"inverse" histogram
in [Matlab]
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From: Jonathan on 23 Feb 2010 04:30 what is the best way to 'invert' an histogram, i.e., x = [1, 1, 2, 2, 2, 2, 3] n = hist(x, 1:3) => n = [2, 4, 1] how can I generate 'x' from 'n' (without loops of course) ? thanks in advance, Jonathan
From: Oleg Komarov on 23 Feb 2010 04:58 Jonathan <jonathan.rubin(a)gmail.com> wrote in message <8bfd8f88-70f4-4ae8-8988-7e4a218b8140(a)x22g2000yqx.googlegroups.com>... > what is the best way to 'invert' an histogram, i.e., > > > x = [1, 1, 2, 2, 2, 2, 3] > > n = hist(x, 1:3) => n = [2, 4, 1] > > how can I generate 'x' from 'n' (without loops of course) ? > > thanks in advance, > Jonathan vakues = 1:3; % Fastest invH1 = cell2mat(arrayfun(@(x,y) repmat(y,1,x), n,values,'un',0)); % Another alternative invH2 = cell2mat(arrayfun(@(x,y) kron(y,ones(1,x)), n,values,'un',0)); % Slowest invH3 = zeros(1, sum(n)); pos = cumsum([0 n]); for ii = 1:numel(values) invH3(pos(ii)+1:pos(ii+1)) = repmat(values(ii),1,n(ii)); end Oleg
From: Jos (10584) on 23 Feb 2010 05:08 Jonathan <jonathan.rubin(a)gmail.com> wrote in message <8bfd8f88-70f4-4ae8-8988-7e4a218b8140(a)x22g2000yqx.googlegroups.com>... > what is the best way to 'invert' an histogram, i.e., > > > x = [1, 1, 2, 2, 2, 2, 3] > > n = hist(x, 1:3) => n = [2, 4, 1] > > how can I generate 'x' from 'n' (without loops of course) ? > > thanks in advance, > Jonathan This process is called run-length decoding. Search the FEX for some excellent tools on this, and especially look at: http://www.mathworks.com/matlabcentral/fileexchange/6436 n = [2 4 1] ; rude([2 4 1],1:numel(n)) % -> [1 1 2 2 2 2 3] hth Jos
From: ImageAnalyst on 23 Feb 2010 09:15 Jonathan: You can get the values of x back (using methods like the others showed you), but you can't get their positions back. This is of course because x1=[1, 1, 2, 2, 2, 2, 3] and x2=[3, 1, 2, 2, 1, 2, 2] both have the same histogram (what you're non-descriptively calling "n"). They have the same elements, just in different positions, and "n" will be identical.
From: Jos (10584) on 23 Feb 2010 13:01
ImageAnalyst <imageanalyst(a)mailinator.com> wrote in message <dcdc97c3-5d01-4575-a42c-8d2bf8cbb57e(a)y7g2000vbb.googlegroups.com>... > Jonathan: > You can get the values of x back (using methods like the others showed > you), but you can't get their positions back. This is of course > because > x1=[1, 1, 2, 2, 2, 2, 3] > and > x2=[3, 1, 2, 2, 1, 2, 2] > both have the same histogram (what you're non-descriptively calling > "n"). They have the same elements, just in different positions, and > "n" will be identical. Right! but thist might be of interest: x = [3 1 2 2 1 2 2] e = [1 2 3] [n, ix] = histc(x,e) x2 = e(ix) hth Jos |