I'm struggling with determining the probability of occurrence of unique elements in thresh_strain matrix (which can be seen below as a 100 x 16). I was trying to use the code at the bottom to do this, but I get an equal probability of occurrence associated with each of the elements, whereas I want the probability of occurrence associated with unique elements in thresh_strain.
function [thresh_strain] = MCsolution()
no_iterations = 100;
thresh_strain = zeros(100, 16);
casechoice =input('Enter 1 for 1st Layup and 2 for 2nd layup:');
for i=1:no_iterations
for j=1:16
J = Nielsennew(casechoice);
thresh_strain(i,j) = J(1, j);
end
end
% [uniqueValues,~,uniqueIndex] = unique(thresh_strain);
% frequency = accumarray(uniqueIndex(:),1)./numel(thresh_strain);
Thanks
It is not really clear from the title and description, but I suppose you may be looking for something like this:
myUniqueValues = unique(myMatrix);
nelements = hist(myMatrix(:),myUniqueValues);
%plot(myUniqueValues,nelements)
Basically calculating how often each unique value occurs. From here getting the corresponding percentage is of course trivial.
Related
The goal is generating six lotto numbers, but obviously they have to be unique. This has to be written in function form though, following is the equivalent using the library:
(randsample(42,6))'
My idea was to create the vector with all possibilities, pick one out at a time through index and making it impossible to pick this one again by grabbing it out before the next one is picked.
function numbers = lottonumbers()
pool = 1:42;
numbers = zeros(1,6);
for i=1:6
for j=42-i
randIndex = round(1+j*rand);
randNumber = pool(randIndex);
numbers(i) = randNumber;
if randIndex==1
pool = pool(2:end);
else if randIndex==length(pool)
pool = pool(1:(end-1));
else
pool = [pool(1:randIndex-1), pool(randIndex+1:end)];
end
end
end
end
Since I'm pretty noob at MATLAB (just noob at programming really) and since I solved it myself while asking the question, I'm just going to leave it here and ask you guys for suggestions (better style, other algorithm...)
Lotto is based on permutations where the order does not play a role.
% p = randperm(n,k) returns a row vector containing k unique integers selected randomly from 1 to n inclusive.
randperm( 42, 6 )
should do the trick.
From the code: "This is sometimes referred to as a K-permutation of 1:N or as sampling without replacement."
Another approach is to use rejection sampling: generate the numbers independently, and if they are not all different start again. This is efficient as long as the chance of numbers not being all different is small.
N = 6;
M = 42;
done = false;
while ~done
result = randi(M,1,N); %// generate N numbers from [1,...,M]
done = all(diff(sort(result))); %// if all are different, we're done
end
I'm trying to code a loop in Matlab that iteratively solves for an optimal vector s of zeros and ones. This is my code
N = 150;
s = ones(N,1);
for i = 1:N
if s(i) == 0
i = i + 1;
else
i = i;
end
select = s;
HI = (item_c' * (weights.*s)) * (1/(weights'*s));
s(i) = 0;
CI = (item_c' * (weights.*s)) * (1/(weights'*s));
standarderror_afterex = sqrt(var(CI - CM));
standarderror_priorex = sqrt(var(HI - CM));
ratio = (standarderror_afterex - standarderror_priorex)/(abs(mean(weights.*s) - weights'*select));
ratios(i) = ratio;
s(i) = 1;
end
[M,I] = min(ratios);
s(I) = 0;
This code sets the element to zero in s, which has the lowest ratio. But I need this procedure to start all over again, using the new s with one zero, to find the ratios and exclude the element in s that has the lowest ratio. I need that over and over until no ratios are negative.
Do I need another loop, or do I miss something?
I hope that my question is clear enough, just tell me if you need me to explain more.
Thank you in advance, for helping out a newbie programmer.
Edit
I think that I need to add some form of while loop as well. But I can't see how to structure this. This is the flow that I want
With all items included (s(i) = 1 for all i), calculate HI, CI and the standard errors and list the ratios, exclude item i (s(I) = 0) which corresponds to the lowest negative ratio.
With the new s, including all ones but one zero, calculate HI, CI and the standard errors and list the ratios, exclude item i, which corresponds to the lowest negative ratio.
With the new s, now including all ones but two zeros, repeat the process.
Do this until there is no negative element in ratios to exclude.
Hope that it got more clear now.
Ok. I want to go through a few things before I list my code. These are just how I would try to do it. Not necessarily the best way, or fastest way even (though I'd think it'd be pretty quick). I tried to keep the structure as you had in your code, so you could follow it nicely (even though I'd probably meld all the calculations down into a single function or line).
