I have an Nx2 matrix say D(k1,k2).I have to compare k1 and k2 from each row and switch accordingly. There is another vector d(i) which has M values.
if k1 and k2 is any one value of d(i) I have to switch.
if D(k1,1)==d(i)&&D(k1,2)==d(i)....
Is there any method to compare all the d(i) elements in the if loop without using a for loop for i?
You can use the ismember function for checking if the vector d contains certain values:
D_in_d = ismember(D,d);
and then you still have to loop to perform the flipping operation on specific rows:
for i=1:size(D,1)
if all(D_in_d(i,:))
D(i,:)=fliplr(D(i,:));
end
end
This is relatively easy to accomplish with matlab's vectorizion without any loops at all.
% A swap logical vector ( 1 if you need to swap that row, 0 otherwise)
swap_logical = ( ismember(D(:,1),d) | ismember(D(:,2),d) );
% Vectorized swapping based on the swap boolian.
Dnew = swap_logical.*D(:,2:-1:1) + ~swap_logical.*D;
Related
I'm trying to generate a 100-by-5 matrix where every line is a permutation of 1..100 (that is, every line is 5 random numbers from [1..100] without repetitions).
So far I've only been able to do it iteratively with a for-loop. Is there a way to do it more efficiently (using fewer lines of code), without loops?
N = 100;
T = zeros(N, 5);
for i = 1:N
T(i, :) = randperm(100, 5);
end
Let
N = 100; % desired number of rows
K = 5; % desired number of columns
M = 100; % size of population to sample from
Here's an approach that's probably fast; but memory-expensive, as it generates an intermediate M×N matrix and then discards N-K rows:
[~, result] = sort(rand(N, M), 2);
result = result(:, 1:K);
There is very little downside to using a loop here, at least in this minimal example. Indeed, it may well be the best-performing solution for MATLAB's execution engine. But perhaps you don't like assigning the temporary variable i or there are other advantages to vectorization in your non-minimal implementation. Consider this carefully before blindly implementing a solution.
You need to call randperm N times, but each call has no dependency on its position in the output. Without a loop index you will need something else to regulate the number of calls, but this can be just N empty cells cell(N,1). You can use this cell array to evaluate a function that calls randperm but ignores the contents (or, rather, lack of contents) of the cells, and then reassemble the function outputs into one matrix with cell2mat:
T = cell2mat(cellfun(#(~) {randperm(100,5)}, cell(N,1)));
This question already has answers here:
Element-wise array replication in Matlab
(7 answers)
Closed 6 years ago.
This is a basic program but since I'm new to MATLAB, I'm not able to figure out the solution.
I have a column vector "Time" in which I want to print value "1" in first 147 cells, followed by "2" in 148 to 2*147 cells and so on. For that, I have written the following script:
Trial>> c=1;
Trial>> k=0;
Trial>> for i = c:146+c
Time(i,1)=1+k;
c=i;
k=k+1;
end
I know I need to iterate the loop over "Time(i,1)=1+k;" before it executes the next statement. I tried using break but that's not supposed to work. Can anyone suggest me the solution to get the desired results?(It was quite simple in C with just the use of curly braces.)
I am sure you don't want to run c=i; in every iteration.
My code should work for you:
x = 10; % Replace 10 by the max number you need in your array.
k = 1;
for i = 1 : x * 147
Time(i, 1) = k;
if rem(i, 147) == 0
k = k + 1;
end
end
This is the prime example of a piece of code that should be vectorized can help you understand vectorization. Your code can be written like this:
n = 147;
reps = 10; %% Replace this by the maximum number you want your matrix to have
Time = reshape(bsxfun(#plus, zeros(n,1), 0:reps), 1, []);
Explanation:
Let A be a column vector (1 column, n rows), and B be a row vector (1 row, m columns.
What bsxfun(#plus, A, B) will do here is to add all elements in A with all elements in B, like this:
A(1)+B(1) A(1)+B(2) A(1)+B(3) ... A(1)+B(m)
A(2)+B(1) A(2)+B(2) ............. A(2)+B(m)
............................................
A(n)+B(1) A(n)+B(2) .............. A(n)+B(m)
Now, for the two vectors we have: zeros(n,1), and 0:reps, this will give us;
0+0 0+1 0+2 0+reps
0+0 0+1 0+2 0+reps
% n rows of this
So, what we need to do now is place each column underneath each other, so that you will have the column with zeros first, then the row with ones, ... and finally the one with reps (147 in your case).
This can be achieved by reshaping the matrix:
reshape(bsxfun(#plus, zeros(n,1), 0:reps), [], 1);
^ ^ ^ ^
| | | Number of rows in the new matrix. When [] is used, the appropriate value will be chosen by Matlab
| | Number of rows in the new matrix
| matrix to reshape
reshape command
Another approach is using kron:
kron(ones(reps+1, 1) * 0:(n-1)
For the record, a review of your code:
You should always preallocate memory for matrices that are created inside loops. In this case you know it will become a matrix of dimensions ((reps+1)*n-by-1). This means you should do Time = zeros((reps+1)*n, 1);. This will speed up your code a lot.
