Let X = [1, 2, 3, 4, 5] and Y = [1, 2, 1, 0, 1] be vectors where X maps into Y.
Now I want to identify the maximum and minimum of Y, which is easy: [value_min, id_min] = min(Y) = [0, 4] and [value_max, id_max] = max(Y) = [2, 2].
Then I want to remove the element from X corresponding to the minimum in Y and expand evenly around the element in X corresponding to the maximum in Y, while keeping the number of points equal. For this example we remove X(4)=[]. Then we expand like X(2)=(X(2) - X(1))/2 and X(3)=(X(3) - X(2))/2 such that X looks like X = [1, 1.5, 2.5, 3, 5]. How can I achieve this? I think there is a general pattwern.
Solution
Now the following snipped should work for any vector of length N. Note that the first and final element are fixed.
[value_max, id_max] = max(Y(2:N-1));
X(id_max) = (X(id_max) - X(id_max-1))/2;
X(id_max+1) = (X(id_max+1) - X(id_max))/2;
[value_min, id_min] = min(Y(2:N-1));
X(id_min)=[];
Here is a solution to your problem but there are a few things you should take care of
% Any Vector should work
X=[1 2 3 4 5];
Y=[1 2 1 0 1];
%We dont need the actual min max
[~,MIN]=min(Y(2:end-1));
[~,MAX]=max(Y(2:end-1));
%you dont look at the first element so the index has to be increased by 1
MIN=MIN+1;
MAX=MAX+1;
X(MIN)=[];%taking out the smallest element
Xnew= [X(1:MAX) X(MAX:end)]; %Extend the vector by taking the MAX value twice
%the mean for 2 elements is A+B/2
Xnew(MAX)=mean(Xnew(MAX-1:MAX)); %the left one and the element next to it
Xnew(MAX+1)=mean(Xnew(MAX+1:MAX+2)); %the right one and the element next ot it
%rewrite X and clear Xnew
X=Xnew;
clear Xnew;
First of all this isnt very efficient, but if its just used to
modify some vectors and not get called a million times a day it will
do the trick.
In your text you say remove the minima then stretch
around the maxima, in your solution metacode it is the other way
around. this will influence the outcome when min and max are next to
each other, so please check which way you prefer.
Y isnt changed in this at all so it cant be performed multiple times on the same vector.
Is N (the length) of any importance later on? if not you can always just refer to "end"
Related
Good day,
I have a question what I want to achieve without the loop if possible. As title says I need to do windowed subtraction of vectors that are not same size and then finding the mean of results.
As example, let say that we have vector a = [2 3 4 5 6] and vector b = [1 2].
Program will have to move window with smaller numbers of elements (in this example vector b) over bigger one (vector a) and make operations on that way so it starts in first two elements in vector a and make subtraction with vector b and then sum results and find mean.
In this example it will just make calculation of subtraction 2-1 = 1, 3-2 = 1, summing results 1+1=2 and divide them with 2 (because vector b is that size). Final result is 1.
Then we move window on second elements of vector a (value 3 and 4 there, or index 2 and 3) and repeat process to the last elements of vector a.
For final result we need to get vector c who consist of elements [1 2 3 4] for this example.
Is this possible to do without looping because I have data sets over 10k of size. Thanks in advance
I can solve it with only one loop, iterating through "b" (two loops in your example).
Declare vectors (as columns! This is needed for matlabs computations to work)
a = [2 3 4 5 6]';
b = [1 2]';
Declare matrix for computed results. Each column represents subtractions of elements in "a" with one of the elements in "b".
c = zeros(length(a)-length(b)+1,length(b));
for k = 1:length(b)
c(:,k) = a(k:length(a)-length(b)+k)-b(k);
end
Now just sum the elements in "c" row wise and divide by length of "b" to get the mean
result = sum(c,2)/length(b);
You can simplify this for your exact example, but this is a generic solution for any vetors "a" and "b", where "b" is the smaller vector.
I'm trying to find an idiomatic way to do this.
Essentially I have an Nx2 matrix of points of the form
A = [3 4; 3 5; 4 5; 4, 6; 7 3]
I'd like my output to be [3 5; 4 6; 7 3]. In other words I would like each unique x value along with the maximum y value associated with that x.
I was hoping there would be some sort of
unique(A, 'rows', 'highestterm', 2)
method for accomplishing this, but couldn't find anything. Can anyone think of a vectorized way to solve this problem? I can do it pretty easily in a for loop, but would like to avoid that if possible.
