I have a matrix, c, and I want to search many times for the index of the positive minimum element
d = min(c(c>0));
[x,y] = find(c == d);
but in the next search I want it to skip the old y.
how to do it?
I want to use x and y in some other calculation.
also I want to find this d minimum just within specific columns in the matrix c like:
j from m+1 to n-1
please help
Define mask = zeros(size(c)); before the loop.
And before finding the minimum use,
newc = c + mask;
d = min(newc(newc>0));
[x,y] = find(newc == d);
mask(:,y) = NaN;
I think you can update the c matrix. I mean:
% In the loop, use it:
[x,y]=find(c==d);
c(:, y) = [];
If c matrix is important, you can use a temporary variable equals to c, instead of using c.
Related
I have been playing with matlab and was trying to calculate absolute relative error using two arrays. However, when I divide the two arrays, my resulting array has the same value throughout the array even though when I calculate the values by hand, they are not all the same. I was wondering why my resulting array shows the same answer for every value in the array.
Here is my code:
function [X] = absrelerror(A, B)
% Calculates absolute relative error for true value A and approximate value B.
A = linspace(sin(-pi/6), sin(pi/6), 50); %True
B = linspace(-pi/6, pi/6, 50); %Approximate
Y = abs((A-B) ./ A); %ARE equation
X = Y * 100; %convert to percent
end
I think you have approached the problem in a wrong way.
Here, the first elements of both A and B are constant. Also, the spacing between two elements of any of the vectors is also constant. Say, they are c and d where c = {A(50) - A(1)}/49 and d = {B(50) - B(1)}/49. Now, the nth value of Y is {A(1)*nc - B(1)*nd}/{A(1)*c} = {A(1)*c - B(1)*c}/A(1) which is constant. So, it's not surprising that MATLAB is giving a constant value in Y.
If I have understood correctly what you are trying to do, then you should do it in the following way:
%A = linspace(sin(-pi/6), sin(pi/6), 50); %True
B = linspace(-pi/6, pi/6, 50); %Approximate
%Y = abs((A-B) ./ A); %ARE equation
%X = Y * 100; %convert to percent
A = sin(B);
X = abs((A-B)./A) * 100;
fprintf('%f ', X)
The output is:
4.719755 4.330958 3.960262 3.607413 3.272170 2.954306 2.653606 2.369868 2.102903 1.852533 1.618593 1.400927 1.199394 1.013862 0.844209 0.690325 0.552111 0.429477 0.322344 0.230643 0.154315 0.093311 0.047592 0.017130 0.001903 0.001903 0.017130 0.047592 0.093311 0.154315 0.230643 0.322344 0.429477 0.552111 0.690325 0.844209 1.013862 1.199394 1.400927 1.618593 1.852533 2.102903 2.369868 2.653606 2.954306 3.272170 3.607413 3.960262 4.330958 4.719755
I'm trying to create a vector of a specified size, the numbers of which would start from value K and increase/decrease with the specified step. I have to only use : operator.
Here's an example:
vector_dimension = 5;
start_value = 1;
step = 4;
last_number = ??
vector = start_value:step:(?last_number?)
% outputs: 1 5 9 13 17
The important thing is that the step can be either positive or negative.
So, basically, I just need to find the largest/smallest value of the range.
I tried doing things like this:
if (step >= 0)
vector = start_num:step:vector_dimension*step-1;
else
vector = start_num:step:(vector_dimension/step)+vector_dimension+1;
end
but this doesn't work correctly for far too many cases.
Does anybody have any ideas on how to implement that?
vector = start_value + (0:vector_dimension-1)*step
Basically everything relies on the following formula. If the step S, the start value s and the number of points n are known, you can evaluate the step S as:
S=(f-s)/(n-1)
where f is the last number (unknown).
By inverting such formula you find f as:
f=S*(n-1)+s
in your case f=4*(5-1)+1=17.
Once f is known you can apply the usual syntax vector=s:S:f.
I have a dwc = [3001 x 2 double] which more or less is a sin function, I have a for loop finding top values in dwc(:,2). Lets say that there is a top value in dwc(531,2) which way is best way or what is easy to take dwc(531,1) and dwc(531,2) and make an M = [num_of_top_points x 2 double]?
