I have a matrix S something like:
1 4 7
2 5 8
3 6 9
Then I make a=complex(S{2},S{3}) and wanted to find the abs(a);. This is not possible in MATLAB as a is not an integer - it is a matrix. How can I get the magnitude of each row of matrix a?
PS: the matrix is read from a text file using textscan() as S = textscan(fileID,'%d %d %d', 'delimiter','\t');.
Second question:
Assuming again hav the following S matrix.
1 4 7 2 1
2 5 8 3 4
3 6 9 6 8
Now I wanted to arrange them in such way that column 2,3 and 4,5 alternate like this:
4 2
7 1
5 3
8 4
6 6
9 8
How can I do that without using a loop?
Thanks.
Going with my assumption in the comments, I'm going to assume that the second column consists of your real component of your matrix while the third column consists of your imaginary components. Your matrix S is actually a cell array of elements. You don't need to use complex then abs. You can simply take each of the columns, individually square them, add them together and take the square root. What I would do is convert the cell array into a 2D matrix, cast it to double to allow for floating point precision when finding the magnitude, and do what I just did above. This is necessary because abs and sqrt will only work for floating-point numbers. Your elements in S are already int32 due to the %d delimiter from textread. In other words:
Smat = double(cell2mat(S));
realComp = Smat(:,2);
imagComp = Smat(:,3);
mag = sqrt(realComp.^2 + imagComp.^2);
mag will thus return the magnitude of each row for you, assuming that the second column is the real component and the third component is the imaginary component as we specified.
However, if you're dead set on using complex and abs, you can do it like so:
Smat = double(cell2mat(S));
imagNumbers = complex(Smat(:,2), Smat(:,3));
mag = abs(imagNumbers);
This should still give you the same results as we talked about above.
Edit
Seeing your edit in your post above, we can achieve that quite easily by subsetting the matrix, then applying reshape to each part of the matrix you want. In other words:
Smat = double(cell2mat(S));
realMat = Smat(:,2:3); %// Grab second and third columns
imagMat = Smat(:,4:5); %// Grab fourth and fifth columns
realCol = reshape(realMat.', [], 1); % // Form the columns like you specified
imagCol = reshape(imagMat.', [], 1);
finalMatrix = [realCol imagCol];
finalMatrix should contain those two columns that you specified above in a single matrix.
Related
Below is the sample code that describe my issue.
ff= [{'1 2 3 4 5'};{'2 2 3 4 2'};{'3 2 3 4 3'};{'4 2 3 4 4'}];
YY=[];
for i=1:length(ff)
xx=str2num(ff{i,1});
YY=[YY;xx];
end
similar to the sample code my real length of ff length is very large and it is taking longer to finish the conversion. is there a way to make it faster?
Your solution is going to be particularly slow since you keep expanding the size of YY every time through the for loop.
To optimize this, you could first convert your cell array of strings into one long string using strjoin. Then you can apply str2num to this entire string at once and reshape the result.
YY = reshape(str2num(strjoin(ff)), [], numel(ff)).'
% 1 2 3 4 5
% 2 2 3 4 2
% 3 2 3 4 3
% 4 2 3 4 4
If your version of MATLAB doesn't have strjoin, you can always replace it with sprintf
YY = reshape(str2num(sprintf('%s ', ff{:})), [], numel(ff)).';
Another option would be to convert each entry of the cell array to numbers using cellfun and num2str and then concatenate the result along the first dimension.
values = cellfun(#num2str, ff, 'UniformOutput', false);
YY = cat(1, values{:});
The first option is about twice as fast since you call num2str only once and the memory needed to store the temporary string created by strjoin is going to be less than the space required to store the same data as a numeric datatype (double).
I would like to create a column vector from the elements of a matrix A of size (3,3) that are not on the diagonal. Thus, I would have 6 elements in that output vector. How can I do this?
Use eye and logical negation, although this is no better than Divakar's original answer, and possibly significantly slower for very large matrices.
