In MATLAB, I have a set of P numbers. I would like to generate a random array of size N from this set.
For the sake of example, let say I have the set {1, 4}. Let say I would like to generate an array of size 5 (e.g., [1 1 4 1 4]).
What I did is this: I generated the following array using randi.
N = 5;
v = randi([1 4],[1 N]);
The problem is that I got a random array which contains values in 1:4 and not in {1, 4}.
I can simply do this but I need a better way.
for i = 1:length(v)
if v(i) ~= 1 || v(i) ~= 4
v(i) = 1; % or v(i) = 4
end
end
I think I am missing a simple hint here.
You should use datasample,
y = datasample(data,k) returns k observations sampled uniformly at random, with replacement, from the data in data.
a = [1,4];
datasample(a,5)
Depending on the usage, you might consider using,
datasample(unique(a),5)
If you don't have the Statistics Toolbox (which contains the datasample function), you can use randi:
N = 5; %// desired number of samples
data = [1 4]; %// data values
sample = data(randi(numel(data),1,N));
And if you use a very old version of Matlab that doesn't have randi, you can employ rand:
sample = data(ceil(numel(data)*rand(1,N)));
Related
I am trying to implement this code so it works as quickly as possible.
Say I have a population of 100 different values, you can think of it as pop = 1:100 or pop = randn(1,100) to keep things simple. I have a vector n which gives me the size of samples I want to get. Say, for example, that n=[1 3 10 6 2]. What I want to do is to take 5 (which in reality is length(n)) different samples of pop, each consisting of n(i) elements without replacement. This means that for my first sample I want 1 element out of pop, for the second sample I want 3, for the third I want 10, and so on.
To be honest, I am not really interested in which elements are sampled. What I want to get is the sum of those elements that are present in the ith-sample. This would be trivial if I implemented it with a loop, but I am trying to avoid using them to keep my code as quick as possible. I have to do this for many different populations and with length(n)being very large.
If I had to do it with a loop, this would be how:
pop = randn(1,100);
n = [1 3 10 6 2];
sum_sample = zeros(length(n),1);
for i = 1:length(n)
sum_sample(i,1) = sum(randsample(pop,n(i)));
end
Is there a way to do this?
The only way to figure out what is fastest for you is to do a comparison of the different methods.
In fact the loop appears to be very fast in this case!
pop = randn(1,100);
n = [1 3 10 6 2];
tic
sr = #(n) sum(randsample(pop,n));
sum_sample = arrayfun(sr,n);
toc %% Returns about 0.004
clear su
tic
for t=numel(n):-1:1
su(t)=sum(randsample(pop,n(t)));
end
toc %% Returns about 0.003
You can create a function handle which choses the random samples and sums these up. Then you can use arrayfun to execute this function for all values of n:
pop = randn(1,100);
n = [1 3 10 6 2];
sr = #(n) sum(randsample(pop,n));
sum_sample = arrayfun(sr,n);
You can do something like this:
pop = randn(1,100);
n = [1 3 10 6 2];
sampled_data_index = randi(length(pop),1,sum(n));
sampled_data = pop(sampled_data_index);
The randi function randomly selects integer values in a specified range that is suitable for indexing. After you have the indices you can use those at once to sample the data from the pop database.
If you want to have unique indices you can replace the randi function with randperm:
sampled_data_index = randperm(length(pop),sum(n));
Finally:
You can have all the sampled values as a cell variable using the following code:
pop = randn(1,100);
n = [1 3 10 6 2];
fun = #(m) pop(randperm(length(pop),m));
C = arrayfun(fun,n,'UniformOutput',0)
Also having the sum of the sampled data:
funs = #(m) sum(pop(randperm(length(pop),m)));
sumC = arrayfun(funs,n)
I have a matrix with constant consecutive values randomly distributed throughout the matrix. I want the indices of the consecutive values, and further, I want a matrix of the same size as the original matrix, where the number of consecutive values are stored in the indices of the consecutive values. For Example
original_matrix = [1 1 1;2 2 3; 1 2 3];
output_matrix = [3 3 3;2 2 0;0 0 0];
I have struggled mightily to find a solution to this problem. It has relevance for meteorological data quality control. For example, if I have a matrix of temperature data from a number of sensors, and I want to know what days had constant consecutive values, and how many days were constant, so I can then flag the data as possibly faulty.
temperature matrix is number of days x number of stations and I want an output matrix that is also number of days x number of stations, where the consecutive values are flagged as described above.
If you have a solution to that, please provide! Thank you.
