What is the fastest way of calculating the maximum value, with it's corresponding index, of each 'slice' of a 3D array?
Say you have A with n slices (here I just made each slice 10 by 10, but this can be changed to any size):
A = rand(10,10,n);
You can reshape it to n-columns matrix, then take the maximum of each column:
[val,ind] = max(reshape(A,[],n),[],1);
The first output val will be an n-element vector with all the maximum values, and the second output ind will be their row index in the reshaped A.
Then you get the size of the slices:
sz = size(A);
and use it to find the row (r) and column (c) of each maximum element in each slice:
[r,c] = ind2sub(sz(1:2),ind)
So in this example (using rand and 10x10x6 array for A) you would get something like this at the end (but with different values):
val =
0.99861 0.98895 0.98681 0.99991 0.96057 0.99176
r =
9 7 3 8 2 9
c =
1 1 8 10 10 5
If you have a matrix A with n layers, you can apply max function in two steps to get a 1 x 1 x n matrix with max of each layer
A = rand(10,10,n);
layer_max = max(max(A,[],1),[],2); % 1 x 1 x n matrix, use squeeze to remove extra dims
layer_max = squeeze(layer_max);
Related
I have a matrix X of size n x m. I resized X to a vector a of length n x m.
How can I know "automatically" that the ith element in vector a corresponds to what element position (coordinates) in X?
I have written the following MATLAB code but I do not know how to continue.
X = rand(10,10);
[n,m] = size(X);
a = reshape(X, [n*m, 1]);
t = zeros(length(a),1);
for i = 1 : length(a)
t(i) = % I want to perform here the sum over the x and y coordinate values of the element in X
% that corresponds to the ith element in vector a.
end
Any help will be very appreciated.
That's what ind2sub does:
[row, col] = ind2sub([m n], i);
However, you may prefer to do it manually:
row = mod(i-1,m)+1;
col = floor((i-1)/m)+1;
This works because Matlab used column-major order for storing array elements. For example, in a 3×4 matrix the order in which the elements are stored in memory is as follows:
1 4 7 10
2 5 8 11
3 6 9 12
So the entry in the 2nd row, 3rd column is the 8th element in (column-major) linear order. When this matrix is reshaped into a vector (or into any other shape) this linear order is preserved. Therefore you can retrieve the original coordinates by divisions and modulus operations. Note also that, since Matlab's indexing is 1-based (as opposed to 0-based), the modulus operations need to be shifted by 1.
I have an array A=[a1,a2,a3, ..., aN] I would like to take a product of each 3 elements:
s1=a1+a2+a3
s2=a4+a5+a6
...
sM=a(N-2)+a(N-1)+aN
My solution:
k=size(A);
s=0;
for n=1:k
s(n)=s(n-2)+s(n-1)+s(n);
end
Error: Attempted to access s(2); index out of bounds because numel(s)=1.
Hoe to fix it?
If you want to sum in blocks, for the general case when the number of elements of A is not necessarily a multiple of the block size, you can use accumarray:
A = [3 8 5 8 2 3 4 7 9 6 4]; % 11 elements
s = 3; % block size
result = accumarray(ceil((1:numel(A))/s).', A(:));
If you want a sliding sum with a given block size, you can use conv:
A = [3 8 5 8 2 3 4 7 9 6 4]; % 11 elements
s = 3; % block size
result = conv(A(:).', ones(1,s), 'valid');
You try to calculate sby using values from s. Dont you mean s(n)=A(n-2)+A(n-1)+A(n);? Also size returns more than one dimension on its own.
That being said, getting the 2 privous values n-2 and n-1 doenst work for n=1;2 (because you must have positive indices). You have to explain how the first two values should be handeled. I assume either 0 for elements not yet exisiting
k=size(A,2); %only the second dimension when A 1xn, or length(A)
s=zeros(1,k); %get empty values instead of appending each value for better performance
s(1)=A(1);
s(2)=A(2)+A(1);
for n=3:k %start at 3
s(n)=A(n-2)+A(n-1)+A(n);
end
or sshoult be 2 values shorter than A.
k=size(A,2);
s=zeros(1,k-2);
for n=1:k-2
s(n)=A(n)+A(n+1)+A(n+2);
end
You initialise s as a scalar with s = 0. Then you try and index it like an array, but it only has a single element.
Your current logic (if fixed) will calculate this:
s(1) = a(1)+a(2)+a(3)
s(2) = a(2)+a(3)+a(4)
...
% 's' will be 2 elements shorter than 'a'
So we need to be a bit wiser with the indexing to get what you describe, which is
s(1) = a(1)+a(2)+a(3)
s(2) = a(4)+a(5)+a(6)
...
% 's' will be a third as big as 'a'
You should pre-allocate s to the right size, like so:
k = numel(A); % Number of elements in 'A'
s = zeros( 1, k/3 ); % Output array, assuming 'k' is divisible by 3
for n = 0:3:k-3
s(n/3+1) = a(n+1) + a(n+2) + a(n+3);
end
You could do this in one line by reshaping the array to have 3 rows, then summing down each column, this assumes that the number of elements in a is divisible by 3, and that a is a row vector...
s = sum( reshape( a, 3, [] ) );
How do I find the index of the 2 maximum values of a 1D array in MATLAB? Mine is an array with a list of different scores, and I want to print the 2 highest scores.
You can use sort, as #LuisMendo suggested:
[B,I] = sort(array,'descend');
This gives you the sorted version of your array in the variable B and the indexes of the original position in I sorted from highest to lowest. Thus, B(1:2) gives you the highest two values and I(1:2) gives you their indices in your array.
