I have a function that processes a row vector. I want to make it generic for any type of input vector. Be it column or row.
One solution i thought is to preserve the existing implementation and check for the input vector for column or row type.
How can i perform this check?
iscolumn() or isrow() functions do not work here!
First, verify that the input IS a vector. isvector can help here. Or, use size, or any of a number of various artifices.
Second, convert your vector to a column vector.
vec = vec(:);
Third, write your code to always expect a column vector, since vec(:) does that.
Finally, save the original shape of your vector and reshape any output vector expected to be the same shape as the input. So your final code should look vaguely like this...
% test for errors
if ~isvector(vec)
error('The sky is falling')
end
% convert to column form always
vecshape = size(vec);
% process the vector
outputvec = ... % do stuff here
% reshape the output to be the same shape as the input
outputvec = reshape(outputvec,vecshape);
Check the size of the vector using size - if it has one column and many rows, your function can call itself with the transposed variant.
Related
I'm currently attempting to select a subset of 0's in a very large matrix (about 400x300 elements) and change their value to 1. I am able to do this, but it requires using a loop where each instance it selects the next value in a randperm vector. In other words, 50% of the 0's in the matrix are randomly selected, one-at-a-time, and changed to 1:
z=1;
for z=1:(.5*numberofzeroes)
A(zeroposition(rpnumberofzeroes(z),1),zeroposition(rpnumberofzeroes(z),2))=1;
z=z+1;
end
Where 'A' is the matrix, 'zeroposition' is a 2-column-wide matrix with the positions of the 0's in the matrix (the "coordinates" if you like), and 'rpnumberofzeros' is a randperm vector from 1 to the number of zeroes in the matrix.
So for example, for z=20, the code might be something like this:
A(3557,2684)=1;
...so that the 0 which appears in this location within A will now be a 1.
It performs this loop thousands of times, because .5*numberofzeroes is a very big number. This inevitably takes a long time, so my question is can this be done without using a loop? Or at least, in some way that takes less processing resources/time?
As I said, the only thing that needs to be done is an entirely random selection of 50% (or whatever proportion) of the 0's changed to 1.
Thanks in advance for the help, and let me know if I can clear anything up! I'm new here, so apologies in advance if I've made any faux pa's.
That's very easy. I'd like to introduce you to my friend sub2ind. sub2ind allows you to take row and column coordinates of a matrix and convert them into linear column-major indices so that you can access multiple values in a matrix simultaneously in a single call. As such, the equivalent code you want is:
%// First access the values in rpnumberofzeroes
vals = rpnumberofzeroes(1:0.5*numberofzeroes, :);
%// Now, use the columns of these to determine which rows and columns we want
%// to access A
rows = zeroposition(vals(:,1), 1);
cols = zeroposition(vals(:,2), 2);
%// Get linear indices via sub2ind
ind1 = sub2ind(size(A), rows, cols);
%// Now set these locations to 1
A(ind1) = 1;
The first statement gets the first half of your matrix of coordinates stored in rpnumberofzeroes. The first column is the row coordinates, the second column is the column coordinates. Notice that in your code, you wish to use the values in zeroposition to access the locations in A. As such, extract out the corresponding rows and columns from rpnumberofzeroes to figure out the right rows and columns from zeroposition. Once that's done, we wish to use these new rows and columns from zeroposition and index into A. sub2ind requires three inputs - the size of the matrix you are trying to access... so in our case, that's A, the row coordinates and the column coordinates. The output is a set of column major indices that are computed for each row and column pair.
The last piece of the puzzle is to use these to index into A and set the locations to 1.
This can be accomplished with linear indexing as well:
% find linear position of all zeros in matrix
ix=find(abs(A)<eps);
% set one half of those, selected at random, to one.
A(ix(randperm(round(numel(ix)*.5)))=1;
I'm looking for a way to select the vector that has the largest sum. Is there a simple way of doing this? I was thinking of writing a loop, but I'm not sure how to loop over a set of vectors.
Thanks for your help!
For the case in which the vectors have the same length (as stated in the comment), I think a simple loop-free way would be to build a matrix from each vector and fetch directly the row (or column) with the largest sum:
clear
clc
RandMat = rand(8,10);
[~,Ind] = max(sum(RandMat,2)); %// Get row index for largest sum. If you want the column, use 1 instead.
MaxRow = RandMat(Ind,:); %// Index in original matrix to get the vector. If you want the column, use RandMat(:,Ind);
If vectors don't have the same length then you would need to pad the missing values with NaN for example to use a regular matrix, otherwise you would need a cell array.
If you prefer a solution in which you don't have to build a matrix then you could loop through each individual vector and store the sum in a variable, then compare the sums at the end. If you would like such a solution please ask!
