Mercurial > hg > octave-nkf
view scripts/plot/hist.m @ 4772:9eed17b2c8d1
[project @ 2004-02-16 17:57:34 by jwe]
author | jwe |
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date | Mon, 16 Feb 2004 17:57:34 +0000 |
parents | 1541c3ed2c93 |
children | 198f3712c692 |
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## 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, 59 Temple Place - Suite 330, Boston, MA ## 02111-1307, USA. ## -*- texinfo -*- ## @deftypefn {Function File} {} hist (@var{y}, @var{x}, @var{norm}) ## Produce histogram counts or plots. ## ## With one vector input argument, plot a histogram of the values with ## 10 bins. The range of the histogram bins is determined by the range ## of the data. ## ## Given a second scalar argument, use that as the number of bins. ## ## Given a second vector argument, use that as the centers of the bins, ## with the width of the bins determined from the adjacent values in ## the vector. ## ## If third argument is provided, the histogram is normalised such that ## the sum of the bars is equal to @var{norm}. ## ## Extreme values are lumped in the first and last bins. ## ## 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{bar} ## Author: jwe function [nn, xx] = hist (y, x, norm) if (nargin < 1 || nargin > 3) usage ("[nn, xx] = hist (y, x, norm)"); endif if (isvector (y)) max_val = max (y); min_val = min (y); else error ("hist: first argument must be a vector"); endif if (nargin == 1) n = 10; delta = (max_val - min_val) / n / 2; x = linspace (min_val+delta, max_val-delta, n); cutoff = x + delta; else ## nargin is either 2 or 3 if (isscalar (x)) n = x; if (n <= 0) error ("hist: number of bins must be positive"); endif delta = (max_val - min_val) / n / 2; x = linspace (min_val+delta, max_val-delta, n); cutoff = x + delta; elseif (isvector (x)) tmp = sort (x); if (any (tmp != x)) warning ("hist: bin values not sorted on input"); x = tmp; endif cutoff = (x(1:end-1) + x(2:end)) / 2; n = length (x); else error ("hist: second argument must be a scalar or a vector"); endif endif if (n < 30) ## The following algorithm works fastest for n less than about 30. chist = [zeros(n,1); length(y)]; for i = 1:n-1 chist(i+1) = sum (y < cutoff(i)); endfor else ## The following algorithm works fastest for n greater than about 30. ## Put cutoff elements between boundaries, integrate over all ## elements, keep totals at boundaries. [s, idx] = sort ([cutoff(:); y(:)]); chist = cumsum(idx>n); chist = [0; chist(idx<n); chist(end)]; endif freq= diff(chist)'; if (nargin == 3) ## Normalise the histogram. freq = freq / length (y) * norm; endif if (nargout > 0) nn = freq; xx = x; else bar (x, freq); endif endfunction