You shouldn't use i and j as variable names in Matlab, as they denote the imaginary unit (sqrt(-1)). You can for instance do: for ii = 1:(n*147) instead.
You don't want c=i inside the loop, when the loop is supposed to go from c to c + 146. That doesn't make much sense.
You can use repmat,
x = 10; % Sequence length (or what ever it can be called)
M = repmat(1:x,147,1); % Replicate array 1:x for 147 columns
M = M(:); % Reshape the matrix so that is becomes a column vector.
I can assume that this is a task to practice for loops, but this will work.
An alternative solution may be to do
n = 147;
reps = 10;
a = ceil( (1:(n*reps)) / n);
You first construct an array with the length you want. Then you divide, and round of upwards. 1 to 147 will then become 1.
I'm a little confused as to how to vectorize this for loop see code below:
array1=[xfreq_orig,yamp_orig,yamp_inv,phase_orig] %frequency, amplitudes, phases to use
t_rebuilt=linspace(0,2*pi,44100)
aa_sig_rebuilt_L=zeros(1,length(t_rebuilt));
aa_sig_combined_L=zeros(1,length(t_rebuilt));
sig_full_L=zeros(1,length(t_rebuilt));
for kk=1:1:numel(xfreq_orig);
aa_sig_rebuilt_L = array1(kk, 2)*cos ((array1(kk,1))*t_rebuilt+(array1(kk, 4)));
aa_sig_combined_L = aa_sig_combined_L + aa_sig_rebuilt_L;
end
sig_full_L=(aa_sig_combined_L/max(abs(aa_sig_combined_L))*.8);
I came up with this as vectorization
Example:
array1=[10,.4,.34,2.32;12,.3,.45,.4];
t_rebuilt=linspace(0,2*pi,44100)
aa_sig_rebuilt_L = array1(kk, 2).*cos ((array1(kk,1)).*t_rebuilt+(array1(kk, 4)));
aa_sig_combined_L = sum(aa_sig_rebuilt_L);
What I don't know how to do is how to get the kk variable to access the rows incrementally
THanks.
One option is to use bsxfun as follows
a = array1;
t = t_rebuilt;
aa_sig_rebuilt_L = bsxfun(#times, a(:,2) , ...
cos( bsxfun(#plus, bsxfun(#times, a(:,1), t), a(:,4)) ));
aa_sig_combined_L = sum(aa_sig_rebuilt_L);
Bear in mind that this will use more memory than the version will a loop (it will use numel(xfreq_orig) times as much memory, as it computes every row of aa_sig_rebuilt_L before summing them, whereas the loop computes each row, adds it to the sum and then discards it).
The function bsxfun is used when you need to perform a binary operation between arrays of different sizes, e.g. a TxN matrix and a Tx1 vector. In this case it would perform the operation between each column of the matrix and the vector.
In your case you have a column vector and a row vector, and the operation is applied to the i'th element of the column vector and the j'th element of the row vector, to get the ij'th element of a matrix.
I want a code the below code more efficient timewise. preferably without a loop.
arguments:
t % time values vector
t_index = c % one of the possible indices ranging from 1:length(t).
A % a MXN array where M = length(t)
B % a 1XN array
code:
m = 1;
for k = t_index:length(t)
A(k,1:(end-m+1)) = A(k,1:(end-m+1)) + B(m:end);
m = m + 1;
end
Many thanks.
I'd built from B a matrix of size NxM (call it B2), with zeros in the right places and a triangular from according to the conditions and then all you need to do is A+B2.
something like this:
N=size(A,2);
B2=zeros(size(A));
k=c:length(t);
B2(k(1):k(N),:)=hankel(B)
ans=A+B2;
Note, the fact that it is "vectorized" doesn't mean it is faster these days. Matlab's JIT makes for loops comparable and sometimes faster than built-in vectorized options.
I have two matrices A and B, both of which are Nx3 matrices.
I'm currently getting the maximum value and index for each row of matrix A using:
[maxA, idx] = max(A, [], 2)
idx(j) indicates which column contained the maximum for row j. Now I'd like to select those same positions from matrix B.
I've currently implemented this using a loop:
for j = 1:numel(idx)
maxB(j) = B(j, idx(j))
end
My current implementation is fast enough, although I prefer to avoid unneeded loops so is there a way to express this without a loop?
You can build a vector of linear indices (I expect B to be the same size as A):
vec_indices = sub2ind(size(A), 1:numel(idx), idx);
Then you can use that vector directly for lookup:
maxB = B(vec_indices)
You can construct the single dimension index into the matrix and get them that way. All multidimensional matrices in matlab can be addressed.
You can use
maxB = B(sub2ind([1:length(idx)]',idx(:)));
In one line:
maxB = B(A == max(A, [], 2) * ones(1, 3));
But this is not safe. It assumes unique values in every row of A.