histogram without hist3 function

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Kamil Tkacz
Kamil Tkacz 2015-5-13
评论： Kamil Tkacz 2015-5-13
Hello folks
I have a problem with hist3 function. I need to draw histogram 3D without using hist function but i need to define some new function. I have 1000x2 matrix of random number from 1.5 to 3.5 range and i dont really know how to do this. I was thinking about using bar3 function fron this code:
data=[1 3 5 7 4 8 0 1 3];
If you want to distribute your data into 10 bins, you would create a new array of size 10:
histArray=zeros(1,10); % prealocate
x=0:1:9;
then, you would run a forloop to count how many times you encounter in a particular value:
for n=1:length(data)
histArray(1,data(n)+1)=histArray(1,data(n)+1)+1; % every time you meet the particular value, you add 1 into to corresponding bin
end
bar(histArray)
but i need this to use not with natural but with integral. I would be very gratefull for any help.
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Walter Roberson
Walter Roberson 2015-5-13
[uniqvals, ia, ib] = unique(x);
histArray = zeros(1, length(uniqvals));
for n=1:length(ib)
histArray(1,ib)=histArray(1,ib)+1;
end
bar(uniquevals, histArray);
Kamil Tkacz
Kamil Tkacz 2015-5-13
Thanks man its working, but if i wanna to get 3D bar plot what should i do?

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回答（1 个）

Image Analyst
Image Analyst 2015-5-13
You can't use b directly as an index. You need to figure out what bin that the b value should be in. Like if b = 3.4423423423 do you want that in bin number 2, 8, or 42 or 300? If you want a bin for each unit integer range, then you can just use floor() or round() on the b values.
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Kamil Tkacz
Kamil Tkacz 2015-5-13
yea but when i take floor() or round() i will get only 1 2 or 3 but thats not a point

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