MATLAB: sample from population randomly many times? - matlab

I am aware of MATLAB's datasample which allows to select k times from a certain population. Suppose population=[1,2,3,4] and I want to uniformly sample, with replacement, k=5 times from it. Then:
datasample(population,k)
ans =
1 3 2 4 1
Now, I want to repeat the above experiment N=10000 times without using a for loop. I tried doing:
datasample(repmat(population,N,1),5,2)
But the output I get is (just a short excerpt below):
1 3 2 1 3
1 3 2 1 3
1 3 2 1 3
1 3 2 1 3
1 3 2 1 3
1 3 2 1 3
1 3 2 1 3
1 3 2 1 3
1 3 2 1 3
Every row (result of an experiment) is the same! But obviously they should be different... It's as though some random seed is not updating between rows. How can I fix this? Or some other method I could use that avoids a for loop? Thanks!

You seem to be confusing the way datasample works. If you read the documentation on the function, if you specify a matrix, it will generate a data sampling from a selection of rows in the matrix. Therefore, if you simply repeat the population vector 10000 times, and when you specify the second parameter of the function - which in this case is how many rows of the matrix to extract, even though the actual row locations themselves are different, the actual rows over all of the matrix is going to be the same which is why you are getting that "error".
As such, I wouldn't use datasample here if it is your intention to avoid looping. You can use datasample, but you'd have to loop over each call and you explicitly said that this is not what you want.
What I would recommend you do is first create your population vector to have whatever you desire in it, then generate a random index matrix where each value is between 1 up to as many elements as there are in population. This matrix is in such a way where the number of columns is the number of samples and the number of rows is the number of trials. Once you create this matrix, simply use this to index into your vector to achieve the desired sampling matrix. To generate this random index matrix, randi is a fine choice.
Something like this comes to mind:
N = 10000; %// Number of trials
M = 5; %// Number of samples per trial
population = 1:4; %// Population vector
%// Generate random indices
ind = randi(numel(population), N, M);
%// Get the stuff
out = population(ind);
Here's the first 10 rows of the output:
>> out(1:10,:)
ans =
4 3 1 4 2
4 4 1 3 4
3 2 2 2 3
1 4 2 2 2
1 2 3 4 2
2 2 3 2 1
4 1 3 2 4
1 4 1 3 1
1 1 2 4 4
1 2 4 2 1
I think the above does what you want. Also keep in mind that the above code generalizes to any population vector you want. You simply have to change the vector and it will work as advertised.

datasample interprets each column of your data as one element of your population, sampling among all columns.
To fix this you could call datasample N times in a loop, instead I would use randi
population(randi(numel(population),N,5))
assuming your population is always 1:p, you could simplify to:
randi(p,N,5)

Ok so both of the current answers both say don't use datasample and use randi instead. However, I have a solution for you with datasample and arrayfun.
>> population = [1 2 3 4];
>> k = 5; % Number of samples
>> n = 1000; % Number of times to execute datasample(population, k)
>> s = arrayfun(#(k) datasample(population, k), n*ones(k, 1), 'UniformOutput', false);
>> s = cell2mat(s);
s =
1 4 1 4 4
4 1 2 2 4
2 4 1 2 1
1 4 3 3 1
4 3 2 3 2
We need to make sure to use 'UniformOutput', false with arrayfun as there is more than one output. The cell2mat call is needed as the result of arrayfun is a cell array.

Related

mean of two matrixes with Nans, element per element

I want to have the nanmean of two matrixes element per element. I don't seem to be able to do it using the function nanmean, since that doesn't sum element per element.
Simplified example:
A= [1 1 1
1 1 1
1 nan 1];
B=[3 3 3
3 3 3
3 3 3];
Result I want :
C= [2 2 2
2 2 2
2 3 2];
So the nan is ignored.
One way to sum element per element I found on https://nl.mathworks.com/matlabcentral/answers/366304-how-to-sum-up-multiple-matrices-element-by-element
A(:,:,1)=randi([1 3],100,100);
A(:,:,2)=randi([1 3],100,100);
A(:,:,3)=randi([1 3],100,100);
A(:,:,4)=randi([1 3],100,100);
B=zeros(size(A,1),size(A,2));
for i=1:size(A,3)
B=B+A(1:size(A,1),1:size(A,2),i);
end
disp(B)
BUT this isn't the nansum. How can I do this not taking into account the nans?
You can concatenate the matrices along the 3rd dimension, then apply nanmean along that dimension:
C = cat(3,A,B);
C = nanmean(C,3);
(Of course you can write these two statements as a single one, I wrote it like this for clarity.)

How to count patterns columnwise in Matlab?

