Subj : Re: any function to handle this kind of counting?
To   : comp.programming
From : Gerry Quinn
Date : Sat Jul 30 2005 03:37 pm

In article <dcdi1t$g7m$1@justice.itsc.cuhk.edu.hk>, a@cuhk.edu.hk 
says...
> Dear all,
> 
> i have generated a 2000 X 1 vector, e.g.

> and then i want to count, say, for "1", there are two A's, one E, ...
> 
> it seems that a for loop can do the job but as we don't know how many 
> instances there are, i.e. don't know A, C, E and then what will occur in 
> this corresponding vector, is there any function which is devoted to handle 
> this kind of counting problem? thx!! 

Well, I guess there are libraries for linked lists, vectors etc.  There 
are also sorting functions that can often make such problems easier.

The following is untested C++ to count them directly.

I will treat both vectors as containing ints though I'll call them 
'nums' and 'lets'.  It's trivial to convert the second to and from 
chars if required (e.g. just use val = letter - 'A'.)  Besides, they 
could be enums for all I know.  The algorithm should be modified if 
speed is at a premium and there are a large number of different letter 
values.

'targ' is the required number, 1 in the example.  One could modify it 
easily to return a complete solution for all numbers as a vector of 
vectors, using a similar methodology.

If the data are in fact C-style arrays rather than vectors, it's just 
as easy to implement, but using an STL vector for the result makes the 
solution very easy in C++: 

#include <vector>
using namespace std;

class Datum
{
public:
  int letter;
  int count;
};

vector< Datum > result;

ASSERT( nums.size() == lets.size() );

for ( int iLet = 0; iLet < lets.size(); iLet++ )
{
  if ( nums[ iLet ] != targ )
  {
    continue;
  }
  int iRes = 0;
  while ( iRes < result.size() )
  {
    if ( result[ iRes ].letter == lets[ iLet ] )
    {
      break;
    }
    iRes++;   
  }    
  if ( iRes == result.size() )
  {
    Datum datum;
    datum.letter = lets[ iLet ];
    datum.count = 0;
    result.push_back( datum );
  }
  result[ iRes ].count++;
}

- Gerry Quinn

.