Can i adjust each thickness of each histogram bar of an image differently ? and can i plot or extract an exact value of the x axis in the histogram ?

I want to make the histogram of an Image but i want to focus on a specific zone ( around the value 2048 in the x axis) . So can i plot the histogram and adjust the bars so i can see clearly this value ? and i want the bars around this value to be narrower then the others .

Can anyone help me please because i have been trying without getting the result that i want. Thank you

Answer 1

[dpb]

The bar chart in Matlab is extraordinarily difficult to do anything with that's "out of the box".

To your specific question, the bars are a single object for each column so the width is a fixed single value for the series. The only way I can think to change this without writing something specific would be to use the ability of bar to treat NaN values as placeholders. To do this you would take the range of bins that are wanted to be different and substitute NaN for their values in one column and then do the obverse in a second vector--keep those values in their bins and replace all other with NaN. Then use bar twice, with hold on between the two calls and you'll have two bar plots on the same axes that can set the properties on to differentiate.

>> y=randn(1,1000);  % sample data
>> [n,x]=hist(y,20); % get hist data
>> ix=[8:12];        % center bins to accentuate in histogram
>> n=[n; nan(size(n))];  % make second set same size
>> n(2,ix)=n(1,ix);      % second is center section values
>> n(1,ix)=nan;          % clear first center section (NaN)
>> hB=bar(x,n(1,:));     % outer portion normal bars
>> hold on               % get ready to add center
>> hB(2)=bar(x,n(2,:),0.5);  % with narrower bars
>>

ADDENDUM Per the comment following IA's alternative, I did a File Exchange search and found <fileexchange/1420-scatterbar3> "SCATTERBAR3(X,Y,Z,WIDTH) draws 3-D bars of height Z at locations X and Y with width WIDTH."

ADDENDUM 2 Following up the discussion there wherein it (finally) dawns on me that 2D histogram is bogus, I think the solution you're looking for is

[n,x]=hist(YourImageArray(:),nBins);

and then follow the above example to emphasize chosen area(s).

[Yassine Zaafouri]

thank you for your time . The Problem is that i'am working with an image so the matrix is 2 dimensional so when i want to apply your idea it doesn't work properly . what do you think ?

[dpb]

You'll have to use hist3 and bar3 then, with the desired bins in the two dimensions set as above. I've got meetings; gotta' run so can't try to do a demo now but same idea should work with some extra effort to keep track of "who's who in the zoo" as far as intersecting bin locations.

Answer 2

[Image Analyst]

As an alternative, if you want, you can color each bar independently, like you can make one particular bar red and the others blue, or whatever. Demo attached.

[dpb]

Looks like about the only alternative w/o quite a lot of effort...I just got back and thought would put a demo together quickly but turns out that bar3 won't accept x, y coordinates in general form so I got stuck there...probably there's a workaround but I've never used it before so would take a fair amount of digging it appears.

As a looksee; I looked to see what hist3 does; it computes its own xy coordinates and uses surf internally; looks like a fair hassle to mung on it, too. :(

I'd suggest OP look on File Exchange; maybe somebody else has previously been frustrated and solved the problem before...

[Yassine Zaafouri]

Yes that's the problem with bar3 and hist3 so i can not use them. I will try something else and if nothing works i will color the particular bar . i have to look also for somehow i can make some bars narrower from the others. thank you for your time and effort gentlemens

[Image Analyst]

You could make one counts bin a NaN or zero so it won't plot. Then plot another bar chart on top of it with just a single bar and where you set the 'BarWidth' property of the bar to something less than 1, lik 0.7 or whatever.

[dpb]

That's essentially my first pass but how do you make it work for the 3D version, IA? I don't see that working for OP's situation???

Yassine, did you notice the link I posted as ADDENDUM to Answer above to a File Exchange submission? It appears it'll solve the problem from the description altho I've not tested it...

[Image Analyst]

I don't see why bar3() or a 2D histogram is needed if you want to display the histogram of an image. Unless you just want it to look cool or something. A regular flat bar chart conveys the count information just fine.

[dpb]

As long as only care about overall, agreed...hmmmm, though. I don't do image processing so was figuring there was a reason to keep the dimensional info, to???

[Image Analyst]

When he said "matrix is 2 dimensional" I think he meant the image, not that the histogram counts array was 2D or that the bars are like solid columns, like a Manhattan cityscape.

[dpb]

Well, I was thinking that, but wasn't really thinking about what a histogram of an image would be (as I say, I've never done any image processing at all more than display one now and then :) )...on some reflection, I grok what you're saying. In fact, when I actually slow down and consider I realize a 2D histogram is nuts for an image as the count would be one for each coordinate. :) So, looks like probably what OP is looking for is the (:) syntax to get the overall image histogram instead of N histograms by column that's the default behavior of hist for 2D arrays.

[Yassine Zaafouri]

Good morning gentlemens, So i tried your code above dpb and i adjusted it to fit with my image matrix and it actually worked . But i want to have more precison so how can i define the centers of the bins ? i have a vector with some defined values and i want to use them as centers . So how can i do that by changing the option nBins with the vector of the centers because i actually do not know how many bins i want to plot in general. When I tried to do that i got an erreur : X and Y must have the same size.

[dpb]

The syntax

n = hist(Y,x)

where x is a vector, uses N=length(x) bins centered at x i. In that case, use your x in bar, too, and don't need to return the centers as I did in the example where I just gave a count.