Swapping from hist to histogram

Hi, under the recommendation of Steven Lord, I am trying to update my code and not use hist as the histogram function and instead use histogram

I previously did this:

[counts,xb]=hist(data(:,3),nbins);   %IMHIST ONLY HANDLES 8 & 16 BIT IMAGES, NOT 12BIT  
figure
ax1=subplot(1,2,1);
bar(xb,counts,'b','EdgeColor','b');
grid on

%Now get the max frequency.
mxC=max(counts);
indx=find(counts==mxC);
xmx=xb(indx);
hold on;
lineH1=plot([xmx xmx],ylim,'r--','LineWidth',1);
hold off

So when I swap over to the histogram function, I can get the y-axis values but not the x-axis values

h=histogram(data(:,3),nbins)
counts=h.Values   % frequency (i.e. y-axis)

How do I get the mode of this histogram and plot it?

Thanks Jason

Answer 1

[Star Strider]

One approach:

data = randi(99, 1, 100);
nbins = 25;
h = histogram(data, nbins);
counts = h.BinCounts;
edges = h.BinEdges;
width = h.BinWidth;
ctrs = edges(1:end-1) + width/2;
MaxCountsV = counts >= max(counts);                         % Logical Vector
Desired_Output = [ctrs(MaxCountsV), counts(MaxCountsV)]     % [BinCentre, Counts]

It’s essentially self-documenting with the variable names. The output displays the bin centre corresponding to the maximum count.

You could make this a bit more efficient if you need to. My objective here is to leave a clear trail so you know how I got the result.

[Jason]

Thankyou

[Star Strider]

My pleasure.

[Jason]

Is it too late to ask how to force the number of bins such that each bin is an integer i.e. 1,2,3 all the way upto the max value in my image?

[Star Strider]

It’s never too late, although I’m not at my computer 24x7 so delays are occasionally inevitable.

From the documentation:

• histogram(X,edges) sorts X into bins with the bin edges specified by the vector, edges. Each bin includes the left edge, but does not include the right edge, except for the last bin which includes both edges.

Specifying the bin edges (ranges) is certainly an option. You will have to experiment to get the result you want.

[Jason]

Thank you

[Star Strider]

My pleasure.

[Jason]

So I have tried to play with this. I have created my edges:

edges=[1:max(data)]
h=histogram(data,edges) ;  
counts=h.Values;
bar(edges,counts,'g','EdgeColor','g');

but when I try to bar chart this,

as the length of edges is 1 more than the counts

Error using bar (line 143)
X must be same length as Y.

[Image Analyst]

Look at the lengths of edges and counts. edges has one more element. So you need to decide if you want the left edge, right edge, or center of the bin. To get the left edges:

bar(edges(1:end-1), counts, 'g', 'EdgeColor', 'g');

[Jason]

Thanks

[Steven Lord]

I happened upon this message while doing searching for another message. About the follow-up question about making each integer a different bin, when you call histogram specify the name-value pair 'BinMethod', 'integers' instead of specifying bin edges and histogram will automatically create edges with one integer per bin (unless your data spans too wide a range, in which case the bins will be wider than 1, as stated in the entry for 'integers' in the documentation of the BinMethod argument.)

[Jason]

thankyou for this.