In Matlab I have a structure AVG of size 1 x 6 which has one field averageNEST that is a cell array also of size 1 x 6.
Each averageNEST contains matrices of varying sizes (in one dimension), so for example
AVG(1).averageNEST{1,1} is of size 281 x 3 x 19 and
AVG(1).averageNEST{1,2} is of size 231 x 3 x 19
The 2nd and 3rd dimensions of the matrices are always 3 and 19, it is only the first dimension that can change.
I want to average over all the matrices contained within AVG(1).averageNEST and obtain one matrix of size X x 3 x 19 where X is the size of the smallest matrix in AVG(1).averageNEST.
Then I want to do this for all 6 averageNEST in AVG - so have a separate averaged matrix for AVG(1), AVG(2) ... AVG(6).
I have tried multiple things including trying to concatenate matrices using the following code:
for i=1:6
min_epoch = epoch + 1;
for ii=1:19
averageNEST(:,:,ii) = [AVG(i).averageNEST(1:min_epoch,:,ii)];
end
end
and then average this but it doesn't work and now I'm really confused about what I'm doing!
Can anyone help?
I am not sure if I understand what you want to do. If you want to keep only the elements up to the size of the the smallest matrix and then average those matrices you can do the following:
averageNEST = cell(size(AVG));
for iAVG = 1:numel(AVG)
nests = AVG(iAVG).averageNEST;
minsize = min(cellfun(#(x) size(x,1), nests));
reducednests = cellfun(#(y) y(1:minsize, :, :), nests, 'UniformOutput', false);
averageNEST{iAVG} = sum(cat(4, reducednests{:}), 4) / numel(nests);
end
Related
I have a situation analogous to the following
z = magic(3) % Data matrix
y = [1 2 2]' % Column indices
So,
z =
8 1 6
3 5 7
4 9 2
y represents the column index I want for each row. It's saying I should take row 1 column 1, row 2 column 2, and row 3 column 2. The correct output is therefore 8 5 9.
I worked out I can get the correct output with the following
x = 1:3;
for i = 1:3
result(i) = z(x(i),y(i));
end
However, is it possible to do this without looping?
Two other possible ways I can suggest is to use sub2ind to find the linear indices that you can use to sample the matrix directly:
z = magic(3);
y = [1 2 2];
ind = sub2ind(size(z), 1:size(z,1), y);
result = z(ind);
We get:
>> result
result =
8 5 9
Another way is to use sparse to create a sparse matrix which you can turn into a logical matrix and then sample from the matrix with this logical matrix.
s = sparse(1:size(z,1), y, 1, size(z,1), size(z,2)) == 1; % Turn into logical
result = z(s);
We also get:
>> result
result =
8
5
9
Be advised that this only works provided that each row index linearly increases from 1 up to the end of the rows. This conveniently allows you to read the elements in the right order taking advantage of the column-major readout that MATLAB is based on. Also note that the output is also a column vector as opposed to a row vector.
The link posted by Adriaan is a great read for the next steps in accessing elements in a vectorized way: Linear indexing, logical indexing, and all that.
there are many ways to do this, one interesting way is to directly work out the indexes you want:
v = 0:size(y,2)-1; %generates a number from 0 to the size of your y vector -1
ind = y+v*size(z,2); %generates the indices you are looking for in each row
zinv = z';
zinv(ind)
>> ans =
8 5 9
I have a vector of certain size and I want to reshape it into a square matrix. Here is an example: Let's say the vector is of size 784. Then I would create a matrix of size 28x28. In Matlab I would do it with the following command:
reshape(x,28,28)
Of course it can be possible that it is not possible to have an exact square matrix. In this case the matrix should as squarish as possible.
How can I do this calculation? That means how can I calculate the values a and b in reshape(x,a,b)?
Start with a equal to the square root of numel(x) rounded down. If that number doesn't divide numel(x), subtract 1 and try again. That way you end with a equal to the closest integer to sqrt(x) (from below) that divides numel(x). b would then be numel(x)/a, but you can simply use [] as the third argument to reshape:
a = floor(sqrt(numel(x)));
while mod(x,a)
a = a-1;
end
result = reshape(x,a,[]);
Example:
x = 1:20;
gives
result =
1 5 9 13 17
2 6 10 14 18
3 7 11 15 19
4 8 12 16 20
One possible approach:
x = rand(1, 784);
divisors = find(rem(numel(x), 1:numel(x)) == 0);
[~, idx] = min(abs(divisors - sqrt(numel(x))));
x = reshape(x, divisors(idx), numel(x) / divisors(idx));
Let me explain:
Suppose you have a vector named x:
x = rand(1, 784);
First, you find the divisors of the size of x:
divisors = find(rem(numel(x), 1:numel(x)) == 0);
Then, you proceed to choose the divisor which is closest to the square root of x's size:
[~, idx] = min(abs(divisors - sqrt(numel(x))));
Finally, you reshape x using that divisor (and the corresponding multiple):
x = reshape(x, divisors(idx), numel(x) / divisors(idx));
It is not a simple problem to find closest factors of an integer. You need to use the MATLAB answers to the question Input an integer, find the two closest integers which, when multiplied, equal the input. From that question if you use the answer that provides the function findIntegerFactorsCloseToSquarRoot, you can use the following code to reshape.
