Plot multiple histogram within 1 function

Hi all! I made some histograms, but now I'm trying to make a function that plots multiple histograms at the same time. I want 1 plot with 4 overlapping histograms of different colors and a legend. Can someone please explain to me how to do this? I tried to make a loop (see bottom) but it doesn't seem to work.
histogram_color = [0 .7 .7];
histogram(error1,'normalization','pdf','edgecolor',histogram_color,'facecolor',histogram_color,'facealpha',.5)
x = -90:1:90;
f1 = fitdist(error1,'kernel');
pdf1 = pdf(f1,x);
l1 = line(x,pdf1,'linestyle','-','color',histogram_color,'linewidth',3);
histogram_color = [0 0 .7];
histogram(error2,'normalization','pdf','edgecolor',histogram_color,'facecolor',histogram_color,'facealpha',.5)
x = -90:1:90;
f2 = fitdist(error2,'kernel');
pdf2 = pdf(f2,x);
l2 = line(x,pdf2,'linestyle','-','color',histogram_color,'linewidth',3);
x = -90:1:90;
for i = 1:length(out)
histogram(out(i),'normalization','pdf','edgecolor',histogram_color,'facecolor',histogram_color,'facealpha',.5);
f(i) = fitdist(out(i),'kernel');
pdf(i) = pdf(f(i),x);
l(i) = line(x,pdf(i),'linestyle','-','color',histogram_color,'linewidth',3);
end

1 Kommentar

Adam Danz
Adam Danz am 14 Okt. 2020
You're missing "hold on"
Otherwise the 2nd histogram replaces the first.

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 Akzeptierte Antwort

Sudhakar Shinde
Sudhakar Shinde am 15 Okt. 2020

0 Stimmen

[~,ErrorLength]= size(out);
for i =1:ErrorLength
hold on
histogram(out(:,i),'normalization','pdf')
hold off
end

Weitere Antworten (2)

Steven Lord
Steven Lord am 14 Okt. 2020

1 Stimme

The "Plot Multiple Histograms" example on the documentation page shows how to superimpose two histogram plots using hold on.
MadjeKoe
MadjeKoe am 15 Okt. 2020

0 Stimmen

S = load('/Users/mkoenraad/Desktop/Scriptie/exercise_2/simulated_serial_bias_dataset.mat');
res = S.res;
targets = res(:,7);
responses = res(:,8);
orx = [3,4,5,6];
tgt = [1,2,3,4];
num = numel(orx);
out = cell(1,num);
for k = 1:num
idx = targets==tgt(k);
resp = responses(idx);
ori = res(idx,orx(k));
err = resp - ori;
err(err<-90) = err(err<-90)+180;
err(err>90) = err(err>90)-180;
out{k} = err;
end
out = [out{:}];
ero1 = out(:,1);
ero2 = out(:,2);
ero3 = out(:,3);
ero4 = out(:,4);

9 Kommentare

MadjeKoe
MadjeKoe am 15 Okt. 2020
Is this ok? I don't know how to only add my 'out' variable
histogram_color = [0 .7 .7];
x = -90:1:90;
[~,ErrorLength]= size(out);
for i =1:ErrorLength
figure(i)
hold on
histogram(out(:,i),'normalization','pdf','edgecolor',histogram_color,'facecolor',histogram_color,'facealpha',.5)'
f = fitdist(out(:,i),'kernel');
pdf = pdf(f,x);
l = line(x,pdf,'linestyle','-','color',histogram_color,'linewidth',3);
hold off
end
Sudhakar Shinde
Sudhakar Shinde am 15 Okt. 2020
yes. ok. Try this code at bottom. it works fine.
MadjeKoe
MadjeKoe am 15 Okt. 2020
I keep on getting this message:
Unable to use a value of type prob.KernelDistribution as an index.
Error in forumans (line 35)
pdf = pdf(f,x);
I don't understand why because it worked before
Try this code:
[~,ErrorLength]= size(out);
for i =1:ErrorLength
hold on
histogram(out(:,i),'normalization','pdf')
hold off
end
This gives output on single graph with different histograms without using kernel. As i dont have toolbox i can not check for
f = fitdist(out(:,i),'kernel');
pdf = pdf(f,x);
MadjeKoe
MadjeKoe am 15 Okt. 2020
Thank you so much!!!! This is exactly want I meant!
Sudhakar Shinde
Sudhakar Shinde am 15 Okt. 2020
Glad to help you.
You only need to "hold on" once and that was mentioned in a comment under the question 13 hours prior to this sub-thread.
hold on % <--- Move here
[~,ErrorLength]= size(out);
for i =1:ErrorLength
% hold on <--- Remove
histogram(out(:,i),'normalization','pdf')
% hold off <--- remove or move out of the loop
end
Adam Danz
Adam Danz am 15 Okt. 2020
Bearbeitet: Adam Danz am 15 Okt. 2020
"I did not work."
I doubt it. It's the same solution as Sudhakar Shinde's and you indicated that it worked. There must have been some other problem (which is likely given the new propblem follwing "clear all").
" I tried 'clear all' & now my whole thing is not working anymore"
'clear all' is rarely necessary. You likely were using a global variable named x and now it's no longer defined. If that's the case, it s a good lesson to never use global variables. Now you need to figure out where x was defined.

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