Random values from histogram

Hi

how i can use one value at the time from histogram and after use that value is deleted and i cannot use that again eg. if i have

x=rand(1,100)

[N,X]=hist(x,15)

bar(X,N,1,'w')

i want to get 100 values one by one

Thanks a lot

[Oleg Komarov]

I don't understand what are you asking.

[niko kauppinen]

sorry I did not asked correctly.

I want to know if I have eg, Radius of something let's say 2-5,and there is different radius 2,3,2,4,5,5,....and so on. and i have PDF of them how i can choose first one of them and after my loop coming back i choose second one and so on....

Answer 1

[Jos (10584)]

Standing on the shoulder of a giant WR, the following run-length based code will create a random list of 10 values that are derived from the results of a histogram

% simplified example
% data obtained by hist
X = [.1 .4 .6 .8] ; % possible values
N = [5  3 0  2] ; % how many times the occurred

% a simple but effective run-length decoding scheme
clear ix ; 
ix([1 cumsum(N(1:end))+1]) = 1 ;
ix = cumsum(ix(1:end-1)) ;
% retrieve the values
X2 = X(N>0) ;
V = X2(ix) ; % all 10 values
RandV = V(randperm(numel(V))) % in random order

Answer 2

[Walter Roberson]

Okay, then, what this is equivalent to is taking each bar count as a run-length encoding of the value associated with that bar (e.g., the bar center), then doing a randperm() of that expanded vector.

I know a couple of people have posted efficient run-length decoding routines; I just don't happen to remember one at the moment.

Answer 3

[Walter Roberson]

x(randperm(1:length(x))

The histogram becomes irrelevant. Each item will occur exactly the number of times it did in the original list.

[Andrew Newell]

That is what niko seems to be asking, but what is the point of a random rearrangement of numbers that are already randomly chosen?

[Walter Roberson]

Hmmm... the bar _counts_ are what are to be chosen from, perhaps? But if so then should the bar count decrease one one of the items from that bar is "used up", or should the bar count itself stay the same but the number of repetitions of it "available" decreases?

Or perhaps it is the bar _index_ that needs to be chosen randomly, in accordance with the counts?

[niko kauppinen]

yes i choose from bar count or PDF and it should decrease one

you are right

can you help me?