Mercurial > hg > octave-nkf
view scripts/plot/hist.m @ 8710:739141cde75a ss-3-1-52
fix typo in Array-f.cc
author | Jaroslav Hajek <highegg@gmail.com> |
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date | Mon, 09 Feb 2009 21:51:31 +0100 |
parents | cadc73247d65 |
children | eb63fbe60fab |
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## Copyright (C) 1994, 1995, 1996, 1997, 1998, 1999, 2000, 2002, 2003, ## 2004, 2005, 2006, 2007 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 3 of the License, 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, see ## <http://www.gnu.org/licenses/>. ## -*- 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. With one matrix input argument, plot a hystogram where ## each bin contains a bar per input column. ## ## 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. ## @seealso{bar} ## @end deftypefn ## Author: jwe function [nn, xx] = hist (y, varargin) if (nargin < 1) print_usage (); 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 real valued"); endif iarg = 1; if (nargin == 1 || ischar (varargin{iarg})) n = 10; x = [0.5:n]'/n; x = x * (max_val - min_val) + ones(size(x)) * min_val; else ## nargin is either 2 or 3 x = varargin{iarg++}; if (isscalar (x)) n = x; if (n <= 0) error ("hist: number of bins must be positive"); endif x = [0.5:n]'/n; x = x * (max_val - min_val) + ones (size (x)) * min_val; elseif (isreal (x)) if (isvector (x)) x = x(:); endif tmp = sort (x); if (any (tmp != x)) warning ("hist: bin values not sorted on input"); x = tmp; endif else error ("hist: second argument must be a scalar or a vector"); endif endif ## Avoid issues with integer types for x and y x = double (x); y = double (y); cutoff = (x(1:end-1,:) + x(2:end,:)) / 2; n = rows (x); y_nc = columns (y); if (n < 30 && columns (x) == 1) ## The following algorithm works fastest for n less than about 30. chist = zeros (n+1, y_nc); for i = 1:n-1 chist(i+1,:) = sum (y <= cutoff(i)); endfor chist(n+1,:) = sum (! isnan (y)); 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 ([y; repmat(cutoff, 1, y_nc)]); len = rows (y); chist = cumsum (idx <= len); chist = [(zeros (1, y_nc)); (reshape (chist(idx > len), rows (cutoff), y_nc)); (chist(end,:) - sum (isnan (y)))]; endif freq = diff (chist); if (nargin > 2 && ! ischar (varargin{iarg})) ## Normalise the histogram. norm = varargin{iarg++}; freq = freq / rows (y) * norm; endif if (nargout > 0) if (arg_is_vector) nn = freq'; xx = x'; else nn = freq; xx = x; endif elseif (size (freq, 2) != 1) bar (x, freq, 0.8, varargin{iarg:end}); else bar (x, freq, 1.0, varargin{iarg:end}); endif endfunction %!test %! [nn,xx]=hist([1:4],3); %! assert(xx, [1.5,2.5,3.5]); %! assert(nn, [2,1,1]); %!test %! [nn,xx]=hist([1:4]',3); %! assert(xx, [1.5,2.5,3.5]); %! assert(nn, [2,1,1]); %!test %! [nn,xx]=hist([1 1 1 NaN NaN NaN 2 2 3],[1 2 3]); %! assert(xx, [1,2,3]); %! assert(nn, [3,2,1]); %!test %! [nn,xx]=hist([[1:4]',[1:4]'],3); %! assert(xx, [1.5;2.5;3.5]); %! assert(nn, [[2,1,1]',[2,1,1]']); %!assert(hist(1,1),1); %!test %! for n = [10, 30, 100, 1000] %! assert(sum(hist([1:n], n)), n); %! assert(sum(hist([1:n], [2:n-1])), n); %! assert(sum(hist([1:n], [1:n])), n); %! assert(sum(hist([1:n], 29)), n); %! assert(sum(hist([1:n], 30)), n); %! endfor %!test %! assert (size (hist(randn(750,240), 200)), [200,240]);