MATLAB:
I am trying to do basis expansion of a huge matrix(1000x15).
For example,
X =
x1 x2
1 4
2 5
3 6
I want to build a new matrix.
Y =
x1 x2 x1*x1 x1*x2 x2*x2
1 4 1 4 16
2 5 4 10 25
3 6 9 18 36
Could any one please suggest a easier way to do this
% your input
A = [1 4; 2 5; 3 6];
% generate pairs
[p,q] = meshgrid(1:size(A,2), 1:size(A,2));
% only retain unique pairs
ii = tril(p) > 0;
% perform element wise multiplication
res = [A A(:,p(ii)) .* A(:,q(ii))];
Using the one-liner from this answer to get the 2-combinations of the indices, you can generate the matrix without the interleaved ordering with
function Y = columnCombo(Y)
comb = nchoosek(1:size(Y,2),2);
Y = [Y , Y.^2 , Y(:,comb(:,1)).*Y(:,comb(:,2))];
end
For the interleaved ordering, I came up with this, possibly sub-optimal, solution:
function Y = columnCombo(Y)
[m,n] = size(Y);
comb = nchoosek(1:n,2);
Y = [Y,zeros(m,n + size(comb,1))];
col = n+1;
for k = 1:n-1
Y(:,col) = Y(:,k).*Y(:,k) ;
ms = comb(comb(:,1)==k,:) ;
ncol = size(ms,1) ;
Y(:,col+(1:ncol)) = Y(:,ms(:,1)).*Y(:,ms(:,2)) ;
col = col + ncol + 1 ;
end
Y(:,end) = Y(:,n).^2;
end
Related
I have created a function to add one to an element as follows;
function xz = addOne(x)
nrow = size(x,1);
ncol = size(x,1);
for K = 1:nrow
for J = 1:ncol
xz = x(:) + 1;
end
end
Example: given 1 the function results in 2:
addOne(1) [2]
I have tried using a matrix as an argument for the function...
x = [1 2 3; 0 0 0; 4 5 6];
x =
1 2 3
0 0 0
4 5 6
addOneWithFors(x)
ans =
2
1
5
3
1
6
4
1
7
How would I update this function to accept a matrix with multiple rows and columns and output it as such instead of just 1 number or a list of elements. Any help would be greatly appreciated.
In Matlab, you don't need a special function to do this. Matlab natively supports adding a scalar to a matrix. For example:
x = [1 2 3; 0 0 0; 4 5 6];
y = x + 1
will produce:
y =
2 3 4
1 1 1
5 6 7
However, if you specifically want to write this out explicitly using for loops, then your addOne() function only needs minor modifications. For example:
function xz = addOne(x)
nrow = size(x,1);
ncol = size(x,2);
xz = zeros(nrow,ncol);
for K = 1:nrow
for J = 1:ncol
xz(K,J) = x(K,J) + 1;
end
end
Note that ncol = size(x,2); has been defined correctly.
I'm trying to generate an n x n matrix like
5 4 3 2 1
4 4 3 2 1
3 3 3 2 1
2 2 2 2 1
1 1 1 1 1
where n = 5 or n 50. I'm at an impasse and can only generate a portion of the matrix. It is Problem 2.14 from Numerical Methods using MATLAB 3rd Edition by Penny and Lindfield. This is the best I have so far:
n = 5;
m = n;
A = zeros(m,n);
for i = 1:m
for j = 1:n
A(i,j) = m;
end
m = m - 1;
end
Any feedback is appreciated.
That was a nice brain-teaser, here’s my solution:
[x,y] = meshgrid(5:-1:1);
out = min(x,y)
Output:
ans =
5 4 3 2 1
4 4 3 2 1
3 3 3 2 1
2 2 2 2 1
1 1 1 1 1
Here's one loop-based approach:
n = 5;
m = n;
A = zeros(m, n);
for r = 1:m
for c = 1:n
A(r, c) = n+1-max(r, c);
end
end
And here's a vectorized approach (probably not faster, just for fun):
n = 5;
A = repmat(n:-1:1, n, 1);
A = min(A, A.');
That's one of the matrices in Matlab's gallery, except that it needs a 180-degree rotation, which you can achieve with rot90:
n = 5;
A = rot90(gallery('minij', n), 2);
I have a 3x3 Matrix and want to save the indices and values into a new 9x3 matrix. For example A = [1 2 3 ; 4 5 6 ; 7 8 9] so that I will get a matrix x = [1 1 1; 1 2 2; 1 3 3; 2 1 4; 2 2 5; ...] With my code I only be able to store the last values x = [3 3 9].