Some features that I'm using in my code:
bsxfun: Learn this! It is amazing how it works and can speed up code, and makes some things easier.
v = rand(n,1);
A = rand(n,4);
% The two lines below compute the same value:
W = bsxfun(#(x,y)x.*y,v,A);
W_= repmat(v,1,4).*A;
bsxfun dot multiplies the v vector with each column of A.
Both W and W_ are matrices the same size as A, but the first will be much faster (usually).
Precalculating dropouts: I made select a matrix, where before it was a vector. This allows me to then form a variable included using logical constructs. The ~(eye(N)) produces an identity matrix and negates it. By logically "and"ing it with select, then the $i$th column is now select, with the $i$th element dropped out.
You were explicitly calculating weights'*s as the denominator in each for-loop. By using the above matrix to calculate this, we can now do a sum(W), where the W is essentially weights.*s in each column.
Take advantage of column-wise operations: the var() and the sqrt() functions are both coded to work along the columns of a matrix, outputting the action for a matrix in the form of a row vector.
Ok. the full thing. Any questions let me know:
% Start with everything selected:
select = true(N);
stop = false; % Stopping flag:
while (~stop)
% Each column leaves a variable out...
included = ~eye(N) & select;
% This calculates the weights with leave-one-out:
W = bsxfun(#(x,y)x.*y,weights,included);
% You can comment out the line below, if you'd like...
W_= repmat(weights,1,N).*included; % This is the same as previous line.
% This calculates the weights before dropping the variables:
V = bsxfun(#(x,y)x.*y,weights,select);
% There's different syntax, depending on whether item_c is a
% vector or a matrix...
if(isvector(item_c))
HI = (item_c' * V)./(sum(V));
CI = (item_c' * W)./(sum(W));
else
% For example: item_c is a matrix...
% We have to use bsxfun() again
HI = bsxfun(#rdivide, (item_c' * V),sum(V));
CI = bsxfun(#rdivide, (item_c' * W),sum(W));
end
standarderror_afterex = sqrt(var(bsxfun(#minus,HI,CM)));
standarderror_priorex = sqrt(var(bsxfun(#minus,CI,CM)));
% or:
%
% standarderror_afterex = sqrt(var(HI - repmat(CM,1,size(HI,2))));
% standarderror_priorex = sqrt(var(CI - repmat(CM,1,size(CI,2))));
ratios = (standarderror_afterex - standarderror_priorex)./(abs(mean(W) - sum(V)));
% Identify the negative ratios:
negratios = ratios < 0;
if ~any(negratios)
% Drop out of the while-loop:
stop = true;
else
% Find the most negative ratio:
neginds = find(negratios);
[mn, mnind] = min(ratios(negratios));
% Drop out the most negative one...
select(neginds(mnind),:) = false;
end
end % end while(~stop)
% Your output:
s = select(:,1);
If for some reason it doesn't work, please let me know.
Because for combinations of large numbers at times matlab replies NaN, the assignment is to write a program to compute combinations of 200 objects taken 90 at a time. Once this works we are to make it into a function y = comb(n,k).
This is what I have so far based on an example we were given of the probability that 2 people in a class have the same birthday.
This is the example:
nMax = 70; %maximum number of people in classroom
nArray = 1:nMax;
prevPnot = 1; %initialize probability
for iN = 1:nMax
Pnot = prevPnot*(365-iN+1)/365; %probability that no birthdays are the same
P(iN) = 1-Pnot; %probability that at least two birthdays are the same
prevPnot = Pnot;
end
plot(nArray, P, '.-')
xlabel('nb. of people')
ylabel('prob. that at least two have same birthday')
grid on
At this point I'm having trouble because I'm more familiar with java. This is what I have so far, and it isn't coming out at all.
k = 90;
n = 200;
nArray = 1:k;
prevPnot = 1;
for counter = 1:k
Pnot = (n-counter+1)/(prevPnot*(n-k-counter+1);
P(iN) = Pnot;
prevPnot = Pnot;
end
The point of the loop I wrote is to separate out each term
i.e. n/k*(n-k), times (n-counter)/(k-counter)*(n-k-counter), and so forth.
I'm also not entirely sure how to save a loop as a function in matlab.
To compute the number of combinations of n objects taken k at a time, you can use gammaln to compute the logarithm of the factorials in order to avoid overflow:
result = exp(gammaln(n+1)-gammaln(k+1)-gammaln(n-k+1));
Another approach is to remove terms that will cancel and then compute the result:
result = prod((n-k+1:n)./(1:k));
So I have a list of 190 numbers ranging from 1:19 (each number is repeated 10 times) that I need to sample 10 at a time. Within each sample of 10, I don't want the numbers to repeat, I tried incorporating a while loop, but computation time was way too long. So far I'm at the point where I can generate the numbers and see if there are repetitions within each subset. Any ideas?