[a, b] = findIntegerFactorsCloseToSquarRoot(numel(x));
reshape(x, a, b);
I suggest you to first check whether the number is prime or not by isprime(784).
Then you can use prime_factors = factor(784) to get the integer factorization of the number. (Depending on the MATLAB version you may use ifactor(784))
The rest needs just a little more work on prime_factors.
Can anybody help me to find out the method to calculate the elements of different sized matrix in Matlab ?
Let say that I have 2 matrices with numbers.
Example:
A=[1 2 3;
4 5 6;
7 8 9]
B=[10 20 30;
40 50 60]
At first,we need to find maximum number in each column.
In this case, Ans=[40 50 60].
And then,we need to find ****coefficient** (k).
Coefficient(k) is equal to 1 divided by quantity of column of matrix A.
In this case, **coefficient (k)=1/3=0.33.
I wanna create matrix C filling with calculation.
Example in MS Excel.
H4 = ABS((C2-C6)/C9)*0.33+ABS((D2-D6)/D9)*0.33+ABS((E2-E6)/E9)*0.33
I4 = ABS((C3-C6)/C9)*0.33+ABS((D3-D6)/D9)*0.33+ABS((E3-E6)/E9)*0.33
J4 = ABS((C4-C6)/C9)*0.33+ABS((D4-D6)/D9)*0.33+ABS((E4-E6)/E9)*0.33
And then (Like above)
H5 = ABS((C2-C7)/C9)*0.33+ABS((D2-D7)/D9)*0.33+ABS((E2-E7)/E9)*0.33
I5 = ABS((C3-C7)/C9)*0.33+ABS((D3-D7)/D9)*0.33+ABS((E3-E7)/E9)*0.33
J5 = ABS((C4-C7)/C9)*0.33+ABS((D4-D7)/D9)*0.33+ABS((E4-E7)/E9)*0.33
C =
0.34 =|(1-10)|/40*0.33+|(2-20)|/50*0.33+|(3-30)|/60*0.33
0.28 =|(4-10)|/40*0.33+|(5-20)|/50*0.33+|(6-30)|/60*0.33
0.22 =|(7-10)|/40*0.33+|(8-20)|/50*0.33+|(9-30)|/60*0.33
0.95 =|(1-40)|/40*0.33+|(2-50)|/50*0.33+|(3-60)|/60*0.33
0.89 =|(4-40)|/40*0.33+|(5-50)|/50*0.33+|(6-60)|/60*0.33
0.83 =|(7-40)|/40*0.33+|(8-50)|/50*0.33+|(9-60)|/60*0.33
Actually A is a 15x4 matrix and B is a 5x4 matrix.
Perhaps,the matrices dimensions are more than this matrices (variables).
How can i write this in Matlab?
Thanks you!
You can do it like so. Let's assume that A and B are defined as you did before:
A = vec2mat(1:9, 3)
B = vec2mat(10:10:60, 3)
A =
1 2 3
4 5 6
7 8 9
B =
10 20 30
40 50 60
vec2mat will transform a vector into a matrix. You simply specify how many columns you want, and it will automatically determine the right amount of rows to transform the vector into a correctly shaped matrix (thanks #LuisMendo!). Let's also define more things based on your post:
maxCol = max(B); %// Finds maximum of each column in B
coefK = 1 / size(A,2); %// 1 divided by number of columns in A
I am going to assuming that coefK is multiplied by every element in A. You would thus compute your desired matrix as so:
cellMat = arrayfun(#(x) sum(coefK*(bsxfun(#rdivide, ...
abs(bsxfun(#minus, A, B(x,:))), maxCol)), 2), 1:size(B,1), ...
'UniformOutput', false);
outputMatrix = cell2mat(cellMat).'
You thus get:
outputMatrix =
0.3450 0.2833 0.2217
0.9617 0.9000 0.8383
Seems like a bit much to chew right? Let's go through this slowly.
Let's start with the bsxfun(#minus, A, B(x,:)) call. What we are doing is taking the A matrix and subtracting with a particular row in B called x. In our case, x is either 1 or 2. This is equal to the number of rows we have in B. What is cool about bsxfun is that this will subtract every row in A by this row called by B(x,:).
Next, what we need to do is divide every single number in this result by the corresponding columns found in our maximum column, defined as maxCol. As such, we will call another bsxfun that will divide every element in the matrix outputted in the first step by their corresponding column elements in maxCol.
Once we do this, we weight all of the values of each row by coefK (or actually every value in the matrix). In our case, this is 1/3.
After, we then sum over all of the columns to give us our corresponding elements for each column of the output matrix for row x.