I have a matrix S in Matlab that looks like the following:
2 2 1 2
2 3 1 1
3 3 1 1
3 4 1 1
3 1 2 1
4 1 3 1
1 1 3 1
I would like to count patterns of values column-wise. I am interested into the frequency of the numbers that follow right after number 3 in any of the columns. For instance, number 3 occurs three times in the first column. The first time we observe it, it is followed by 3, the second time it is followed by 3 again and the third time it is followed by 4. Thus, the frequency for the patters observed in the first column would look like:
3-3: 66.66%
3-4: 33.33%
3-1: 0%
3-2: 0%
To generate the output, you could use the convenient tabulate
S = [
2 2 1 2
2 3 1 1
3 3 1 1
3 4 1 1
3 1 2 1
4 1 3 1
1 1 3 1];
idx = find(S(1:end-1,:)==3);
S2 = S(2:end,:);
tabulate(S2(idx))
Value Count Percent
1 0 0.00%
2 0 0.00%
3 4 66.67%
4 2 33.33%
Here's one approach, finding the 3's then looking at the following digits
[i,j]=find(S==3);
k=i+1<=size(S,1);
T=S(sub2ind(size(S),i(k)+1,j(k))) %// the elements of S that are just below a 3
R=arrayfun(#(x) sum(T==x)./sum(k),1:max(S(:))).' %// get the number of probability of each digit
I'm going to restate your problem statement in a way that I can understand and my solution will reflect this new problem statement.
For a particular column, locate the locations that contain the number 3.
Look at the row immediately below these locations and look at the values at these locations
Take these values and tally up the total number of occurrences found.
Repeat these for all of the columns and update the tally, then determine the percentage of occurrences for the values.
We can do this by the following:
A = [2 2 1 2
2 3 1 1
3 3 1 1
3 4 1 1
3 1 2 1
4 1 3 1
1 1 3 1]; %// Define your matrix
[row,col] = find(A(1:end-1,:) == 3);
vals = A(sub2ind(size(A), row+1, col));
h = 100*accumarray(vals, 1) / numel(vals)
h =
0
0
66.6667
33.3333
Let's go through the above code slowly. The first few lines define your example matrix A. Next, we take a look at all of the rows except for the last row of your matrix and see where the number 3 is located with find. We skip the last row because we want to be sure we are within the bounds of your matrix. If there is a number 3 located at the last row, we would have undefined behaviour if we tried to check the values below the last because there's nothing there!
Once we do this, we take a look at those values in the matrix that are 1 row beneath those that have the number 3. We use sub2ind to help us facilitate this. Next, we use these values and tally them up using accumarray then normalize them by the total sum of the tallying into percentages.
The result would be a 4 element array that displays the percentages encountered per number.
To double check, if we look at the matrix, we see that the value of 3 follows other values of 3 for a total of 4 times - first column, row 3, row 4, second column, row 2 and third column, row 6. The value of 4 follows the value of 3 two times: first column, row 6, second column, row 3.
In total, we have 6 numbers we counted, and so dividing by 6 gives us 4/6 or 66.67% for number 3 and 2/6 or 33.33% for number 4.
If I got the problem statement correctly, you could efficiently implement this with MATLAB's logical indexing and an approach that is essentially of two lines -
%// Input 2D matrix
S = [
2 2 1 2
2 3 1 1
3 3 1 1
3 4 1 1
3 1 2 1
4 1 3 1
1 1 3 1]
Labels = [1:4]'; %//'# Label array
counts = histc(S([false(1,size(S,2)) ; S(1:end-1,:) == 3]),Labels)
Percentages = 100*counts./sum(counts)
Verify/Present results
The styles for presenting the output results listed next use MATLAB's table for a well human-readable format of data.
Style #1
>> table(Labels,Percentages)
ans =
Labels Percentages
______ ___________
1 0
2 0
3 66.667
4 33.333
Style #2
You can do some fancy string operations to present the results in a more "representative" manner -
>> Labels_3 = strcat('3-',cellstr(num2str(Labels','%1d')'));
>> table(Labels_3,Percentages)
ans =
Labels_3 Percentages
________ ___________
'3-1' 0
'3-2' 0
'3-3' 66.667
'3-4' 33.333
Style #3
If you want to present them in descending sorted manner based on the percentages as listed in the expected output section of the question, you can do so with an additional step using sort -
>> [Percentages,idx] = sort(Percentages,'descend');
>> Labels_3 = strcat('3-',cellstr(num2str(Labels(idx)','%1d')'));
>> table(Labels_3,Percentages)
ans =
Labels_3 Percentages
________ ___________
'3-3' 66.667
'3-4' 33.333
'3-1' 0
'3-2' 0
Bonus Stuff: Finding frequency (counts) for all cases
Now, let's suppose you would like repeat this process for say 1, 2 and 4 as well, i.e. find occurrences after 1, 2 and 4 respectively. In that case, you can iterate the above steps for all cases and for the same you can use arrayfun -
%// Get counts
C = cell2mat(arrayfun(#(n) histc(S([false(1,size(S,2)) ; S(1:end-1,:) == n]),...
1:4),1:4,'Uni',0))
%// Get percentages
Percentages = 100*bsxfun(#rdivide, C, sum(C,1))
Giving us -
Percentages =
90.9091 20.0000 0 100.0000
9.0909 20.0000 0 0
0 60.0000 66.6667 0
0 0 33.3333 0
Thus, in Percentages, the first column are the counts of [1,2,3,4] that occur right after there is a 1 somewhere in the input matrix. As as an example, one can see column -3 of Percentages is what you had in the sample output when looking for elements right after 3 in the input matrix.
If you want to compute frequencies independently for each column:
S = [2 2 1 2
2 3 1 1
3 3 1 1
3 4 1 1
3 1 2 1
4 1 3 1
1 1 3 1]; %// data: matrix
N = 3; %// data: number
r = max(S(:));
[R, C] = size(S);
[ii, jj] = find(S(1:end-1,:)==N); %// step 1
count = full(sparse(S(ii+1+(jj-1)*R), jj, 1, r, C)); %// step 2
result = bsxfun(#rdivide, count, sum(S(1:end-1,:)==N)); %// step 3
This works as follows:
find is first applied to determine row and col indices of occurrences of N in S except its last row.
The values in the entries right below the indices of step 1 are accumulated for each column, in variable count. The very convenient sparse function is used for this purpose. Note that this uses linear indexing into S.
To obtain the frequencies for each column, count is divided (with bsxfun) by the number of occurrences of N in each column.
The result in this example is
result =
0 0 0 NaN
0 0 0 NaN
0.6667 0.5000 1.0000 NaN
0.3333 0.5000 0 NaN
Note that the last column correctly contains NaNs because the frequency of the sought patterns is undefined for that column.

