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Re: Modifications to hist.m
From: |
Paul Kienzle |
Subject: |
Re: Modifications to hist.m |
Date: |
Sat, 15 Mar 2003 21:43:01 -0500 |
User-agent: |
Mozilla/5.0 (Windows; U; Win 9x 4.90; en-US; rv:1.3a) Gecko/20021212 |
Andy Adler wrote:
I propose the following patch to hist.m;
it results in about 2.5x speedup.
At one point a rewrote hist to not have any
loops. I was doing a whole lot of really large
histograms (e.g., 100000 values drawn from
a poisson distribution --- I had to rewrite the
poisson generator too ;-)
Please check it out and tell me if it is any faster
than what you've got. I think it might be slower
if you have a histogram with a lot of empty bins.
Thanks,
Paul Kienzle
address@hidden
## 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 (is_vector (y))
max_val = max (y);
min_val = min (y);
else
error ("hist: first argument must be a vector");
endif
if (nargin == 1)
x = 10;
endif
if (is_scalar (x))
n = x;
if (n <= 0 || n != fix(n))
error ("hist: number of bins must be a positive integer");
endif
delta = (max_val - min_val) / n / 2;
x = linspace (min_val+delta, max_val-delta, n);
freq = zeros(1,n);
q = sort(y(:).');
L = length(q);
if (q(1) == q(L))
freq(n) = L;
else
q = (q - q(1))/(q(L)-q(1))/(1+eps); # set y-range to [0,1)
q = fix(q*n); # split into n bins
same = ( q == [q(2:L),-Inf] ); # true if neighbours are in the same bin
q = q(~same); # q lists the 'active' bins (0-origin)
f = cumsum(same); # cumulative histogram
f = f(~same);
f = [f(1), diff(f)] + 1; # cumulative histogram -> histogram
## we need to add 1 since we did not count the point at the
## boundary between bins (it was turned into zero)
## distribute f to the active bins, leaving the remaining bins empty
freq(q+1) = f;
endif
elseif (is_vector (x))
tmp = sort (x);
if (any (tmp != x))
warning ("hist: bin values not sorted on input");
x = tmp;
endif
n = length(x);
cutoff = ( x(1:n-1) + x(2:n) ) / 2; # find bin boundaries
[s, idx] = sort ( [cutoff(:); y(:)] ); # put elements between boundaries
chist = cumsum(idx>n); # integrate over all elements
chist = [chist(idx<n); chist(length(chist))]; # keep totals at boundaries
freq = [chist(1); diff(chist) ]; # differentiate for histogram
else
error ("hist: second argument must be a scalar or a vector");
endif
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