As we wish to do this for all of the rows, going from 1, 2, 3, ... up to as many rows as we have in B, we apply arrayfun that will substitute values of x going from 1, 2, 3... up to as many rows in B. For each value of x, we will get a numCol x 1 vector where numCol is the total number of columns shared by A and B. This code will only work if A and B share the same number of columns. I have not placed any error checking here. In this case, we have 3 columns shared between both matrices. We need to use UniformOutput and we set this to false because the output of arrayfun is not a single number, but a vector.
After we do this, this returns each row of the output matrix in a cell array. We need to use cell2mat to transform these cell array elements into a single matrix.
You'll notice that this is the result we want, but it is transposed due to summing along the columns in the second step. As such, simply transpose the result and we get our final answer.
Good luck!
Dedication
This post is dedicated to Luis Mendo and Divakar - The bsxfun masters.
Assuming by maximum number in each column, you mean columnwise maximum after vertically concatenating A and B, you can try this one-liner -
sum(abs(bsxfun(#rdivide,bsxfun(#minus,permute(A,[3 1 2]),permute(B,[1 3 2])),permute(max(vertcat(A,B)),[1 3 2]))),3)./size(A,2)
Output -
ans =
0.3450 0.2833 0.2217
0.9617 0.9000 0.8383
If by maximum number in each column, you mean columnwise maximum of B, you can try -
sum(abs(bsxfun(#rdivide,bsxfun(#minus,permute(A,[3 1 2]),permute(B,[1 3 2])),permute(max(B),[1 3 2]))),3)./size(A,2)
The output for this case stays the same as the previous case, owing to the values of A and B.
I have a <206x193> matrix A. It contains the values of a parameter at 206 different locations at 193 time steps. I am interested in the maximum value at each location over all times as well as the corresponding indices. I have another matrix B with the same dimensions of A and I'm interested in values for each location at the time that A's value at that location was maximal.
I've tried [max_val pos] = max(A,[],2), which gives the right maximum values, but A(pos) does not equal max_val.
How exactly does this function work?
I tried a smaller example as well. Still I don't understand the meaning of the indices....
>> H
H(:,:,1) =
1 2
3 4
H(:,:,2) =
5 6
7 8
>> [val pos] = max(H,[],2)
val(:,:,1) =
2
4
val(:,:,2) =
6
8
pos(:,:,1) =
2
2
pos(:,:,2) =
2
2
The indices in idx represent the index of the max value in the corresponding row. You can use sub2ind to create a linear index if you want to test if A(pos)=max_val
A=rand(206, 193);
[max_val, idx]=max(A, [], 2);
A_max=A(sub2ind(size(A), (1:size(A,1))', idx));
Similarly, you can access the values of B with:
B_Amax=B(sub2ind(size(A), (1:size(A,1))', idx));
From your example:
H(:,:,2) =
5 6
7 8
[val pos] = max(H,[],2)
val(:,:,2) =
6
8
pos(:,:,2) =
2
2
The reason why pos(:,:,2) is [2; 2] is because the maximum is at position 2 for both rows.
max is a primarily intended for use with vectors. In normal mode, even the multi-dimensional arrays are treated as a series of vectors along which the max function is applied.
So, to get the values in B at each location at the time where A is maximum, you should
// find the maximum values and positions in A
[c,i] = max(A, [], 2);
// iterate along the first dimension, to retrieve the corresponding values in B
C = [];
for k=1:size(A,1)
C(k) = B(k,i(k));
end
You can refer to #Jigg's answer for a more concise way of creating matrix C
One simple question..
Let's suppose in Matlab we have a variable: numEntries = 8
I know "ones" is used to create "1's matrixes" but what's the result of the following?
W = ones(1:numEntries);
And,if possible,where I can find a good reference about matrixes in Matlab? I looked in official documentation but it is not so clear to me.
Thank you very much
ones creates a matrix of ones. ones works for any number of dimensions, however if you specify only one argument, you'll get a square matrix of ones of that size. Otherwise, you'll get a N-dimensional matrix of ones, N being the number of arguments you give to ones.
If numEntries is 8, 1:numEntries creates a 1x8 vector 1 2 3 4 5 6 7 8. This vector being given as input to ones, you'll get a 8-dimensional matrix of ones, in which the size of 1st dimension (vertical) is 1, the size of 2nd dimension (horizontal) is 2, the size of third dimension is 3, and so on, until the 8th dimension, that has size of 8.
numEntries = 8;
W = ones(1:numEntries);
% To check sizes of each dimension of W:
size(W)
ans =
1 2 3 4 5 6 7 8
% To calculate the total number of elements in W:
prod(size(W))
ans =
40320
% Edit: to get the number of elements in W, numel(W) is faster than prod(size(W)) :
numel(W)
ans =
40320
It's an N-D (N-dimensional) array where N = numEntries.
More specifically, it's a N-D array of size 1 x 2 x ... x N filled with all ones.