For this kind of problems, I made my own utility function runlength:
function RL = runlength(M)
% calculates length of runs of consecutive equal items along columns of M
% work along columns, so that you can use linear indexing
% find locations where items change along column
jumps = diff(M) ~= 0;
% add implicit jumps at start and end
ncol = size(jumps, 2);
jumps = [true(1, ncol); jumps; true(1, ncol)];
% find linear indices of starts and stops of runs
ijump = find(jumps);
nrow = size(jumps, 1);
istart = ijump(rem(ijump, nrow) ~= 0); % remove fake starts in last row
istop = ijump(rem(ijump, nrow) ~= 1); % remove fake stops in first row
rl = istop - istart;
assert(sum(rl) == numel(M))
% make matrix of 'derivative' of runlength
% don't need last row, but needs same size as jumps for indices to be valid
dRL = zeros(size(jumps));
dRL(istart) = rl;
dRL(istop) = dRL(istop) - rl;
% remove last row and 'integrate' to get runlength
RL = cumsum(dRL(1:end-1,:));
It only works along columns since it uses linear indexing. Since you want do something similar along rows, you need to transpose back and forth, so you could use it for your case like so:
>> original = [1 1 1;2 2 3; 1 2 3];
>> original = original.'; % transpose, since runlength works along columns
>> output = runlength(original);
>> output = output.'; % transpose back
>> output(output == 1) = 0; % see hitzg's comment
>> output
output =
3 3 3
2 2 0
0 0 0
I have a 3 dimensional (or higher) array that I want to aggregate by another vector. The specific application is to take daily observations of spatial data and average them to get monthly values. So, I have an array with dimensions <Lat, Lon, Day> and I want to create an array with dimensions <Lat, Lon, Month>.
Here is a mock example of what I want. Currently, I can get the correct output using a loop, but in practice, my data is very large, so I was hoping for a more efficient solution than the second loop:
% Make the mock data
A = [1 2 3; 4 5 6];
X = zeros(2, 3, 9);
for j = 1:9
X(:, :, j) = A;
A = A + 1;
end
% Aggregate the X values in groups of 3 -- This is the part I would like help on
T = [1 1 1 2 2 2 3 3 3];
X_agg = zeros(2, 3, 3);
for i = 1:3
X_agg(:,:,i) = mean(X(:,:,T==i),3);
end
In 2 dimensions, I would use accumarray, but that does not accept higher dimension inputs.
Before getting to your answer let's first rewrite your code in a more general way:
ag = 3; % or agg_size
X_agg = zeros(size(X)./[1 1 ag]);
for i = 1:ag
X_agg(:,:,i) = mean(X(:,:,(i-1)*ag+1:i*ag), 3);
end
To avoid using the for loop one idea is to reshape your X matrix to something that you can use the mean function directly on.
splited_X = reshape(X(:), [size(X_agg), ag]);
So now splited_X(:,:,:,i) is the i-th part
that contains all the matrices that should be aggregated which is X(:,:,(i-1)*ag+1:i*ag)) (like above)
Now you just need to find the mean in the 3rd dimension of splited_X:
temp = mean(splited_X, 3);
However this results in a 4D matrix (where its 3rd dimension size is 1). You can again turn it into 3D matrix using reshape function:
X_agg = reshape(temp, size(X_agg))
I have not tried it to see how much more efficient it is, but it should do better for large matrices since it doesn't use for loops.
How does one create a vector that is composed of a random sampling of two other vectors?
For example
Vector 1 [1, 3, 4, 7], Vector 2 [2, 5, 6, 8]
Random Vector [random draw from vector 1 or 2 (value 1 or 2), random draw from vector 1 or 2 (value 3 or 5)... etc]
Finally, how can one ask matlab to repeat this process n times to draw a distribution of results?
Thank you,
There are many ways you could do this. One possibility is:
tmp=round(rand(size(vector1)))
res = tmp.*vector1 + (1-tmp).*vector2
To get one mixed sample, you may use the idea of the following code snippet (not the optimal one, but maybe clear enough):
a = [1, 3, 4, 7];
b = [2, 5, 6, 8];
selector = randn(size(a));
sample = a.*(selector>0) + b.*(selector<=0);
For n samples put the above code in a for loop:
for k=1:n
% Sample code (without initial "samplee" assignments)
% Here do stuff with the sample
end;
More generally, if X is a matrix and for each row you want to take a sample from a column chosen at random, you can do this with a loop:
y = zeros(size(X,1),1);
for ii = 1:size(X,1)
y(ii) = X(ii,ceil(rand*size(X,2)));
end
You can avoid the loop using clever indexing via sub2ind:
idx_n = ceil(rand(size(X,1),1)*size(X,2));
idx = sub2ind(size(X),(1:size(X,1))',idx_n);
y = X(idx);
If I understand your question, you are choosing two random numbers. First you decide whether to select vector 1 or vector 2; next you pick an element from the chosen vector.
