How to Create a Equidistant histogram
显示 更早的评论
% The edge of each bin
Xedges = [10,22,36,49,55,67,77];
Yedges = [21,30,55,70,85,93];
% Obtain the sample count in each bin
X1 = rand(1, 10000) * 100;
Y1 = rand(1, 10000) * 100;
X2 = rand(1, 10000) * 100;
Y2 = rand(1, 10000) * 100;
N(:,:,1) = histcounts2(X1, Y1, Xedges, Yedges);
N(:,:,2) = histcounts2(X2, Y2, Xedges, Yedges);
N = squeeze(mean(N,3));
Hi,
I have a matrix N representing the number of samples in each bin, and I know the boundary (Xedges and Yedges) for each bin. How can I create a histogram for matrix N?
I want the box values (in red) to represent the sample count in the corresponding x1-x2 and y1-y2 bin, and the axis scales to correspond to the interval boundaries. Even though the intervals for each bin are unevenly spaced, I would like the resulting graph to have consistent grid sizes.
11 个评论
If you want to create random integers, why don't you use "randi" instead of "rand" ?
And N is a vector, not a matrix. So I don't know what kind of histogram you want to plot.
Maybe you mean
N = (N1+N2)/2
instead of
N = mean(N1+N2)
?
peng peng
2023-10-8
Torsten
2023-10-8
Did you experiment with "histogram2" ?
peng peng
2023-10-8
Yes, I tried it.
However, I only know how to plot every N, but I don't know how to plot averaged N.
h1=histogram2(X1, Y1,Xedges,Yedges,'DisplayStyle','tile','ShowEmptyBins','on');
h1=histogram2(X2, Y2,Xedges,Yedges,'DisplayStyle','tile','ShowEmptyBins','on');
and I don't know hwo to make every bin have the same width and height, thoug the range of every bin is not same
Torsten
2023-10-8
h=histogram2((X1+X2)/2, (Y1+Y2)/2,Xedges,Yedges,'DisplayStyle','tile','ShowEmptyBins','on');
And manipulating the bin width is cheating :-)
peng peng
2023-10-8
peng peng
2023-10-8
I cannot change the distances between the edges, but I think the histogram is correct.
And I think you shouldn't change the bin sizes in the graphics because they are the main reason for the colors.
% The edge of each bin
Xedges = [10,22,36,49,55,67,77];
Yedges = [21,30,55,70,85,93];
% Obtain the sample count in each bin
X1 = rand(1, 10000) * 100;
Y1 = rand(1, 10000) * 100;
X2 = rand(1, 10000) * 100;
Y2 = rand(1, 10000) * 100;
N(:,:,1) = histcounts2(X1, Y1, Xedges, Yedges);
N(:,:,2) = histcounts2(X2, Y2, Xedges, Yedges);
N = squeeze(mean(N,3));
X = [];
Y = [];
for i = 1:numel(Xedges)-1
for j = 1:numel(Yedges)-1
x = (Xedges(i)+Xedges(i+1))/2;
y = (Yedges(j)+Yedges(j+1))/2;
X = [X;x*ones(round(N(i,j)),1)];
Y = [Y;y*ones(round(N(i,j)),1)];
end
end
histogram2(X,Y,Xedges,Yedges,'DisplayStyle','tile','ShowEmptyBins','on');
colorbar
peng peng
2023-10-8
Torsten
2023-10-8
N1 = histcounts2((X1+X2)/2, (Y1+Y2)/2,Xedges,Yedges);
N(:,:,1) = histcounts2(X1, Y1, Xedges, Yedges);
N(:,:,2) = histcounts2(X2, Y2, Xedges, Yedges);
N2 = squeeze(mean(N,3));
I didn't use this method in my last response, but thanks for your appreciation.
peng peng
2023-10-8
采纳的回答
Well, the prior has some things of academic interest so I won't delete it, but actually when got a moment to poke around, it turns out the histogram is a Child of an ordinary axes so you can do what you want with the ticks and tick labels after all...just not on the histogram itself but its parent. So, everything above is ok, just a couple more steps to finish from where left off..
% The edge of each bin
Xedges = [10,22,36,49,55,67,77];
Yedges = [21,30,55,70,85,93];
% Obtain the sample count in each bin
X1 = randi(100, [10000,1]);
X2 = randi(100, [10000,1]);
Y1 = randi(100, [10000,1]);
Y2 = randi(100, [10000,1]);
N(:,:,1)=histcounts2(X1, Y1, Xedges, Yedges);
N(:,:,2)=histcounts2(X2, Y2, Xedges, Yedges);
N = squeeze(mean(N,3));
X=0:numel(Xedges)-1;
Y=0:numel(Yedges)-1;
hHG2=histogram2('XBinEdges',X,'YBinEdges',Y,'BinCounts',N,'DisplayStyle','tile','ShowEmptyBins','on');
hAx=hHG2.Parent; % get the parent axes holding the histogram
xticks(hAx,X), xticklabels(hAx,Xedges) % per Bruno's admonition, make sure working on intended axes
yticks(hAx,Y), yticklabels(hAx, Yedges)
3 个评论
peng peng
2023-10-8
dpb
2023-10-8
It isn't possible for a histogram to have negative counts and, unfortunately for this purpose, Mathworks enforces that the bin counts have to be nonnegative. If, indeed, N would have negative values in your application then histogram is totally out as a solution -- as, of course, would be @doc:histcounts2 as well.
peng peng
2023-10-9
更多回答（2 个）
% The edge of each bin
Xedges = [10,22,36,49,55,67,77];
Yedges = [21,30,55,70,85,93];
% Obtain the sample count in each bin
X1 = randi(100, [10000,1]);
X2 = randi(100, [10000,1]);
Y1 = randi(100, [10000,1]);
Y2 = randi(100, [10000,1]);
N(:,:,1)=histcounts2(X1, Y1, Xedges, Yedges);
N(:,:,2)=histcounts2(X2, Y2, Xedges, Yedges);
N = squeeze(mean(N,3));
X=0:numel(Xedges)-1;
Y=0:numel(Yedges)-1;
hHG2=histogram2('XBinEdges',X,'YBinEdges',Y,'BinCounts',N,'DisplayStyle','tile','ShowEmptyBins','on');
will put the mean N counts in uniformly-spaced bins; however, you can't then change the bin edges displayed on the axes without changing the spacing which goes with it; unlike a regular axes for which the x,y tick display values are not inextricably tied to the x,y tick values. So, histogram2() doesn't look like a real good match here...
This doesn't get you all the way home, but I've got other commitments at the moment and I don't have a clear solution in mind at the moment -- heatmap would work to get uniform mesh and counts, but it labels the bin centers, not the edges.
You may have to make the labels invisible on the histogram2 axes and then overlay another axes to label...
figure
% using IA's idea...
imagesc(X,Y,flipud(N.')) % images are from upper LH corner instead...
xl=xlim; yl=ylim;
xticks(linspace(xl(1),xl(2),numel(X)))
xticklabels(Xedges)
yticks(linspace(yl(1),yl(2),numel(Y)))
yticklabels(Yedges)
Colors seem a little washed out in comparison, but gets the specifics requested. I'd forgotten how @doc:imagesc expands the pixels to an input x,y instead of just displaying a single pixel...
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