A = [1 2 3 ; 4 5 6 ; 7 8 9];
x=[];
for i = 1:size(A)
for j = 1:size(A)
x =[i j A(i,j)]
end
end
Thanks for your help
Vectorized approach
Here's one way to do it that avoids loops:
A = [1 2 3 ; 4 5 6 ; 7 8 9];
[ii, jj] = ndgrid(1:size(A,1), 1:size(A,2)); % row and column indices
vv = A.'; % values. Transpose because column changes first in the result, then row
x = [jj(:) ii(:) vv(:)]; % result
Using your code
You're only missing concatenation with previous x:
A = [1 2 3 ; 4 5 6 ; 7 8 9];
x = [];
for i = 1:size(A)
for j = 1:size(A)
x = [x; i j A(i,j)]; % concatenate new row to previous x
end
end
Two additional suggestions:
Don't use i and j as variable names, because that shadows the imaginary unit.
Preallocate x instead of having it grow in each iteration, to increase speed.
The modified code is:
A = [1 2 3 ; 4 5 6 ; 7 8 9];
x = NaN(numel(A),3); % preallocate
n = 0;
for ii = 1:size(A)
for jj = 1:size(A)
n = n + 1; % update row counter
x(n,:) = [ii jj A(ii,jj)]; % fill row n
end
end
I developed a solution that works much faster. Here is the code:
% Generate subscripts from linear index
[i, j] = ind2sub(size(A),1:numel(A));
% Just concatenate subscripts and values
x = [i' j' A(:)];
Try it out and let me know ;)
i have a vector and a scalar as input in a problem , the problem is to compute the largest product of n consecutive number of the vector and output the product and the index of the element of the vector that is the first term of the product.
E.g vector =[ 1 2 3 4 5 6] , n=3
I'm supposed to get '3' (i e n) consecutive number of vector whose product is the largest .
in this case it will be 4*5*6
so output will be 120 and 4 as the index.
now if vector has fewer than 'n' elements, the function outputon returns 0 and -1 as the output.
please i need ideas on how to achieve this
You can make a simple implementation with loops:
a = [1 2 3 4 5 6]
w = 3
n = length(a)
maximum = -1
for i = 1:n-w
p = prod(a(i:i+w))
if (p > maximum)
maximum = p
end
end
maximum
Or, you can use the nlfilter from the image processing palette.
a = []
w = 3
if (length(a) >= w)
products = nlfilter(a, [1 w], #(x) prod(x))
res = max(products)
else
res = -1
end
You can create the moving window from the answer of raryeng for the question Matrix with sliding window elements and then appyl cumprod on the columns and take the max with its index.
myvec = [1 2 2 1 3 1];
n = 3;
ind = bsxfun(#plus, 1:n, (0:1:length(myvec)-n).')';
M = cumprod(myvec(ind));
[val,its_ind] = max(M(end,:));
You can encapsulate this with an if condition checking whether the length of myvec is larger than
You can compute the element-wise logarithm of your vector, so multiplication becomes addition; and sliding addition is just convolution with a window of ones:
function [m, p] = f(v, n)
if numel(v) < n
m = -1;
p = 0;
else
c = conv(log(v(:)), ones(n,1), 'valid'); % convolution with vector of n ones
[~, m] = max(c); % starting index of maximizing window
p = prod(v(m+(0:n-1))); % corresponding product
end
Examples:
>> v = [1 2 3 4 5 6]; n = 3;
>> [m, p] = f(v,n)
m =
4
p =
120
>> v = [1 2 3 4 5 6]; n = 7;
>> [m, p] = f(v,n)
m =
-1
p =
0
>> v = [1 4 6 2 5 3]; n = 3;
>> [m, p] = f(v,n)
m =
3
p =
60
I have a set of values (say 20 values) in an array .
A = [1:20]
I want to divide it in to subsets of known size.
If I want to divide it in to 4 sets of size 5 then I can use
y = reshape(A,5,4)'
But my problem is when I dont have matching multiples of sizes.
Say I want to divide array in to sets of 3. So there will be 7 sets but the last set will be short.
what exactly I want is
a1= [1 2 3]
a2= [4 5 6]
a3= [7 8 9]
a4= [10 11 12]
a5= [13 14 15]
a6= [16 17 18]
a7= [19 20]
How can I do this kind of subgrouping to a vector in MATLAB?
You can use
y = mat2cell(A,1, diff([0:n:numel(A)-1 numel(A)]));
Then a1=y{1} and so on.
You'll need to write your own function for this. For example:
A = 1:20;
n = length(A);
x = 3;
y = ceil(n/x);
out = cell(y,1);
for i = 1:y
startIdx = x*(i-1)+1;
endIdx = min(startIdx + x - 1,n);
out{i} = A(startIdx:endIdx);
end
Then you can access each row in the cell array:
a1 = out{1};
a2 = out{2};
...
A = 1:20;
n = 3;
% y = number of sets
y = ceil(length(A) / n);
ai = num2cell(reshape(A(1:n*y),m,n)');
if length(A) > n * y
ai = [ai; A(n*y+1:end)];
end