I don't know of any single call, like you hoped. But, it can be done fairly tightly (and fully vectorized) like the code below.
%sort by first and then second column
A = sortrows(A,[1 2]);
%find each change in the first column of A
inds = find(diff(A(:,1)) > 0);
%add the last point...because find(diff) doesn't get the last point
inds(end+1) = size(A,1);
%get just those rows that meet the desired criteria
A = A(inds,:);
So, this works by sorting the data and looking for the values in the first column that don't repeat. If there are repeated values, this code grabs the last of the repeating values. Finally, because we sorted by both columns via sortrows(A,[1 2]), the last entry for a repeating value will have the biggest corresponding value from the 2nd column. I think that this hits all of your requirements.
Using accumarray and unique:
[r1, ~, u] = unique(A(:,1));
r2 = accumarray(u, A(:,2), [], #max);
result = [r1 r2];
How does one create a vector that is composed of a random sampling of two other vectors?
For example
Vector 1 [1, 3, 4, 7], Vector 2 [2, 5, 6, 8]
Random Vector [random draw from vector 1 or 2 (value 1 or 2), random draw from vector 1 or 2 (value 3 or 5)... etc]
Finally, how can one ask matlab to repeat this process n times to draw a distribution of results?
Thank you,
There are many ways you could do this. One possibility is:
tmp=round(rand(size(vector1)))
res = tmp.*vector1 + (1-tmp).*vector2
To get one mixed sample, you may use the idea of the following code snippet (not the optimal one, but maybe clear enough):
a = [1, 3, 4, 7];
b = [2, 5, 6, 8];
selector = randn(size(a));
sample = a.*(selector>0) + b.*(selector<=0);
For n samples put the above code in a for loop:
for k=1:n
% Sample code (without initial "samplee" assignments)
% Here do stuff with the sample
end;
More generally, if X is a matrix and for each row you want to take a sample from a column chosen at random, you can do this with a loop:
y = zeros(size(X,1),1);
for ii = 1:size(X,1)
y(ii) = X(ii,ceil(rand*size(X,2)));
end
You can avoid the loop using clever indexing via sub2ind:
idx_n = ceil(rand(size(X,1),1)*size(X,2));
idx = sub2ind(size(X),(1:size(X,1))',idx_n);
y = X(idx);
If I understand your question, you are choosing two random numbers. First you decide whether to select vector 1 or vector 2; next you pick an element from the chosen vector.
The following code takes advantage of the fact that vector1 and vector2 are the same length:
N = 1000;
sampleMatrix = [vector1 vector2];
M = numel(sampleMatrix);
randIndex = ceil(rand(1,N)*M); % N random numbers from 1 to M
randomNumbers = sampleMatrix(randIndex); % sample N times from the matrix
You can then display the result with, for instance
figure; hist(randomNumbers); % draw a histogram of numbers drawn
When vector1 and vector2 have different elements, you run into a problem. If you concatenate them, you will end up picking elements from the longer vector more often. One way around this is to create random samplings from both arrays, then choose between them:
M1 = numel(vector1);
M2 = numel(vector2);
r1 = ceil(rand(1,N)*M1);
r2 = ceil(rand(1,N)*M2);
randMat = [vector1(r1(:)) vector2(r2(:))]; % two columns, now pick one or the other
randPick = ceil(rand(1,N)*2);
randomNumbers = [randMat(randPick==1, 1); randMat(randPick==2, 2)];
On re-reading, maybe you just want to pick "element 1 from either 1 or 2", then "element 2 from either 1 or 2", etc for all the elements of the vector. In that case, do
N=numel(vector1);
randPick = ceil(rand(1,N)*2);
randMat=[vector1(:) vector2(:)];
randomNumbers = [randMat(randPick==1, 1); randMat(randPick==2, 2)];
This problem can be solved using the function datasample.
Combine both vectors into one and apply the function. I like this approach more than the handcrafted versions in the other answers. It gives you much more flexibility in choosing what you actually want, while being a one-liner.
I have a 3x3 matrix, A. I also compute a value, g, as the maximum eigen value of A. I am trying to change the element A(3,3) = 0 for all values from zero to one in 0.10 increments and then update g for each of the values. I'd like all of the other matrix elements to remain the same.