>> A = magic(4)
A =
16 2 3 13
5 11 10 8
9 7 6 12
4 14 15 1
>> A(~eye(size(A)))
ans =
5
9
4
2
7
14
3
10
15
13
8
12
Use this to get such a column vector, assuming A is the input matrix -
column_vector = A(eye(size(A))==0)
If you don't care about the order of the elements in the output, you can also use a combination of setdiff and diag -
column_vector = setdiff(A,diag(A))
You can also use linear indexing to access the diagonal elements and null them. This will automatically reshape itself to a single vector:
A(1:size(A,1)+1:end) = [];
Bear in mind that this will mutate the original matrix A. If you don't want this to happen, make a copy of your matrix then perform the above operation on that copy. In other words:
Acopy = A;
Acopy(1:size(A,1)+1:end) = [];
Acopy will contain the final result. You need to create a vector starting from 1 and going to the end in increments of the rows of the matrix A added with 1 due to the fact that linear indices are column-major, so the linear indices used to access a matrix progress down each row first for a particular column. size(A,1) will allow us to offset by each column and we add 1 each time to ensure we get the diagonal coefficient for each column in the matrix.
Assuming that the matrix is square,
v = A(mod(0:numel(A)-1, size(A,1)+1) > 0).';
In MATLAB I have a very large matrix (matrix A). Now I would like to find the row-index of the row which contain certain values in the second column. These values - which I'm looking for in Matrix A - are stored in anonther Matrix (Matrix B) with consists out of a row (800 numbers).
Besides I would like to redo this calculation for the same matrix A, but for ten different matrices, with different sizes (which contain the values I'm looking for in different columns of matrix A).
Because of the sizes of the matrix I think i need a loop to extract the row in matrix A which contain te value of Matrix B. How can I do this?
regards,
V
Thanks for the quick response! Indeed the problem is maybe a bit complex to answer without an example, and indeed duplicate entries cause some problems. Therefore hereby an example
For example I have a -simplified- matrix A:
1 2 3 4
9 9 9 9
4 3 2 1
And a -simplified- matrix (row) B: [9 3]
And a -simplified- matrix (row) C: [9 2]
Then I would like to get matrix D and matrix E.These matrices should contain in the first column the numbers from the original matrix D(or E) and in the second column the corresponding row-location of this value in matrix A.
So, matrix D =
9 2
3 3
matrix E =
9 2
2 3
As represented in this example matrix B and matrix C can contain data which is present in several column of matrix A (like the nine). However, martix B should "search" in column 2 of matrix A. Likewise, should matrix C "search" in column 3 of Matrix A, resulting in matrix D and E as given in the example.
As mentionned by Shai in his comment, your question is quite vague and a lot of special case could arise (duplicate entries, relative size of A and B, etc.). But in all generality I tried a small piece of code that seems to do what you want. There are certainly quicker ways of doing it, and certainly more information on your problem could help optimize this.
colA=2;
% Example
nmax=10;
nA=5;
A=randi(nmax,[nA nA]);
nB=3;
B=randi(nmax,[1 nB]);
% Find rows
rows=cell(size(B));
for i=1:numel(B)
rows(i)={find(A(:,colA)==B(i))};
end
The input / output was:
A =
3 7 8 5 4
9 7 3 7 5
8 2 9 9 8
9 5 9 7 9
3 3 4 6 8
B =
1 7 5
rows =
[0x1 double] [1;2] [4]
Assuming you have two vectors, largeDataIndex (the second column of your matrix) and interestingIndex (your b) and you want the following:
For each value of interestingIndex , find the position in largeDataIndex
Then an easy method would be this:
result = zeros(size(interestingIndex))
for i = 1:length(result)
result(i) = find(interestingIndex(i) == largeDataIndex)
end
Note that this assumes there is always just one entry that matches, otherwise you should define result as a cell array rather than a vector.
I would like to average every 3 values of an vector in Matlab, and then assign the average to the elements that produced it.
Examples:
x=[1:12];
y=%The averaging operation;
After the operation,
y=
[2 2 2 5 5 5 8 8 8 11 11 11]
Therefore the produced vector is the same size, and the jumping average every 3 values replaces the values that were used to produce the average (i.e. 1 2 3 are replaced by the average of the three values, 2 2 2). Is there a way of doing this without a loop?
