Re: histc in Octave?

[Aakash Dalwani]

Francesco Potorti` wrote:
> > "help histc
> > HISTC Histogram count.
> >    N = HISTC(X,EDGES), for vector X, counts the number of values in X
> >    that fall between the elements in the EDGES vector (which must contain
> >    monotonically non-decreasing values).  N is a LENGTH(EDGES) vector
> >    containing these counts.
> >
> >    N(k) will count the value X(i) if EDGES(k) <= X(i) < EDGES(k+1).  The
> >    last bin will count any values of X that match EDGES(end).  Values
> >    outside the values in EDGES are not counted."
> >     
>
> The hist function does a very similar function, even if it is not
> equal.  If you have
>   histc (x, e)
> in Matlab, you can obtain very similar results by using
>   hist (x, (e(1:len-1)+e(2:len))/2)
> in Octave.  However, if you want exactly the same results, you could use
> this one, that I adapted quickly (15 minutes, no debugging, no guarantee)
> from the Octave 2.9 hist function:
>
> ## Copyright (C) 1996, 1997 John W. Eaton
> ##
> ## This file is part of Octave.
> ##
> ## Octave is free software; you can redistribute it and/or modify it
> ## under the terms of the GNU General Public License as published by
> ## the Free Software Foundation; either version 2, or (at your option)
> ## any later version.
> ##
> ## Octave is distributed in the hope that it will be useful, but
> ## WITHOUT ANY WARRANTY; without even the implied warranty of
> ## MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the GNU
> ## General Public License for more details.
> ##
> ## You should have received a copy of the GNU General Public License
> ## along with Octave; see the file COPYING.  If not, write to the Free
> ## Software Foundation, Inc., 51 Franklin Street, Fifth Floor, Boston, MA
> ## 02110-1301, USA.
>
> ## -*- texinfo -*-
> ## @deftypefn {Function File} {} histc (@var{y}, @var{edges})
> ## Produce histogram counts or plots.
> ##
> ## Plot a histogram of the values in @var{y}, using @var{edges} as the
> ## edges of the bins: element @var{k} in the output vector counts
> ## elements of @var{y} satisfying @math{y(k) <= x < y(k+1)}.  The last
> ## element of the output vector will also count elements of @var{y}
> ## equal to @var{edges(end)}.
> ##
> ## With two output arguments, produce the values @var{nn} and @var{xx} such
> ## that @code{bar (@var{xx}, @var{nn})} will plot the histogram.
> ## @end deftypefn
> ##
> ## @seealso{hist}
>
> function freq = histc (y, x)
>
>   if (nargin < 1 || nargin > 2)
>     usage ("hist (y, x)");
>   endif
>
>   arg_is_vector = isvector (y);
>
>   if (rows (y) == 1)
>     y = y(:);
>   endif
>
>   if (isreal (y))
>     max_val = max (y);
>     min_val = min (y);
>   else
>     error ("hist: first argument must be a vector");
>   endif
>
>   if (nargin > 1)
>     ## nargin is 2
>     if (isreal (x))
>       if (isvector (x))
>         x = x(:);
>       endif
>       tmp = sort (x);
>       if (any (tmp != x))
>         warning ("hist: edge values not sorted on input");
>         x = tmp;
>       endif
>     else
>       error ("hist: second argument must be a vector");
>     endif
>   endif
>
>   n = rows (x) - 1;
>   freq = zeros (n, columns (y));
>   for i = 1:n
>     freq(i,:) = sum (y <= x(i) && y < x(i+1));
>   endfor
>   freq(n,:) += sum (y == x(end));
>
> endfunction
>   

Thanks for your time and prompt reply. Your gesture is highly appreciated :)

However, hist (x, (e(1:len-1)+e(2:len))/2), gives different results, 
because even though the hist documentation says that it takes the second 
vector as the center points, but I the actual behaviour is that for the 
first bin it counts values from -inf to e(1), so then it doesn't 
actually count values around the center point.

The function didn't work, it gave me all zeros :) but then you had said 
no guarantee.


I figured the solution though to serve my purpose:
Based on the documentation of Matlab's histc, the following piece of 
code does exactly the same thing as histc(X,e)

% X is the vector, e is a vector with the boundaries
freq = zeros(length(e),1);

for i=1:length(X)
 for k=1:length(e)-1
   if(e(k) <= X(i) && X(i) < e(k+1))
     freq(k)=freq(k)+1;
   end
 end

 if ( X(i)== e(end))
   freq(end) = freq(end)+1;
 end
end


John, if you think it is valuable to add this function, let me know

Once again, thanks for your time and reply, I really appreciate your 
gesture,
Thanks,
Aakash