I thought a for loop would be the way to do this, but I do not know how to update only one element in a matrix without storing this update as one increasingly larger matrix. If I call the element at A(3,3) = p (thereby creating a new matrix Atry) I am able (below) to get all of the values from 0 to 1 that I desired. I do not know how to update Atry to get all of the values of g that I desire. The state of the code now will give me the same value of g for all iterations, as expected, as I do not know how to to update Atry with the different values of p to then compute the values for g.
Any suggestions on how to do this or suggestions for jargon or phrases for me to web search would be appreciated.
A = [1 1 1; 2 2 2; 3 3 0];
g = max(eig(A));
% This below is what I attempted to achieve my solution
clear all
p(1) = 0;
Atry = [1 1 1; 2 2 2; 3 3 p];
g(1) = max(eig(Atry));
for i=1:100;
p(i+1) = p(i)+ 0.01;
% this makes a one giant matrix, not many
%Atry(:,i+1) = Atry(:,i);
g(i+1) = max(eig(Atry));
end
This will also accomplish what you want to do:
A = #(x) [1 1 1; 2 2 2; 3 3 x];
p = 0:0.01:1;
g = arrayfun(#(x) eigs(A(x),1), p);
Breakdown:
Define A as an anonymous function. This means that the command A(x) will return your matrix A with the (3,3) element equal to x.
Define all steps you want to take in vector p
Then "loop" through all elements in p by using arrayfun instead of an actual loop.
The function looped over by arrayfun is not max(eig(A)) but eigs(A,1), i.e., the 1 largest eigenvalue. The result will be the same, but the algorithm used by eigs is more suited for your type of problem -- instead of computing all eigenvalues and then only using the maximum one, you only compute the maximum one. Needless to say, this is much faster.
First, you say 0.1 increments in the text of your question, but your code suggests you are actually interested in 0.01 increments? I'm going to operate under the assumption you mean 0.01 increments.
Now, with that out of the way, let me state what I believe you are after given my interpretation of your question. You want to iterate over the matrix A, where for each iteration you increase A(3, 3) by 0.01. Given that you want all values from 0 to 1, this implies 101 iterations. For each iteration, you want to calculate the maximum eigenvalue of A, and store all these eigenvalues in some vector (which I will call gVec). If this is correct, then I believe you just want the following:
% Specify the "Current" A
CurA = [1 1 1; 2 2 2; 3 3 0];
% Pre-allocate the values we want to iterate over for element (3, 3)
A33Vec = (0:0.01:1)';
% Pre-allocate a vector to store the maximum eigenvalues
gVec = NaN * ones(length(A33Vec), 1);
% Loop over A33Vec
for i = 1:1:length(A33Vec)
% Obtain the version of A that we want for the current i
CurA(3, 3) = A33Vec(i);
% Obtain the maximum eigen value of the current A, and store in gVec
gVec(i, 1) = max(eig(CurA));
end
EDIT: Probably best to paste this code into your matlab editor. The stack-overflow automatic text highlighting hasn't done it any favors :-)
EDIT: Go with Rody's solution (+1) - it is much better!
I have a 2-D matrix A(value, label). I want to find the label that has the largest and second largest sum of values. For example:
A = (1, 1;
2, 1;
3, 2;
4, 2;
5, 3)
In this case the result should be largest = 2, second largest = 3. How can I do this in MATLAB?
[b,m,n]=unique(a(:,2));
[val, idx]= sort(accumarray(n,a(:,1)),'descend');
b(idx(1:2))
Output is:
ans =
2
3
Something like this should do the trick.
A = [1, 1;
2, 1;
3, 2;
4, 2;
5, 3];
labels = unique(A(:,2)); % Pull out unique labels
for i = 1:numel(labels)
idx = (A(:,2) == labels(i)); % Find elements which match current label
s(i,1) = sum(A(idx,1)); % Sum them
end
r = sortrows([s labels], -1); % Sort by decreasing sum
r(1,2); % Label corresponding to largest sum
r(2,2); % Label corresponding to second largest sum
EDIT accumarray is a built-in function that will do this for you. Although I find the documentation on it somewhat cryptic.
Since your question isn't super clear and i don't get what sum you are referring to I'm just gonna guess that you are aiming for something like this
q=sortrows(A,-1);
q=q(1:2,:);
which will give the two labels (right column) with the largest values (left column) in q.
If this wasn't what you were looking for please comment.
EDIT: Missread which column that contained labels, corrected