What are these lines of code doing?
x0 = rand(n,2)
x0(:,1)=W*x0(:,1)
x0(:,2)=H*x0(:,2)
x0=x0(:)
Is this just one big column vector?
I'd encourage you to take a MATLAB Tutorial as indexing arrays is a fundamental skill. Also see Basic Concepts in MATLAB. Line-by-line descriptions are below to get you started.
What are these lines of code doing?
Let's take this line by line.
1. This line uses rand() to generate an n x 2 matrix of uniform random numbers (~U(0,1)).
x0 = rand(n,2) % Generate nx2 matrix of U(0,1) random numbers
2. Multiply the first column by W
In this case, x0(:,1) means take all rows of x0 (the colon in the first argument) and the 1st column (the 1). Here, the * operator indicates W is a scalar or an appropriately sized array for feasible matrix multiplication (my guess is a scalar). The notation .* can be used for element-by-element multiplication; see here and here for more details.
x0(:,1)=W*x0(:,1) % Multiply (all rows) 1st column by W
3. Multiply the first column by H.
Using similar logic as #2.
x0(:,2)=H*x0(:,2) % Multiply (all rows) 2nd column by H
4. Force column
The x0(:) takes the array x0 and forces all elements into a single column.
From the documentation for colon:
A(:) reshapes all elements of A into a single column vector. This has
no effect if A is already a column vector.
A related operation is forcing a row vector by combining this with the transpose operator.
For example, try the following: x0(:).'
x0 = x0(:) % Force Column
x0 = x0(:).' % Force Row
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How does MATLAB's colon operator work?
Combination of colon-operations in MATLAB
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I am a newbie to Matlab and am trying to figure out code. One thing that keeps returning and confusing me is the ./ sign. I tried googling and stackoverflowing it, but I can not find any documentation on it. Just googling ./ in combination with Matlab does not get me there.
To my understanding the code in the bottom does: Take the first record of the contry.farm.potatoes and divide this by all country.farm.tractors values summed. The result however is always a 0.x numbers. So from 0 to 1. Does that mean that the ./ sign makes sure it is a percentage?
country.farm.potatoes(1,:)./sum(country.farm.tractors,1)
General explanation
the following code does the following:
(1) takes the first row of country.farm.potatoes
(2) generate a vector such that the i'th coordinate is the sum of the i'th column of country.farm.tractors
(3) divides each coordinate of (1) by the corresponding coordinate in (2).
(:,1) syntax
Given a matrix M of size m,n, the syntax '(:,1)' extracts the first row of the matrix.
It generates a row vector of size 1xn.
M(1,:)
sum(M,1) syntax
Given a matrix M of size m,n, the syntax sum(M,1) generates a row vecotr of size 1xn,
s.t each coordinate j is the sum of the j'th column of the matrix.
sum(M,1)
A ./ B syntax
Given two matrices or vectors A,B (of same size), the syntax 'C=A./B' yields a per coordinate division result.
This means that C(i,j)=A(i,j)/B(i,j)
C = A./B
for example:
[9,6,2] ./ [3,3,2]
ans = 3 2 1
I am trying to create a loop code in MATLAB that "fills up" the elements in an empty column vector of size l x 1, called m. As I don't have very much experience in MATLAB, I am unsure if this is the correct way to go about it.
Note: Seeing as i pertains to the complex quantity in matlab, I denote the i-th element of an array as the ii-th element.
l=length(A); %The number of rows in the empty vector we seek as our output;
%so as to preallocate space for this vector.
q=eigencentrality(A);%An lx1 column vector whose ii-th elements are used in the loop.
l1=max(eig(A)); %A scalar used in the loop.
CS=sg_centrality(A); %%An lx1 column vector whose ii-th elements are used in the loop.
%Now for the actual loop that will "fill up" each ii-th entry
%of our empty vector, m.
m=NaN(l,1); %create the empty vector to be "filled up".
for ii=1:l
m(ii,:)=log(q(ii)^2)*sinh(l1)/CS(ii)^1/2;%this is the form that I want each entry
%of m to have. Note how the ii-th element
%of m depends on the corresponding ii-th
%element of CS and q!
end
Is this the right way to go about "filling up" such an empty column vector, m, whose entries depend on the corresponding elements of two other vectors as above?
Cheers!