I'm trying to implement a 1 dimensional DFT without using Matlab built-in functions such as fft(). This is my code
function [Xk] = dft1(xn)
N=length(xn);
n = 0:1:N-1; % row vector for n
k = 0:1:N-1; % row vecor for k
WN = exp(-1j*2*pi/N); % Twiddle factor (w)
nk = n'*k; % creates a N by N matrix of nk values
WNnk = WN .^ nk; % DFT matrix
Xk = (WNnk*xn );
when i run the code after using the following commands:
I = imread('sample.jpg')
R = dft1(I)
I get this particular error:
Error using *
MTIMES is not fully supported for
integer classes. At least one input
must be scalar.
To compute elementwise TIMES, use
TIMES (.*) instead.
Can someone please help me to figure out how to solve this problem
Note: I am still in the very beginning level of learning Matlab
thank you very much
You just need to cast the data to double, then run your code again. Basically what the error is saying is that you are trying to mix classes of data together when applying a matrix multiplication between two variable. Specifically, the numerical vectors and matrices you define in dft1 are all of a double type, yet your image is probably of type uint8 when you read this in through imread. This is why you're getting that integer error because uint8 is an integer class and you are trying to perform matrix multiplication with this data type with those of a double data type. Bear in mind that you can mix data types, so long as one number is a single number / scalar. This is also what the error is alluding to. Matrix multiplication of varaibles that are not floating point (double, single) is not supported in MATLAB so you need to make sure that your image data and your DFT matrices are the same type before applying your algorithm.
As such, simply do:
I = imread('sample.jpg');
R = dft1(double(I));
Minor Note
This code is quite clever, and it (by default) applies the 1D DFT to all columns of your image. The output will be a matrix of the same size as I where each column is the 1D DFT result of each column from I.
This is something to think about, but should you want to apply this to all rows of your image, you would simply transpose I before it goes into dft1 so that the rows become columns and you can operate on these new "columns". Once you're done, you simply have to transpose the result back so that you'll get your output from dft1 shaped such that the results are applied on a per row basis. Therefore:
I = imread('sample.jpg');
R = dft1(double(I.')).';
Hope this helps! Good luck!
I am trying to add a number of vectors in a Matrix where each row represents a vector, but it gives me "Subscripted assignment dimension mismatch." error. The main problem is that each vector has a different size. I tried to add zeros at the end of the short vectors but I couldn't do it. Any Help.
Example:
%signal is a vector of data.
[x(1,:),y(1,:)] = findpeaks(signal1);
[x(2,:),y(2,:)] = findpeaks(signal2); %error as the peaks count in signal 2 is not the same as in signal 1.
OK, given two vectors of unequal length,
A=rand(1,10)
B=rand(1,5)
the proper way to deal with this is to use a cell array
D={A;B}
And then you can get whatever elements you want like this, for example:
D{1}(1:3) %// A(1:3)
If you don't want to use cells, you can add rows using this little function that adds row vector M to matrix F
addRow=#(F,M) [F NaN(size(F,1),size(M,2)-size(F,2));M NaN(1,size(F,2)-size(M,2))]
you would use it like this:
F=A
F=addRow(F,B)
I am trying to take a matrix and normalize the values in each cell around the average for that column. By normalize I mean subtract the value in each cell from the mean value in that column i.e. subtract the mean for Column1 from the values in Column1...subtract mean for ColumnN from the values in ColumnN. I am looking for script in Matlab. Thanks!
You could use the function mean to get the mean of each column, then the function bsxfun to subtract that from each column:
M = bsxfun(#minus, M, mean(M, 1));
Additionally, starting in version R2016b, you can take advantage of the fact that MATLAB will perform implicit expansion of operands to the correct size for the arithmetic operation. This means you can simply do this:
M = M-mean(M, 1);
Try the mean function for starters. Passing a matrix to it will result in all the columns being averaged and returns a row vector.
Next, you need to subtract off the mean. To do that, the matrices must be the same size, so use repmat on your mean row vector.
a=rand(10);
abar=mean(a);
abar=repmat(abar,size(a,1),1);
anorm=a-abar;
or the one-liner:
anorm=a-repmat(mean(a),size(a,1),1);
% Assuming your matrix is in A
m = mean(A);
A_norm = A - repmat(m,size(A,1),1)
As has been pointed out, you'll want the mean function, which when called without any additional arguments gives the mean of each column in the input. A slight complication then comes up because you can't simply subtract the mean -- its dimensions are different from the original matrix.
So try this:
a = magic(4)
b = a - repmat(mean(a),[size(a,1) 1]) % subtract columnwise mean from elements in a
repmat replicates the mean to match the data dimensions.