Assuming I have a set of lines/curves and want to find at once where each one intersects with a selected one. Is it possible to vectorize this operation in matlab assuming I now all the equations that define the lines/curves?
lines to intersect with:
y_i=m_i*x+b_i; % i integer
master line
y=M*x+B;
Like in the figure I show.
I know I can do this one to one by:
M*x_inter+B=m_i*x_inter+b_i;
y_inter=M*x_inter+B;
and then put this in a for loop, But since the actual use of this is against hundreds of lines it would be more efficient to vectorize the operation.
You can just join all your m_i and b_i in column vectors and then solve it at once. Lets say you have 3 lines:
nlines = 3;
Bcol = ones(nlines ,1)*B; %your B value put into a column
Mcol = ones(nlines ,1)*M;
x_inter = (b - Bcol)./( Mcol - m); %this b contains all your b_i values, the same for m
y_inter = Mcol .*x_inter + Bcol
Obviously x_inter and y_inter wil be arrays and not single values
Related
So I want to concatenate an m x n matrix to obtain a 1 x mn matrix. The matrix I want to concatenate are generated from a while loop. Although the number of columns will always be 3, I however cannot tell how many rows there will be for each iteration. Also, the row sizes for each iteration may not always be the same.
The code runs in cases where the row sizes were all equal to 6, but in cases where they aren't equal I get an error:
Error using vertcat Dimensions of matrices being concatenated are not consistent.
parts of the code are as follows:
A = [];
B = [];
searchArea = 2;
for ii = 1: numel(velocity)
Do ....
while area(ii,:) < searchArea
Do ....
% COLLATE vectors for A
A = [A; [Ax(ii), Ay(ii), Az(ii)]];
Do ...
end
%# Copy the A into new variable (B) and Reshape into row vector so as to associate each row to its corresponding velocity
B = [B; reshape(A.',1,[])];
A = [];
end
Could someone please advice me on what I am doing wrong here. I would clarify further if there be need. Thanks guys!
If it's your intent that B ends up being a row vector, then you need to change this:
B = [B; reshape(A.',1,[])]; % Does vertical concatenation
to this:
B = [B reshape(A.',1,[])]; % Does horizontal concatenation (note there's no semicolon)
so that each row vector gotten from reshaping A gets added to the end of the row instead of as a new row (as the semicolon indicates).
I have a matrix:
1|2|3|4
4|5|6|7
7|8|9|10
10|11|12|13
I want to multiply the indices of this matrix with indices of another matrix of different size:
7|8|9
9|10|10
10|11|11
for these two matrices I have used the following for loops:
for x=1:4
for y=1:4
for m=1:3
for n=1:3
c=(m*x+n*y);
end
end
end
end
Is there any way to rewrite the above code without using loops? If the indices of each element can be generated in the above matrices, I think it can be done. Please help
mx = m'*x;
mx = mx(:);
ny = n'*y;
ny = ny(:);
mxe = repmat(mx, [length(ny), 1]);
nye = repmat(ny, [length(mx), 1]);
c = mxe+nye;
This will result in c containing all the values that get put in during that loop you have there (note that in your loop, value gets assigned and overwritten).
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 need some help to vectorize the following operation since I'm a little confused.
So, I have a m-by-2 matrix A and n-by-1 vector b. I want to create a n-by-1 vector c whose entries should be the values of the second column of A whose line is given by the line where the correspondent value of b would fall...
Not sure if I was clear enough. Anyway, the code below does compute c correctly so you can understand what is my desired output. However, I want to vectorize this function since my real n and m are in the order of many thousands.
Note that values of bare non-integer and not necessarily equal to any of those in the first column of A (these ones could be non-integers too!).
m = 5; n = 10;
A = [(0:m-1)*1.1;rand(1,m)]'
b = (m-1)*rand(n,1)
[bincounts, ind] = histc(b,A(:,1))
for i = 1:n
c(i) = A(ind(i),2);
end
All you need is:
c = A(ind,2);
B = randn(1,25,10);
Z = [1;1;1;2;2;3;4;4;4;3];
Ok, so, I want to find the locations where Z=1(or any numbers that are equal to each other), then average across each of the 25 points at these specific locations. In the example you would end with a 1*25*4 array.
Is there an easy way to do this?
I'm not the most versed in Matlab.
First things first: break down the problem.
Define the groups (i.e. the set of unique Z values)
Find elements which belong to these groups
Take the average.
Once you have done that, you can begin to see it's a pretty standard for loop and "Select columns which meet criteria".
Something along the lines of:
B = randn(1,25,10);
Z = [1;1;1;2;2;3;4;4;4;3];
groups = unique(Z); %//find the set of groups
C = nan(1,25,length(groups)); %//predefine the output space for efficiency
for gi = 1:length(groups) %//for each group
idx = Z == groups(gi); %//find it's members
C(:,:,gi) = mean(B(:,:,idx), 3); %//select and mean across the third dimension
end
If B = randn(10,25); then it's very easy because Matlab function usually works down the rows.
Using logical indexing:
ind = Z == 1;
mean(B(ind,:));
If you're dealing with multiple dimensions use permute (and reshape if you actually have 3 dimensions or more) to get yourself to a point where you're averaging down the rows as above:
B = randn(1,25,10);
BB = permute(B, [3,2,1])
continue as above