Consider two n by n-1 matrix and an n by 1 vector (for example lets call them in order A, B and v). Elements of v are zero or one. If element v(m,1) is equal to one, I want to replace elements A(1:m-1,m-1) by B(1:m-1,m-1) and elements A(m+1:n,m) by B(m+1:n,m).
How can I do that? Could anyone help? To make the question more clear, consider below example.
example:
A=[1,2,3;4,5,6;7,8,9;12,13,14]
B=[3,4,5;6,7,8;9,10,11;6,5,3]
v=[0,1,0,1]
Result should be:
result= [3,2,5;4,5,8;7,10,11;12,5,14]
Here is an alternative, using logical indexing:
temp = A;
ind = 1:size(v,2);
for k = ind(v==1)
if k<=size(A,2)+1
A(1:k-1,k-1) = B(1:k-1,k-1);
B(1:k-1,k-1) = temp(1:k-1,k-1);
if k<size(A,2)
A(k+1:end,k) = B(k+1:end,k);
B(k+1:end,k) = temp(k+1:end,k);
end
end
end
You can acheive the desired result using find, which returns the indexes on non-zero elements, and a for-loop:
R = A; % assuming you've set A, B and v already.
n = size(A,1);
v1 = find(v);
for i=1:length(v1)
m=v1(i);
if m>1
R(1:m-1,m-1)=B(1:m-1,m-1);
end
if m<n
R(m+1:end,m)=B(m+1:end,m);
end
end
As I point out in a comment, v has to have length n-1 if v(n-1)=1 otherwise m+1:end is not a valid index range.
Edited to make second assignment optionally as per comments.
Related
I have two matrices A and B. A(:,1) corresponds to an x-coordinate, A(:,2) corresponds to a y-coordinate, and A(:,3) corresponds to a certain radius. All three values in a row describe the same circle. Now let's say...
A =
[1,4,3]
[8,8,7]
[3,6,3]
B =
[1,3,3]
[1, 92,3]
[4,57,8]
[5,62,1]
[3,4,6]
[9,8,7]
What I need is to be able to loop through matrix A and determine if there are any rows in matrix B that are "similar" as in the x value is within a range (-2,2) of the x value of A (Likewise with the y-coordinate and radius).If it satisfies all three of these conditions, it will be added to a new matrix with the values that were in A. So for example I would need the above data to return...
ans =
[1,4,3]
[8,8,7]
Please help and thank you in advance to anyone willing to take the time!
You can use ismembertol.
result = A(ismembertol(A,B,2,'ByRows',1,'DataScale',1),:)
Manual method
A = [1,4,3;
8,8,7;
3,6,3];
B = [1,3,3;
1,92,3;
4,57,8;
5,62,1;
3,4,6;
9,8,7]; % example matrices
t = 2; % desired threshold
m = any(all(abs(bsxfun(#minus, A, permute(B, [3 2 1])))<=t, 2), 3);
result = A(m,:);
The key is using permute to move the first dimension of B to the third dimension. Then bsxfun computes the element-wise differences for all pairs of rows in the original matrices. A row of A should be selected if all the absolute differences with respect to any column of B are less than the desired threshold t. The resulting variable m is a logical index which is used for selecting those rows.
Using pdist2 (Statistics and Machine Learning Toolbox)
m = any(pdist2(A, B, 'chebychev')<=t, 2);
result = A(m,:);
Ths pdist2 function with the chebychev option computes the maximum coordinate difference (Chebychev distance, or L∞ metric) between pairs of rows.
With for loop
It should work:
A = [1,4,3;
8,8,7;
3,6,3]
B = [1,3,3;
1,92,3;
4,57,8;
5,62,1;
3,4,6;
9,8,7]
index = 1;
for i = 1:size(A,1)
C = abs(B - A(i,:));
if any(max(C,[],2)<=2)
out(index,:) = A(i,:);
index = index + 1
end
end
For each row of A, computes the absolute difference between B and that row, then checks if there exists a row in which the maximum is less than 2.
Without for loop
ind = any(max(abs(B - permute(A,[3 2 1])),[],2)<=2);
out = A(ind(:),:);
Sorry about the bad title, I'm struggling to word this question well. Basically what I want to do is extract elements from a 2d matrix, from row by row, taking out a number of elements (N) starting at a particular column (k). In for loops, this would look like.
A = magic(6);
k = [2,2,3,3,4,4]; % for example
N = 3;
for j = 1:length(A)
B(j,:) = A(j,k(j):k(j)+N-1);
end
I figure there must be a neater way to do it than that.