The following code takes advantage of the fact that vector1 and vector2 are the same length:
N = 1000;
sampleMatrix = [vector1 vector2];
M = numel(sampleMatrix);
randIndex = ceil(rand(1,N)*M); % N random numbers from 1 to M
randomNumbers = sampleMatrix(randIndex); % sample N times from the matrix
You can then display the result with, for instance
figure; hist(randomNumbers); % draw a histogram of numbers drawn
When vector1 and vector2 have different elements, you run into a problem. If you concatenate them, you will end up picking elements from the longer vector more often. One way around this is to create random samplings from both arrays, then choose between them:
M1 = numel(vector1);
M2 = numel(vector2);
r1 = ceil(rand(1,N)*M1);
r2 = ceil(rand(1,N)*M2);
randMat = [vector1(r1(:)) vector2(r2(:))]; % two columns, now pick one or the other
randPick = ceil(rand(1,N)*2);
randomNumbers = [randMat(randPick==1, 1); randMat(randPick==2, 2)];
On re-reading, maybe you just want to pick "element 1 from either 1 or 2", then "element 2 from either 1 or 2", etc for all the elements of the vector. In that case, do
N=numel(vector1);
randPick = ceil(rand(1,N)*2);
randMat=[vector1(:) vector2(:)];
randomNumbers = [randMat(randPick==1, 1); randMat(randPick==2, 2)];
This problem can be solved using the function datasample.
Combine both vectors into one and apply the function. I like this approach more than the handcrafted versions in the other answers. It gives you much more flexibility in choosing what you actually want, while being a one-liner.
I have a non-fixed dimensional matrix M, from which I want to access a single element.
The element's indices are contained in a vector J.
So for example:
M = rand(6,4,8,2);
J = [5 2 7 1];
output = M(5,2,7,1)
This time M has 4 dimensions, but this is not known in advance. This is dependent on the setup of the algorithm I'm writing. It could likewise be that
M = rand(6,4);
J = [3 1];
output = M(3,1)
so I can't simply use
output=M(J(1),J(2))
I was thinking of using sub2ind, but this also needs its variables comma separated..
#gnovice
this works, but I intend to use this kind of element extraction from the matrix M quite a lot. So if I have to create a temporary variable cellJ every time I access M, wouldn't this tremendously slow down the computation??
I could also write a separate function
function x= getM(M,J)
x=M(J(1),J(2));
% M doesn't change in this function, so no mem copy needed = passed by reference
end
and adapt this for different configurations of the algorithm. This is of course a speed vs flexibility consideration which I hadn't included in my question..
BUT: this is only available for getting the element, for setting there is no other way than actually using the indices (and preferably the linear index). I still think sub2ind is an option. The final result I had intended was something like:
function idx = getLinearIdx(J, size_M)
idx = ...
end
RESULTS:
function lin_idx = Lidx_ml( J, M )%#eml
%LIDX_ML converts an array of indices J for a multidimensional array M to
%linear indices, directly useable on M
%
% INPUT
% J NxP matrix containing P sets of N indices
% M A example matrix, with same size as on which the indices in J
% will be applicable.
%
% OUTPUT
% lin_idx Px1 array of linear indices
%
% method 1
%lin_idx = zeros(size(J,2),1);
%for ii = 1:size(J,2)
% cellJ = num2cell(J(:,ii));
% lin_idx(ii) = sub2ind(size(M),cellJ{:});
%end
% method 2
sizeM = size(M);
J(2:end,:) = J(2:end,:)-1;
lin_idx = cumprod([1 sizeM(1:end-1)])*J;
end
method 2 is 20 (small number of index sets (=P) to convert) to 80 (large number of index sets (=P)) times faster than method 1. easy choice
For the general case where J can be any length (which I assume always matches the number of dimensions in M), there are a couple options you have:
You can place each entry of J in a cell of a cell array using the num2cell function, then create a comma-separated list from this cell array using the colon operator:
cellJ = num2cell(J);
output = M(cellJ{:});
You can sidestep the sub2ind function and compute the linear index yourself with a little bit of math:
sizeM = size(M);
index = cumprod([1 sizeM(1:end-1)]) * (J(:) - [0; ones(numel(J)-1, 1)]);
output = M(index);
Here is a version of gnovices option 2) which allows to process a whole matrix of subscripts, where each row contains one subscript. E.g for 3 subscripts:
J = [5 2 7 1
1 5 2 7
4 3 9 2];
sizeM = size(M);
idx = cumprod([1 sizeX(1:end-1)])*(J - [zeros(size(J,1),1) ones(size(J,1),size(J,2)-1)]).';