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Text File Read Question

Posted on 2013-05-17
3
450 Views
Last Modified: 2013-05-23
Hello,
I have a humongous text file with elements like this... (This is an example only and NOT the real file)...Note that in each item... First half before the underscore is the major item name and the other half is the minor descriptions.

item1_data1
item1_data2
item1_data3
.............
item2_data1
item2_data2
item2_data3
.............
item3_data1
item3_data2
item3_data3
............

There are about 75000 items in this file.

I am writing a C++ class to pickup only the major name of each item. I.E. My result from the above humongous file should be...

item1
item2
item3
.......

I know thare are tons of techniques out there. What is the real efficient method I should use so that my result is produced in nano seconds :-) (serioulsly efficiency is extreamly important for me)
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Comment
Question by:prain
3 Comments
 
LVL 86

Expert Comment

by:jkr
ID: 39175364
You'll have to read the entire file anyway, so there's not much room for improvement. The simplest way I can imagine would be to

#include <fstream>
#include <string>
#include <list>

using namespace std;

//...

string line;
size_t pos;
list<string> items;
ifstream is("file.txt");

if (!is.is_open()) {

  // error, no such file
}

while (!is.eof()) {

  getline(is,line);

  if (string::npos == (pos = line.find('_'))) {

    // error, malformed line w/o underscore
  }

  items.push_back(line.substr(0,pos));
}

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LVL 33

Accepted Solution

by:
sarabande earned 200 total points
ID: 39185173
the nanoseconds is not realistic, even for fast ssd or flash storage, each file access to a new file not currently in cache would need milliseconds to position at file-begin and read all blocks to memory, manage the cache, handle overhead of the filesystem, schedule the thread, ... if the file was not stored contiguously, all times must be multiplied with the number of file pieces. same applies to debug mode which also would generate extra times.

generally if you read a file in total to memory in binary mode you normally can halve the reading times for a file of significant size.

#include <sys/stat.h>
#include <fstream>
#include <string>
...
struct stat filestatus = { 0 };
if (stat(szfilepath, &filestatus) == 0)
{
     std::ifstream file(szfilepath, std::ios::binary | std::ios::in);
     if (file)
     {
           std::string buf(filestatus.st_size+1, '\0');
           if (file.read(&buf[0], filestatus.st_size))
           {
                  std::string crlf = "\r\n";  
                  std::string line;
                  buf += crlf; // add carriagereturn-linefeed for easier parsing
                  size_t pos, lpos = 0;
                  while ((pos = buf.find(crlf, lpos)) != std::string::npos)
                  {
                      if (pos > lpos)
                      {
                           line = buf.substr(lpos, pos - lpos);
                           // here you have one line extracted
                           ...
                      }
                      lpos = pos + crlf.length();

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the above code would need contiguous memory of 75,000 times (average line length + 2). That can be a problem on a low-memory or busy system. if that could be a problem you might think of reading the file in - say - 64k chunks.

 
At the ... you could extract the first string part from line and add it to a std::set<std::string> container. a set would add a new string only if it is not already in the set.    

std::set<std::string> items;
...
       size_t undl = line.find('_');
       if (undl != std::string::npos)
       {
           line.resize(undl);   // truncate line string
           items.insert(line);

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finally, you could iterate the set to get all entries (sorted alphabetically).

std::set<std::string>::iterator i;
for (i = items.begin(); i != items.end(); ++i)
{
      std::string & item = *i;  // get a reference of the current item in set 

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note, a std::set or a std::map would handle duplicates. a std::list or std::vector would not. if using one of the latter you would need to sort after filling and remove or skip duplicates after sorting. the last method could be faster if there are only few duplicates.

Sara
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Author Closing Comment

by:prain
ID: 39192341
Thanks
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