N=[];
for i=1:10
N=[N randperm(19)];
end
B=[];
for j=1:10
if length(unique(N(j*10-9:j*10)))<10
B=[B 1];
end
end
sum(B)
Below is an updated version of the code. this might be a little more clear in showing what I want. (19 targets taken 10 at a time without repetition until all 19 targets have been repeated 10 times)
nTargs = 19;
pairs = nchoosek(1:nTargs, 10);
nPairs = size(pairs, 1);
order = randperm(nPairs);
values=randsample(order,19);
targs=pairs(values,:);
Alltargs=false;
while ~Alltargs
targs=pairs(randsample(order,19),:);
B=[];
for i=1:19
G=length(find(targs==i))==10;
B=[B G];
end
if sum(B)==19
Alltargs=true;
end
end
Here are some very simple steps to do this, basically you just shuffle the vector once, and then you grab the last 10 unique values:
v = repmat(1:19,1,10);
v = v(randperm(numel(v)));
[a idx]=unique(v);
result = unique(v);
v(idx)=[];
The algorithm should be fairly efficient, if you want to do the next 10, just run the last part again and combine the results into a totalResult
You want to sample the numbers 1:19 randomly in blocks of 10 without repetitions. The Matlab function 'randsample' has an optional 'replacement' argument which you can set to 'false' if you do not want repetitions. For example:
N = [];
replacement = false;
for i = 1:19
N = [N randsample(19,10,replacement)];
end
This generates a 19 x 10 matrix of random integers in the range [1,..,19] without repetitions within each column.
Edit: Here is a solution that addresses the requirement that each of the integers [1,..,19] occurs exactly 10 times, in addition to no repetition within each column / sample:
nRange = 19; nRep = 10;
valueRep = true; % true while there are repetitions
nLoops = 0; % count the number of iterations
while valueRep
l = zeros(1,nRep);
v = [];
for m = 1:nRep
v = [v, randperm(nRange,nRange)];
end
m1 = reshape(v,nRep,nRange);
for n = 1:nRep
l(n) = length(unique(m1(:,n)));
end
if all(l == nRep)
valueRep = false;
end
nLoops = nLoops + 1;
end
result = m1;
For the parameters in the question it takes about 300 iterations to find a result.
I think you should approach this constructively.
It's easy to initially find a 19 groups that fulfill your conditions just by rearranging the series 1:19: series1 = repmat(1:19,1,10); and rearranged= reshape(series1,10,19)
then shuffle the values
I would select two random columns copy them and switch the values at two random positions
then make a test if it fulfills your condition - like: test = #(x) numel(unique(x))==10 - if yes replace your columns
just keep shuffling till your time runs out or you are happy
of course you might come up with more efficient shuffling or testing
I was given another solution through the MATLAB forum that works pretty well (Credit to Niklas Nylen over on the MATLAB forum). Computation time is pretty low too. It basically shuffles the numbers until there are no repetitions within every 10 values. Thanks all for your help.
y = repmat(1:19,1,10);
% Run enough iterations to get the output random enough, I selected 100000
for ii = 1:100000
% Select random index
index = randi(length(y)-1);
% Check if it is allowed to switch places
if y(index)~=y(min(index+10, length(y))) && y(index+1)~=y(max(1,index-9))
% Make the switch
yTmp = y(index);
y(index)=y(index+1);
y(index+1)=yTmp;
end
end
I have a 101x82 size matrix called A. I am trying to minimize an objective function obj_fun, whose value is computed indirectly using A.
Now in order to minimize this objective function obj_fun, I need to perturb the values of A. I want to check if obj_fun is going down in values or not. If not, then I need to do perturb/change values of A to a certain percentage such that it minimizes obj_fun. Keep on perturbing/changing values of A until we get minimum obj_fun. My average value of A before any perturbation is ~ 1.1529e+003.
Does any one have suggestion how can I do this? Also, I care a bit about speed i.e. the method/algorithm should not be too slow. Thanks.
You can add random Gaussian noise to A:
A = 0; % seed value for A with something more interesting than 0
best = obj_fun(A);
for iter = 1:max_iter % max_iter should be the maximum number of iterations
newA = A + normrnd(0, 1, size(A));
newObj = obj_fun(A);
if( newObj < best )
best = newObj;
A = newA;
end
end