I'll go for an O(k*n) solution, where k is the number of maximum values you're looking for, rather than O(n log n):
x = [3 2 5 4 7 3 2 6 4];
y = x; %// make a copy of x because we're going to modify it
[~, m(1)] = max(y);
y(m(1)) = -Inf;
[~, m(2)] = max(y);
m =
5 8
This is only practical if k is less than log n. In fact, if k>=3 I would put it in a loops, which may offend the sensibilities of some. ;)
To get the indices of the two largest elements: use the second output of sort to get the sorted indices, and then pick the last two:
x = [3 2 5 4 7 3 2 6 4];
[~, ind] = sort(x);
result = ind(end-1:end);
In this case,
result =
8 5
I have a matlab problem to solve. In have two vectores that limit my space, x_low and x_high. The matrix pos needs to have values within this spaces and each column of the matrix has different bounds given by the two vectores. Now my problem is that randi gives valus between two integers but i need to change the bounds for each columns. There is another way to use randi or a different matlab function to do this?
I know there are better codes to do this but i'm starting to use matlab and i know to do it this way, any aid is welcome
x_low = [Io_low, Iirr_low, Rp_low, Rs_low, n_low]; % vector of constant values
x_high = [Io_high, Iirr_high, Rp_high, Rs_high, n_high]; % vector of constant values
pos = rand(particles, var);
var = length(x_high);
for i = 1: particles % rows
for k = 1: var %columns
if pos(i, k) < x_low(k) || pos(i, k) > x_high(k) % if the position is out of bounder
pos(i, k) = randi(x_low(k), x_high(k), 1); % fill it with a particle whithin the bounderies
end
end
end
If I understand correctly, you need to generate a matrix with integer values such that each column has different lower and upper limits; and those lower and upper limits are inclusive.
This can be done very simply with
rand (to generate random numbers between 0 and 1 ),
bsxfun (to take care of the lower and upper limits on a column basis), and
round (so that the results are integer values).
Let the input data be defined as
x_low = [1 6 11]; %// lower limits
x_high = [3 10 100]; %// upper limits
n_rows = 7; %// number of columns
Then:
r = rand(n_rows, numel(x_low)); %// random numbers between 0 and 1
r = floor(bsxfun(#times, r, x_high-x_low+1)); %// adjust span and round to integers
r = bsxfun(#plus, r, x_low); %// adjust lower limit
gives something like
r =
2 7 83
3 6 93
2 6 22
3 10 85
3 7 96
1 10 90
2 8 57
If you need to fill in values only at specific entries of matrix pos, you can use something like
ind = bsxfun(#lt, pos, x_low) | bsxfun(#gt, pos, x_high); %// index of values to replace
pos(ind) = r(ind);
This a little overkill, because the whole matrixd r is generated only to use some of its entries. To generate only the needed values the best way is probably to use loops.
You can use cellfun for this. Something like:
x_low = [Io_low, Iirr_low, Rp_low, Rs_low, n_low];
x_high = [Io_high, Iirr_high, Rp_high, Rs_high, n_high];
pos = cell2mat(cellfun(#randi, mat2cell([x_low' x_high'], ones(numel(x_low),1), 1), repmat({[particles 1]}, [numel(x_low) 1)])))';
Best,
Given a matrix A, how do I get the elements (and their indices) larger than x in a specific range?
e.g.
A = [1:5; 2:6; 3:7; 4:8; 5:9]
A =
1 2 3 4 5
2 3 4 5 6
3 4 5 6 7
4 5 6 7 8
5 6 7 8 9
And for instance I want all elements larger than 5 and appear in the range A(2:4,3:5). I should get:
elements:
6 , 6 , 7 , 6 , 7 , 8
indices:
14, 18, 19, 22, 23, 24
A(A>5) would give me all entries which are larger than 5.
A(2:4,3:5) would give all elements in the range 2:4,3:5.
I want some combination of the two. Is it possible or the only way is to put the needed range in another array B and only then perform B(B>5)? Obviously 2 problems here: I'd lose the original indices, and it will be slower. I'm doing this on a large number of matrices.
Code. I'm trying to avoid matrix multiplication, so this may look a bit odd:
A = [1:5; 2:6; 3:7; 4:8; 5:9];
[r,c] = meshgrid(2:4,3:5);
n = sub2ind(size(A), r(:), c(:));
indices = sort(n(A(n) > 5)); %'skip sorting if not needed'
values = A(indices);
Explanation. The code converts the Cartesian product of the subscripts to linear indices in the A matrix. Then it selects the indices that respect the condition, then it selects the values.
However, it is slow.
Optimization. Following LuisMendo's suggestion, the code may be sped up by replacing the sub2ind-based linear index calculation with a handcrafted linear index calculation:
A = [1:5; 2:6; 3:7; 4:8; 5:9];
%'For column-first, 1-based-index array memory '
%'layout, as in MATLAB/FORTRAN, the linear index '
%'formula is: '
%'L = R + (C-1)*NR '
n = bsxfun(#plus, (2:4), (transpose(3:5) - 1)*size(A,1));
indices = n(A(n) > 5);
values = A(indices);
If you only need the values (not the indices), it can be done using the third output of find and matrix multiplication. I don't know if it will be faster than using a temporary array, though:
[~, ~, values] = find((A(2:4,3:5)>5).*A(2:4,3:5));
Assuming you need the linear indices and the values, then if the threshold is positive you could define a mask. This may be a good idea if the mask can be defined once and reused for all matrices (that is, if the desired range is the same for all matrices):
mask = false(size(A));
mask(2:4,3:5) = true;
indices = find(A.*mask>5);
values = A(indices);
its a little clunky, but:
R = 2:4;
C = 3:5;
I = reshape(find(A),size(A))
indicies = nonzeros(I(R,C).*(A(R,C)>5))
values = A(indicies)