Binarisation
in [Matlab]
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From: Satish on 15 Mar 2010 09:22 Hi all, I have a plot from X=1:10 and Y=[-1.6209 1.3306 0.7849 0.1681 -0.3521 -0.6280 -0.6779 -0.9908 -1.6466 -2.1471 ]. I would like to convert the resulting plot into a matrix where corresponding point in y-axis is replaced by 1 and rest with 0. resulting matrix can be like z=1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 ........ Hope i will get the code to do like this. Thank you
From: dpb on 15 Mar 2010 10:09 Satish wrote: > Hi all, > > I have a plot from X=1:10 and Y=[-1.6209 1.3306 0.7849 0.1681 > -0.3521 -0.6280 -0.6779 -0.9908 -1.6466 -2.1471 ]. > I would like to convert the resulting plot into a matrix where > corresponding point in y-axis is replaced by 1 and rest with 0. > > resulting matrix can be like > z=1 0 0 0 0 0 0 0 0 > 1 0 0 0 0 0 0 0 0 > 0 1 0 0 0 0 0 0 0 > 0 0 1 0 0 0 0 0 0 > 0 0 0 1 0 0 0 0 0 > 0 0 0 0 1 0 0 0 0 > ........ .... If you mean lineprinter-type plotting, there are many historical routines; I'd presume there would be quite a few that have code posted that could be found on the net that would show the way. Short on time; the general way I'd approach it in quick 'n dirty fashion would be to scale the values to the range of the width of the array and use round() and/or floor() to convert to an integer that would be the index for each entry. --
From: Satish on 15 Mar 2010 10:39 dpb <none(a)non.net> wrote in message <hnlf5j$2qk$1(a)news.eternal-september.org>... > Satish wrote: > > Hi all, > > > > I have a plot from X=1:10 and Y=[-1.6209 1.3306 0.7849 0.1681 > > -0.3521 -0.6280 -0.6779 -0.9908 -1.6466 -2.1471 ]. > > I would like to convert the resulting plot into a matrix where > > corresponding point in y-axis is replaced by 1 and rest with 0. > > > > resulting matrix can be like > > z=1 0 0 0 0 0 0 0 0 > > 1 0 0 0 0 0 0 0 0 > > 0 1 0 0 0 0 0 0 0 > > 0 0 1 0 0 0 0 0 0 > > 0 0 0 1 0 0 0 0 0 > > 0 0 0 0 1 0 0 0 0 > > ........ > ... > > If you mean lineprinter-type plotting, there are many historical > routines; I'd presume there would be quite a few that have code posted > that could be found on the net that would show the way. > > Short on time; the general way I'd approach it in quick 'n dirty fashion > would be to scale the values to the range of the width of the array and > use round() and/or floor() to convert to an integer that would be the > index for each entry. > > --Hi dpb, Yeah i used scale the value before and used the rounding function as of below but that give the result for positive values of Y. But here I have negative values so I can't scale it. If you have any idea than plz tell me. Also you have told me about many historical routines for this kind of plotting but i didn't find the link. So it would be helpful for me if you can give some links related to this. My previous code was as below:(but in previous code there was not any negative value of Y so it was working but it stops working now as it face negative value) X=1:10; Y=[-1.6209 1.3306 0.7849 0.1681 -0.3521 -0.6280 -0.6779 -0.9908 -1.6466 -2.1471 ] temp=zeros(); for i=1:length(y) value=round((y(i)/max(y))*10); temp(value,i)=1; end
From: dpb on 15 Mar 2010 16:06 Satish wrote: .... > ... But here I have negative values so I can't scale it. .... Of course you can; it's simply a linear transformation -- you have two points (min/max values) you want to scale to another range. That boils down to solving two equations in two unknowns... --
From: Satish on 16 Mar 2010 03:21
dpb <none(a)non.net> wrote in message <hnm43d$7u9$1(a)news.eternal-september.org>... > Of course you can; it's simply a linear transformation -- you have two > points (min/max values) you want to scale to another range. That boils > down to solving two equations in two unknowns... > > -- Hi dpb, Thanks , Finally i made it with linear transformation using following formula. value=((y(i) + abs(min(y)))/(max(y)-min(y)))*100 If you have also any other formula than let me know. |