For the following loop, what do I do?
j = 0;
for i = 2:size(dwcL01,1)-1
if dwcL01(i,2) > dwcL01(i-1,2) && dwcL01(i,2) > dwcL01(i+1,2)
j = j+1;
?? = dwcL01(i,:);
end
end
This is how you complete your loop
j = 0;
M = [];
for i = 2:size(dwcL01,1)-1
if dwcL01(i,2) > dwcL01(i-1,2) && dwcL01(i,2) > dwcL01(i+1,2)
j = j+1;
M(j, :) = dwcL01(i, :);
end
end
But you could do this much more efficiently by vectorizing
%//Some example data
x = -4*pi:0.5:4*pi;
y = cos(x);
dwcL01 = [x(:), y(:)]; %// (:) just makes it a column
%// Finding the peaks using diff and sign. Note that I add the first element to the beginning as diff reduces the size by one so this prevents offsetting
F = diff(sign(diff([dwcL01(1,2);dwcL01(:,2)]))) < 0;
M = [dwcL01(F,:)', dwcL01(F,:)'];
plot(x, y, M(:,1), M(:,2), '*r')
How that works is first we find the difference of each element consecutive element pair. Now when the sign changes, that means we've hit a max or min. If the sign change is negative then the gradient went from positive to negative which is a max. So I use diff(sign()) to find the points where the sign changes and then > 0 to create a logical matrix with false everywhere expect for the max. Then I use logical indexing to extract the max.
You could append it to a matrix (let's call it dwcL01_max) - this isn't the fastest way because the matrix size changes each loop but it works:
dwcL01_max = [dwcL01_max dwcL01(i,:)];
The other option would be to use the builtin findpeaks (from the signal proc toolbox)
[~, dwcL01_peaks] = findpeaks(dwcL01(:,2));
dwcL01_max = dwcL01(dwcL01_peaks, :);
By "practically equivalent", I mean that their distances are of order epsilon apart (or 0.000001). Equality in MATLAB often doesn't really work for long floating numbers.
If I simply do abs(A1 - A2) < 0.000001, it's not going to work since size(A1) != size(A2)
You can get the answer by calculating distance between two vectors using MATLAB's pdist2 function.
dist=pdist2(A1,A2);
minDist=min(dist,[],2);
indices_A1=minDist<=0.000001;
desired_A1=A1(indices_A1);
Not tested, but should work.
Since you care about a distance from any element to any element, you can create a distance matrix from the vectorized matrices and probe that for the distance threshold. Example:
A = rand(10, 4); % (example) matrix A
B = rand(3, 5); % matrix B of different size
minDist = 0.005;
Solution: Repeat vectorized matrices, column- and row-wise to get same size matrices and apply matrix-based distance estimation:
Da = repmat(A(:), 1, length(B(:))); % size 40 x 15
Db = repmat(B(:)', length(A(:)), 1); % size 40 x 15
DD = Da - Db;
indA = any(abs(DD) < minDist, 2);
The use of any() will give you logical indices to any value of A that is close to any value of B). You can directly index/return the elements of A using indA.
The matrix DD (as #Shai also points out) can be equivalently estimated through bsxfun
DD = bsxfun(#minus, A(:), B(:)');
In addition: you can map from row-index (corresponding to A elements) back to matrix A indices using:
[iA, jA] = ind2sub(size(A), indA);
Assert/test that all the returned values are less than minDist,e.g. using:
for k = 1:length(iA);
d(k) = min(min(abs(A(iA(k), jA(k)) - B)));
end
all(d < minDist)
(tested in Octave 3.6.2)
I want to store data coming from for-loops in an array. How can I do that?
sample output:
for x=1:100
for y=1:100
Diff(x,y) = B(x,y)-C(x,y);
if (Diff(x,y) ~= 0)
% I want to store these values of coordinates in array
% and find x-max,x-min,y-max,y-min
fprintf('(%d,%d)\n',x,y);
end
end
end
Can anybody please tell me how can i do that. Thanks
Marry
So you want lists of the x and y (or row and column) coordinates at which B and C are different. I assume B and C are matrices. First, you should vectorize your code to get rid of the loops, and second, use the find() function:
Diff = B - C; % vectorized, loops over indices automatically
[list_x, list_y] = find(Diff~=0);
% finds the row and column indices at which Diff~=0 is true
Or, even shorter,
[list_x, list_y] = find(B~=C);
Remember that the first index in matlab is the row of the matrix, and the second index is the column; if you tried to visualize your matrices B or C or Diff by using imagesc, say, what you're calling the X coordinate would actually be displayed in the vertical direction, and what you're calling the Y coordinate would be displayed in the horizontal direction. To be a little more clear, you could say instead
[list_rows, list_cols] = find(B~=C);
To then find the maximum and minimum, use
maxrow = max(list_rows);
minrow = min(list_rows);
and likewise for list_cols.
If B(x,y) and C(x,y) are functions that accept matrix input, then instead of the double-for loop you can do
[x,y] = meshgrid(1:100);
Diff = B(x,y)-C(x,y);
mins = min(Diff);
maxs = max(Diff);
min_x = mins(1); min_y = mins(2);
max_x = maxs(1); max_y = maxs(2);
If B and C are just matrices holding data, then you can do
Diff = B-C;
But really, I need more detail before I can answer this completely.
So: are B and C functions, matrices? You want to find min_x, max_x, but in the example you give that's just 1 and 100, respectively, so...what do you mean?