问题描述
我目前正在用 C++ 编写一个程序,其中包括读取大量大文本文件.每个都有 ~400.000 行,在极端情况下每行 4000 个或更多字符.只是为了测试,我使用 ifstream 和 cplusplus.com 提供的实现读取了其中一个文件.花了大约 60 秒,这太长了.现在我想知道,有没有一种直接的方法可以提高阅读速度?
I am currently writing a program in c++ which includes reading lots of large text files. Each has ~400.000 lines with in extreme cases 4000 or more characters per line. Just for testing, I read one of the files using ifstream and the implementation offered by cplusplus.com. It took around 60 seconds, which is way too long. Now I was wondering, is there a straightforward way to improve reading speed?
我使用的代码或多或少是这样的:
edit: The code I am using is more or less this:
string tmpString;
ifstream txtFile(path);
if(txtFile.is_open())
{
while(txtFile.good())
{
m_numLines++;
getline(txtFile, tmpString);
}
txtFile.close();
}
编辑 2:我读取的文件只有 82 MB.我主要说可以达到4000,因为我认为可能需要知道才能进行缓冲.
edit 2: The file I read is only 82 MB big. I mainly said that it could reach 4000 because I thought it might be necessary to know in order to do buffering.
编辑 3:感谢大家的回答,但鉴于我的问题,似乎没有太大的改进空间.我必须使用 readline,因为我想计算行数.将 ifstream 实例化为二进制也不会使读取速度更快.我会尽可能地将它并行化,这至少应该可行.
edit 3: Thank you all for your answers, but it seems like there is not much room to improve given my problem. I have to use readline, since I want to count the number of lines. Instantiating the ifstream as binary didn't make reading any faster either. I will try to parallelize it as much as I can, that should work at least.
编辑 4:显然我可以做一些事情.非常感谢 sehe 在这方面投入了这么多时间,我非常感谢!=)
edit 4: So apparently there are some things I can to. Big thank you to sehe for putting so much time into this, I appreciate it a lot! =)
推荐答案
更新:请务必查看初始答案下方的(令人惊讶的)更新
Updates: Be sure to check the (surprising) updates below the initial answer
内存映射文件对我很有用1:
Memory mapped files have served me well1:
#include <boost/iostreams/device/mapped_file.hpp> // for mmap
#include <algorithm> // for std::find
#include <iostream> // for std::cout
#include <cstring>
int main()
{
boost::iostreams::mapped_file mmap("input.txt", boost::iostreams::mapped_file::readonly);
auto f = mmap.const_data();
auto l = f + mmap.size();
uintmax_t m_numLines = 0;
while (f && f!=l)
if ((f = static_cast<const char*>(memchr(f, '
', l-f))))
m_numLines++, f++;
std::cout << "m_numLines = " << m_numLines << "
";
}
这应该很快.
如果它可以帮助您测试这种方法,这里有一个版本 使用 mmap
直接而不是使用 Boost:
In case it helps you test this approach, here's a version using mmap
directly instead of using Boost: see it live on Coliru
#include <algorithm>
#include <iostream>
#include <cstring>
// for mmap:
#include <sys/mman.h>
#include <sys/stat.h>
#include <fcntl.h>
const char* map_file(const char* fname, size_t& length);
int main()
{
size_t length;
auto f = map_file("test.cpp", length);
auto l = f + length;
uintmax_t m_numLines = 0;
while (f && f!=l)
if ((f = static_cast<const char*>(memchr(f, '
', l-f))))
m_numLines++, f++;
std::cout << "m_numLines = " << m_numLines << "
";
}
void handle_error(const char* msg) {
perror(msg);
exit(255);
}
const char* map_file(const char* fname, size_t& length)
{
int fd = open(fname, O_RDONLY);
if (fd == -1)
handle_error("open");
// obtain file size
struct stat sb;
if (fstat(fd, &sb) == -1)
handle_error("fstat");
length = sb.st_size;
const char* addr = static_cast<const char*>(mmap(NULL, length, PROT_READ, MAP_PRIVATE, fd, 0u));
if (addr == MAP_FAILED)
handle_error("mmap");
// TODO close fd at some point in time, call munmap(...)
return addr;
}
<小时>
更新
通过查看 GNU coreutils wc
的源代码,我发现了我可以从中挤出的最后一点性能.令我惊讶的是,使用以下改编自 wc
的(大大简化的)代码 运行上述内存映射文件的时间约为 84%:
Update
The last bit of performance I could squeeze out of this I found by looking at the source of GNU coreutils wc
. To my surprise using the following (greatly simplified) code adapted from wc
runs in about 84% of the time taken with the memory mapped file above:
static uintmax_t wc(char const *fname)
{
static const auto BUFFER_SIZE = 16*1024;
int fd = open(fname, O_RDONLY);
if(fd == -1)
handle_error("open");
/* Advise the kernel of our access pattern. */
posix_fadvise(fd, 0, 0, 1); // FDADVICE_SEQUENTIAL
char buf[BUFFER_SIZE + 1];
uintmax_t lines = 0;
while(size_t bytes_read = read(fd, buf, BUFFER_SIZE))
{
if(bytes_read == (size_t)-1)
handle_error("read failed");
if (!bytes_read)
break;
for(char *p = buf; (p = (char*) memchr(p, '
', (buf + bytes_read) - p)); ++p)
++lines;
}
return lines;
}
<小时>
1 见例如这里的基准:如何解析空格分隔在 C++ 中快速浮动?
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