I hope that makes sense.
Thanks.
I would go this way:
Reshape the vector so that it is a 3×x matrix:
x=[1:12];
xx=reshape(x,3,[]);
% xx is now [1 4 7 10; 2 5 8 11; 3 6 9 12]
after that
yy = sum(xx,1)./size(xx,1)
and now
y = reshape(repmat(yy, size(xx,1),1),1,[])
produces exactly your wanted result.
Your parameter 3, denoting the number of values, is only used at one place and can easily be modified if needed.
You may find the mean of each trio using:
x = 1:12;
m = mean(reshape(x, 3, []));
To duplicate the mean and reshape to match the original vector size, use:
y = m(ones(3,1), :) % duplicates row vector 3 times
y = y(:)'; % vector representation of array using linear indices
I was trying to get the min values of a matrix before the max values of the matrix occurred. I have two matrices: matrix data and matrix a. Matrix a is a subset of matrix data and is composed of the max values of matrix data. I have the following code but obviously doing something wrong.
edit:
Matrix a are the max values of matrix data. I derived it from:
for x=1:size(data,1)
a(x)=max(data(x,:));
end
a=a'
clear x
matrix b code:
for x=1:size(data,1)
b(x)=min(data(x,(x<data==a)));
end
b=b'
clear x
matrix data matrix a matrix b
1 2 3 4 4 1
6 5 4 7 7 4
9 6 12 5 12 6
I need all the min values that occurred before to matrix a occurred in matrix data
Short and simple:
[a,idxmax] = max(data,[],2);
b = arrayfun(#(ii) min(data(ii,1:idxmax(ii))), 1:size(data,1));
which is the same as
b=NaN(1,size(data,1)); % preallocation!
for ii=1:size(data,1)
b(ii) = min(data(ii,1:idxmax(ii)));
end
Ignore maximum itself
If you want minimum of everything really before (and not including the maximum), it's possible that the maximum is the first number, and you try taking minimum of an empty matrix. Solution then is to use cell output, which can be empty:
b = arrayfun(#(ii) min(data(ii,1:idxmax(ii)-1)), 1:size(data,1),'uni',false);
Replace empty cells with NaN
If you want to replace empty cells to Nan and then back to a matrix use this:
b(cellfun(#isempty,b))={NaN};
b=cell2mat(b);
or simply use the earlier version and replace b(ii) with NaN when it is equal to a(ii) same outcome:
b = arrayfun(#(ii) min(data(ii,1:idxmax(ii))), 1:size(data,1));
b(b'==a) = NaN
Example:
data=magic(4)
16 2 3 13
5 11 10 8
9 7 6 12
4 14 15 1
outputs:
a' = 16 11 12 15
b =
16 5 6 4
and
b =[1x0 double] [5] [6] [4]
for the 2nd solution using cell output and ignoring the maximum itself also.
And btw:
for x=1:size(data,1)
a(x)=max(data(x,:));
end
a=a'
clear x
can be replaced with
a=max(data,[],2);
It's not pretty but this is the only way I found so far of doing this kind of thing without a loop.
If loops are ok I would recommend Gunther Struyf answer as the most compact use of matlab's in-built array looping function, arrayfun.
Some of the transposition etc may be superfluous if you're wanting column mins instead of row...
[mx, imx] = max(data');
inds = repmat(1:size(data,2), [size(data,1),1]);
imx2 = repmat(imx', [1, size(data,2)]);
data2 = data;
data2(inds >= imx2) = inf;
min(data2');
NOTE: if data is not needed we can remove the additional data2 variable, and reduce the line count.
So to demonstrate what this does, (and see if I understood the question correctly):
for input
>> data = [1,3,-1; 5,2,1]
I get minima:
>> min(data2')
ans = [1, inf]
I.e. it only found the min values before the max values for each row, and anything else was set to inf.
In words:
For each row get index of maximum
Generate matrix of column indices
Use repmat to generate a matrix, same size as data where each row is index of maximum
Set data to infinity where column index > max_index matrix
find min as usual.