You could use bsxfun to create an array of indices to use. Then combine this with the row numbers and pass it to sub2ind.
inds = sub2ind(size(A), repmat(1:size(A, 1), 3, 1), bsxfun(#plus, k, (0:(N-1))')).';
B = A(inds);
Or alternately without sub2ind (but slightly more cryptic).
B = A(bsxfun(#plus, 1:size(A,1), ((bsxfun(#plus, k, (0:(N-1)).')-1) * size(A,1))).');
Here's one approach using bsxfun's masking capability and thus logical indexing -
C = (1:size(A,2))';
At = A.';
B = reshape(At(bsxfun(#ge,C,k) & bsxfun(#lt,C,k+N)),N,[]).';
I have the following vector u:
u=[a1,a2,a3,a4,b1,b2,b3,b4,c1,c2,c3,c4];
I want to permute the elements of of u to make the following vector, uNew:
uNew=[a1,b1,c1,a2,b2,c2,a3,b3,c3,a4,b4,c4];
I can think of no way of doing this other than with a for loop:
uNew=[];
for i=1:4
uNew=[uNew,u(i:4:end)];
end
But I'm hoping that a built-in function exits? Thanks!
Reshape it to a matrix which, read column by column, contains the order you want:
n=3 % number of categories (a,b,c)
u2=reshape(u,[],n).'
then convert it back to a vector:
u2=u2(:);
I'd use Daniel's approach. But just to provide an alternative:
m = 4; %// size of each initial "block"
[~, ind] = sort(mod(0:numel(u)-1,m)+1);
uNew = u(ind);
Note that this works because, as per sort's documentation,
The ordering of the elements in B (output) preserves the order of any equal elements in A (input)
Another approach:
N = 4;
uNew = u(mod(0:numel(u)-1, N) * N + floor((0:numel(u)-1)/N) + 1);
I'm new to MATLAB, and programming in general, and I am having difficulty accomplishing what I am sure is a very, very simple task:
I have a list of vectors v_i for i from 1 to n (n in some number), all of the same size k. I would like to create a vector v that is a "concatenation" (don't know if this is the correct terminology) of these vectors in increasing order: what I mean by this is that the first k entries of v are the k entries of v_1, the k+1 to 2k entries of v are the k entries of v_2 etc. etc. Thus v is a vector of length nk.
How should I create v?
To put this into context, here is function I've began writing (rpeakindex will just a vector, roughq would be the vector v I mentioned before):
function roughq = roughq(rpeakindex)
for i from 1 to size(rpeakindex) do
v_i = [rpeakindex(i)-30:1:rpeakindex(i)+90]
end
Any help is appreciated
Let's try two things.
First, for concatenating vectors there are a couple of methods here, but the simplest would be
h = horzcat(v_1, v_2);
The bigger problem is to enumerate all vectors with a "for" loop. If your v_n vectors are in a cell array, and they are in fact v{i}, then
h= [];
for j=1:n
h = horzcat(h, v{i});
end
Finally, if they only differ by name, then call them with
h=[];
for j=1:n
h= horzcat(h, eval(sprintf('v_%d',j));
end
Let the arrays (vectors) be:
v_1=1:10;
v_2=11:20;
v_3=21:30;
v_4=31:40;
and so on.
If they are few (e. g. 4), you can directly set then as input in the cat function:
v=cat(2,v_1,v_2,v_3,v_4)
or the horzcat function
v=horzcat(v_1,v_2,v_3,v_4)
otherwise you can use the eval function within a loop
v1=[];
for i=1:4
eval(['v1=[v1 v_' num2str(i) ']'])
end
Hope this helps.
Concatenating with horzcat is definitely an option, but since these vectors are being created in a function, it would be better to concatenate these vectors automatically in the function itself rather than write out horzcat(v1,v2,....vn) manually.
Given the function mentioned in the question, I would suggest something like this:
function v = roughq(rpeakindex)
v = zeros(121,length(rpeakindex)); %// create a 2D array of all zeros
for i = 1:size(rpeakindex)
v(:,i) = [rpeakindex(i)-30:1:rpeakindex(i)+90]; %// set result to ith column of v
end
v = v(:)'; %'//reshape v to be a single vector with the columns concatenated
end
Here's a simplified example of what's going on:
N = 3;
v = zeros(5,N);
for i = 1:N
v(:,i) = (1:5)*i;
end
v = v(:)';
Output:
v =
1 2 3 4 5 2 4 6 8 10 3 6 9 12 15
You may want to read up on MATLAB's colon operator to understand the v(:) syntax.
If you mean 2d matrix, you are using for holding vectors and each row hold vector v then you can simply use the reshape command in matlab like below:
V = [] ;
for i = 1:10
V(i,:) = randi (10,1 ,10) ;
end
V_reshpae = reshape (V, 1, numel(V)) ;
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);