matlab indexing with multiple condition

I can't figure out how to create a vector based on condition on more than one other vectors. I have three vectors and I need values of one vector if values on other vectors comply to condition.
As an example below I would like to choose values from vector a if values on vector b==2 and values on vector c==0 obviously I expect [2 4]
a = [1 2 3 4 5 6 7 8 9 10];
b = [1 2 1 2 1 2 1 2 1 2];
c = [0 0 0 0 0 1 1 1 1 1]
I thought something like:
d = a(b==2) & a(c==0)
but I have d = 1 1 1 1 1 not sure why.
It seems to be basic problem but I can find solution for it.
In your case you can consider using a(b==2 & c==0)
Use ismember to find the matching indices along the rows after concatenating b and c and then index to a.
Code
a(ismember([b;c]',[2 0],'rows'))
Output
ans =
2
4
You may use bsxfun too for the same result -
a(all(bsxfun(#eq,[b;c],[2 0]'),1))
Or you may just tweak your method to get the correct result -
a(b==2 & c==0)

In a 100 X 2 matrix, how do divide each element in column 2 by a constant?

I have a 100 by 2 matrix. And I'm trying to figure out how to divide all terms in the second column by a constant.
For example, let's say I have this matrix.
[1 2;
3 4;
5 6]
I want to divide the 2nd column by 2.
[1 2/2;
3 4/2;
5 6/2]
So my final matrix will be.
[1 1;
3 2;
5 3]
Thank you.
If your matrix is M then:
M(:,2)=M(:,2)./2;
will divide all terms in the second column by a constant (2). By the way, because the value you divide with is a constant you can also write / instead of ./
If you'd like to assemble a new matrix and not overwrite the first one just write something like this:
A=[M(:,1) M(:,2)./2]
I'm not sure how natan 's equations should be read, but I'd multiply the first matrix
1 2
3 4
5 6
by the matrix
1 0
0 .5
The resulting matrix is
1 1
3 2
5 3

How to combine vectors of different length in a cell array into matrix in MATLAB

How to efficiently combined cell array vectors with different length into a matrix, filling the vectors to max length with 0s or NaNs? It would be a nice option for cell2mat().
For example, if I have
C = {1:3; 1:5; 1:4};
I'd like to get either
M = [1 2 3 0 0
1 2 3 4 5
1 2 3 4 0];
or
M = [1 2 3 NaN NaN
1 2 3 4 5
1 2 3 4 NaN];
EDIT:
For a cell of row vectors as in your case, this will pad vectors with zeros to form a matrix
out=cell2mat(cellfun(#(x)cat(2,x,zeros(1,maxLength-length(x))),C,'UniformOutput',false))
out =
1 2 3 0 0
1 2 3 4 5
1 2 3 4 0
A similar question was asked earlier today, and although the question was worded slightly differently, my answer basically does what you want.
Copying the relevant parts here, a cell of uneven column vectors can be zero padded into a matrix as:
out=cell2mat(cellfun(#(x)cat(1,x,zeros(maxLength-length(x),1)),C,'UniformOutput',false));
where maxLength is assumed to be known. In your case, you have row vectors, which is just a slight modification from this.
If maxLength is not known, you can get it as
maxLength=max(cellfun(#(x)numel(x),C));