this is my problem, for example i have an equation x + y =2, so using matlab i want to know how to determine all the possible combination of values of x and y when you add it, and will give sum of 2 (ex: x1 = 0.98, y1 =0.12; x2=0.94 y2=0.16, and etc)
i think i need to use for loop?
for x = 2-y
end
for y =2-x
end
Values of x and y
x y
0 2
0.1 1.9
0.2 1.8
0.3 1.7
0.4 1.6
0.5 1.5
0.6 1.4
0.7 1.3
0.8 1.2
0.9 1.1
1 1
so guys i need your help thanks
To get all possible combinations of x and y between 0 and 2 with a step size of 0.1 you don't even need a for loop. You can create a vector x which contains all possible x values and then calculate the corresponding y's:
x = 0:0.1:2; % Create a vector of values between 0 and 2 in steps 0f 0.1
y = 2 - x;
This will give you two (row) vectors containing all possible combinations which add up to 2.
Related
I have 3 sequences in a cell-array:
Input_cell= {'ABCD','ACD', 'ABD'}
S1= 'ABCD' % which means A<B<C<D
S2= 'ACD' % which means A<C<D % missing B in the full string of 'ABCD'
S3= 'ABD' % which means A<B<D % missing C in the full string of 'ABCD'
I want to convert each of the strings in the Input_cell into a matrix M (i-by-j) which has to satisfy these conditions:
M(i,j) and M(j,i) are random
M(i,i) = 0.5
M(i,j) + M(j,i) = 1
M(i,j) < M(j,i) For example if A<B then M(A,B) < M(B,A)
For example if we have S1 = 'ABCD' (which means A<B<C<D), the M1 matrix will be expected as follows:
A B C D
A 0.5 0.3 0.2 0.1
B 0.7 0.5 0 0.4
C 0.8 1 0.5 0.1
D 0.9 0.6 0.9 0.5
If we have S2 = 'ACD' (which means A<C<D), missing B in the full string of 'ABCD', we will put the value 0.5 in every position of B in the matrix, the M2 matrix will be expected as follows:
A B C D
A 0.5 0.5 0.2 0.1
B 0.5 0.5 0.5 0.5
C 0.8 0.5 0.5 0.1
D 0.9 0.5 0.9 0.5
If we have S3 = 'ABD' (which means A<B<D), missing C in the full string of 'ABCD', we will put the value 0.5 in every position of C in the matrix, the M3 matrix will be expected as follows:
A B C D
A 0.5 0.4 0.5 0.1
B 0.6 0.5 0.5 0.3
C 0.5 0.5 0.5 0.5
D 0.9 0.7 0.5 0.5
How to create that kind of above matrices from a given cell-array of sequences?
Firstly you need to work out how to do this just for a single sequence:
Create a matrix of random numbers between 0.5 and 1:
M = 0.5*rand(4) + 0.5;
Set the main diagonal to equal 0.5
M(logical(eye(4))) = 0.5;
Set the upper triangle of M equal to 1 - the lower triangle:
M(triu(true(4))) = 1 - M(tril(true(4))); %// Note the main diagonal doesn't matter...
Work out which letter is missing and set the row and column equal to 0.5 accordingly:
fullSeq = 'abcd';
idx = find(fullSeq == setdiff(fullSeq, 'abd'));
%// at this point you'll need to check if idx is empty first...
M(:,idx) = 0.5;
M(idx,:) = 0.5;
And now that you can do it for one matrix, just loop over your cell array or else encapsulate this into a function and use cellfun.
Let's say I have a 2 by 9 matrix. I want to replace the 2 by 3 matrices inside this matrix with respect to descending sort of a(2,3), a(2,6), and a(2,9) elements. For example:
a =
0.4 0.4 0.5 0.6 0.2 0.2 0.6 0.2 0.6
0.5 0.8 0.9 0.9 0.6 0.6 0.1 0.2 0.8
[b i] = sort(a(2,3:3:end),2,'descend')
b =
0.9 0.8 0.6
i =
1 3 2
So, I want to have the following matrix:
a =
0.4 0.4 0.5 0.6 0.2 0.6 0.6 0.2 0.6
0.5 0.8 0.9 0.1 0.2 0.8 0.9 0.6 0.6
Try converting to a cell matrix first and then using your i to rearrange the cells
[b i] = sort(a(2,3:3:end),2,'descend')
A = mat2cell(a, 2, 3*ones(1,3));
cell2mat(A(i))
If for whatever reason you don't want to convert the whole of a into a cell matrix, you can do it by extending your indexing vector i to index all the columns. In your case you'd need:
I = [1,2,3,7,8,9,4,5,6]
which you could generate using a loop or else use bsxfun to get
[1 7 4
2 8 5
3 9 6]
and then "flatten" using reshape:
I = reshape(bsxfun(#plus, 3*s-2, (0:2)'), 1, [])
and then finally
a(:,I)
Typically, when a 2d matrix is separated into blocks, best practice ist to use more dimensions:
a=reshape(a,size(a,1),3,[]);
Now you can access each block via a(:,:,1)
To sort use:
[~,idx]=sort(a(2,3,:),'descend')
a=a(:,:,idx)
If you really need a 2d matrix, change back:
a=reshape(a,2,[])
sortrows-based approach:
n = 3; %// number of columns per block
m = size(a,1);
a = reshape(sortrows(reshape(a, m*n, []).', -m*n).', m, []);
This works by reshaping each block into a row, sorting rows according to last column, and reshaping back.
I am new to matlab.
I have a categorical input predictor(X) and the set of past results (Y, binary).
I would like to convert it to numeric variable in the following method.
For each category calculate the average of Y and replace the value with the average.
for example:
X Y X'
1 1 1
2 0 0
3 1 0.5
1 1 1
2 0 0
3 0 0.5
Please help.
you are looking for accumarray with mean function with Y as vals and X as subs
Xprime = accumarray( X, Y, [], #mean );
Xprime = Xptime( X );
I have a matrix containing X coordinates, Y coordinates and V (value to be plotted).
X and Y values can be non integer.
I want to plot V values with a contour plot, but this type of plot only accepts a matrix as input.
My idea is to change X and Y non-integers coordinates into integers so V can be plotted with contour function.
Any idea how can I change the X and Y values to integers? (I don't care about the X an Y coordinates in this case)
Here's an example of a matrix:
H=[-0.5 0 20; 0 0 15 ; 0.5 0 40; -0.5 0.5 18; 0 0.5 35; 0.5 0.5 10; -0.5 1 3; 0 1 9; 0.5 1 20]
Just arrange your values to a matrix.
do something like:
[xcoord,ix] = unique(H(:,1));
[ycoord,iy] = unique(H(:,2));
matrix = nan(numel(xcoord),numel(ycoord))
and then sort your values in, the ix and iy are your integers...
This question already has answers here:
How do I double the size of a vector in MATLAB with interpolation?
(2 answers)
Closed 8 years ago.
I have a row vector like
x = [ 0.125 0.25 0.5 0.75 1];
And I would like to expand it to 100 points with interpolation between points. Ho can i do it so that at the end I have equally spaced points but a length of 100 points?
Thanks
Solution
xi = [0 25 50 75 100];
yi = [0.125 0.25 0.5 0.75 1];
x = 1:1:100;
y = interp1(xi, yi, x);
Solution is y.
Explanation:
I consider your vector [0.125 0.25 0.5 0.75 1] as the result of a function F such that F(xi) = yi where xi = [0 25 50 75 100] and yi = [0.125 0.25 0.5 0.75 1].
I create x a 100-size vector using the same interval as xi;
I compute the interpolation of x based on the relationship between xi and yi;