned long> & logTime, std::vector<std::string> & lines, std::ofstream & file, int & counter)
{
std::string line;
boost::smatch what;
int index = 0;
for (int i = 0; i < vifs.size(); ++i)
{
checkRegex(vifs, logTime, lines, i, counter);
}
index = min_element(logTime.begin(), logTime.end()) - logTime.begin();
file << lines[index] << " --> " << vifs[index].second << "\n";
while (true)
{
checkRegex(vifs, logTime, lines, index, counter);
index = min_element(logTime.begin(), logTime.end()) - logTime.begin();
if (0 == counter)
break;
file << lines[index] << " --> " << vifs[index].second << "\n";
}
}
int main()
{
clock_t begin = clock();
int cnt = std::count_if(fs::directory_iterator(path),fs::directory_iterator(),static_cast<bool(*)(const fs::path&)>(fs::is_regular_file));
std::vector<std::pair<std::ifstream, std::string>> vifs(cnt);
int index = 0;
boost::smatch what;
std::string file;
for (fs::directory_iterator d(path); d != fs::directory_iterator(); ++d)
{
if (fs::is_regular_file(d->path()))
{
file = d->path().filename().string();
if (boost::regex_search(file, what, nameFileEx))
{
vifs[index++] = std::make_pair(std::ifstream(d->path().string()), what[1]);
}
}
}
std::vector<unsigned long> logTime(cnt, ULONG_MAX);
std::vector<std::string> lines(cnt);
std::ofstream filename(path + "\\TestLog.txt");
readAllFilesUntilLine7(vifs);
mergeFiles(vifs, logTime, lines, filename, cnt);
filename.close();
clock_t end = clock();
double elapsed_secs = double(end - begin) / CLOCKS_PER_SEC;
std::cout << "Elapsed time = " << elapsed_secs << "\n";
return 0;
}
它完全按照它应该做的但它很慢。要合并82个文件,大小范围从1 KB到250 MB,并创建一个超过6000000行的最终文件,需要70分钟。
如何加快算法速度?非常感谢任何帮助!
更新
我也用堆实现了这个版本:
Data.h:
#pragma once
#include <string>
class Data
{
public:
Data(DWORD index,
const std::string & line,
ULONG time);
~Data();
inline const ULONG getTime() const {return time; }
inline const DWORD getIndex() const { return index; }
inline const std::string getLine() const { return line; }
private:
DWORD index;
std::string line;
ULONG time;
};
class Compare
{
public:
bool operator()(const Data & lhs, const Data & rhs) { return lhs.getTime() > rhs.getTime(); };
};
Data.cpp:
#include "stdafx.h"
#include "Data.h"
Data::Data(DWORD i_index,
const std::string & i_line,
ULONG i_time)
: index(i_index)
, line(i_line)
, time(i_time)
{
}
Data::~Data()
{
}
Main.cpp的:
#include "stdafx.h"
#include "boost/regex.hpp"
#include "boost/lexical_cast.hpp"
#include "boost\filesystem.hpp"
#include <string>
#include <fstream>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <climits>
#include <ctime>
#include <queue>
#include "Data.h"
namespace fs = boost::filesystem;
static const boost::regex expression(R"(^(?:(?:TIME\s=\s\d{4}\/\d{2}\/\d{2}\s)|(?:@))([0-9:.,]+))");
static const boost::regex nameFileEx(R"(^[\d\-\_]+(\w+\s?\w+|\w+))");
static const std::string path("E:\\2017-02-01");
//static const std::string path("E:\\TestLog");
unsigned long time2Milleseconds(const std::string & time)
{
int a, b, c, d;
if (sscanf_s(time.c_str(), "%d:%d:%d,%d", &a, &b, &c, &d) >= 3)
return a * 3600000 + b * 60000 + c * 1000 + d;
}
void initializeHeap(std::ifstream & ifs, std::priority_queue<Data, std::vector<Data>, Compare> & myHeap, const int index)
{
ULONG time;
std::string line;
boost::smatch what;
bool match = false;
while (!match && std::getline(ifs, line))
{
if (boost::regex_search(line, what, expression))
{
time = time2Milleseconds(what[1]);
myHeap.push(Data(index, line, time));
match = true;
}
}
}
void checkRegex(std::vector<std::pair<std::ifstream, std::string>> & vifs, std::priority_queue<Data, std::vector<Data>, Compare> & myHeap, ULONG time, const int index)
{
std::string line;
boost::smatch what;
if (std::getline(vifs[index].first, line))
{
if (boost::regex_search(line, what, expression))
{
time = time2Milleseconds(what[1]);
}
myHeap.push(Data(index, line, time));
}
}
void mergeFiles(std::vector<std::pair<std::ifstream, std::string>> & vifs, std::priority_queue<Data, std::vector<Data>, Compare> & myHeap, std::ofstream & file)
{
int index = 0;
ULONG time = 0;
while (!myHeap.empty())
{
index = myHeap.top().getIndex();
time = myHeap.top().getTime();
file << myHeap.top().getLine() << " --> " << vifs[index].second << "\n";
myHeap.pop();
checkRegex(vifs, myHeap, time, index);
}
}
int main()
{
clock_t begin = clock();
int cnt = std::count_if(fs::directory_iterator(path), fs::directory_iterator(), static_cast<bool(*)(const fs::path&)>(fs::is_regular_file));
std::priority_queue<Data, std::vector<Data>, Compare> myHeap;
std::vector<std::pair<std::ifstream, std::string>> vifs(cnt);
int index = 0;
boost::smatch what;
std::string file;
for (fs::directory_iterator d(path); d != fs::directory_iterator(); ++d)
{
if (fs::is_regular_file(d->path()))
{
file = d->path().filename().string();
if (boost::regex_search(file, what, nameFileEx))
{
vifs[index] = std::make_pair(std::ifstream(d->path().string()), what[1]);
initializeHeap(vifs[index].first, myHeap, index);
++index;
}
}
}
std::ofstream filename(path + "\\TestLog.txt");
mergeFiles(vifs, myHeap, filename);
filename.close();
clock_t end = clock();
double elapsed_secs = double(end - begin) / CLOCKS_PER_SEC;
std::cout << "Elapsed time = " << elapsed_secs << "\n";
return 0;
}
完成所有这些工作后,我意识到昨天我在Debug中运行程序。启动Release中的两个实现,我得到了以下结果:
因此,或者我对堆结构的实现没有进行优化,或者两个实现在运行时是相同的。
我还能做些什么来加快执行速度吗?
答案 0 :(得分:2)
这可以更快地完成并且内存不足。先考虑一下:
N
行在内存中。)N
行,输出它。如果M
是输出文件的长度(即所有日志的总长度),那么简单的实现将是O(N * M)
。
但是,通过使用堆可以改善上述情况,从而将时间缩短到O(M log N)
。也就是说,将N
内存中的元素放在堆上。从顶部弹出以输出最小元素。然后,当您读取新行时,只需将该行放回堆中。