You can do this completely vectorized. Vectorization is the act of processing data chunks at a time rather than individually as you're doing in your code. In fact, this is one of MATLAB's main advantages. You can replace that for loop with:
m = log(q.^2).*(sinh(l1)./CS).^1/2;
The .* and .^ operators are known as element-wise operators. This means that each value in each of q, li and CS will contribute to the corresponding position in the output. There's no need to use a loop.
Check out this MathWorks note on vectorization here: http://www.mathworks.com/help/matlab/matlab_prog/vectorization.html
You can vectorize all the operations, without using a for loop. This should work:
m=log(q.^2).*(sinh(l1)./CS).^1/2;
Note, that the dots denote elementwise operations. Usually this is much faster than using a loop. Also, just as a side note, you can then drop the preallocation.
Please see the following issue:
P=rand(4,4);
for i=1:size(P,2)
for j=1:size(P,2)
[r,p]=corr(P(:,i),P(:,j))
end
end
Clearly, the loop will cause the number of correlations to be doubled (i.e., corr(P(:,1),P(:,4)) and corr(P(:,4),P(:,1)). Does anyone have a suggestion on how to avoid this? Perhaps not using a loop?
Thanks!
I have four suggestions for you, depending on what exactly you are doing to compute your matrices. I'm assuming the example you gave is a simplified version of what needs to be done.
First Method - Adjusting the inner loop index
One thing you can do is change your j loop index so that it only goes from 1 up to i. This way, you get a lower triangular matrix and just concentrate on the values within the lower triangular half of your matrix. The upper half would essentially be all set to zero. In other words:
for i = 1 : size(P,2)
for j = 1 : i
%// Your code here
end
end
Second Method - Leave it unchanged, but then use unique
You can go ahead and use the same matrix like you did before with the full two for loops, but you can then filter the duplicates by using unique. In other words, you can do this:
[Y,indices] = unique(P);
Y will give you a list of unique values within the matrix P and indices will give you the locations of where these occurred within P. Note that these are column major indices, and so if you wanted to find the row and column locations of where these locations occur, you can do:
[rows,cols] = ind2sub(size(P), indices);
Third Method - Use pdist and squareform
Since you're looking for a solution that requires no loops, take a look at the pdist function. Given a M x N matrix, pdist will find distances between each pair of rows in a matrix. squareform will then transform these distances into a matrix like what you have seen above. In other words, do this:
dists = pdist(P.', 'correlation');
distMatrix = squareform(dists);
Fourth Method - Use the corr method straight out of the box
You can just use corr in the following way:
[rho, pvals] = corr(P);
corr in this case will produce a m x m matrix that contains the correlation coefficient between each pair of columns an n x m matrix stored in P.
Hopefully one of these will work!
this works ?
for i=1:size(P,2)
for j=1:i
Since you are just correlating each column with the other, then why not just use (straight from the documentation)
[Rho,Pval] = corr(P);
I don't have the Statistics Toolbox, but according to http://www.mathworks.com/help/stats/corr.html,
corr(X) returns a p-by-p matrix containing the pairwise linear correlation coefficient between each pair of columns in the n-by-p matrix X.
I have the following Matlab code snippet that I'm having to translate to VBScript. However, I'm not understanding why the last line is even necessary.
clear i
for i = 1:numb_days
doy(i) = floor(dt_daily(i) - datenum(2012,12,31,0,0,0));
end
doy = doy';
Looking over the rest of the code, this happens in a lot of other places where there are single dimension arrays (?) being transposed in place. I'm a newbie when it comes to both these languages, as well as posting a question on Stack, as I'm a sleuth when it comes to finding answers, just not in this case. Thanks in advance.
All "arrays" in MATLAB have at least two dimensions, and can be treated as having any number of dimensions you wish. The transpose operator here is converting between a row (size [1 N] array) and a column (size [N 1] array). This can be significant when it comes to either concatenating the arrays, or performing other operations.
Conceptually, the dimension vector of a MATLAB array has as many trailing 1s as is required to perform an operation. This means that you can index any MATLAB array with any number of subscripts, providing you don't exceed the bounds, like so:
x = magic(4); % 4-by-4 square matrix
x(2,3,1,1,1) % pick an element
One final note: the ' operator is the complex-conjugate transpose CTRANSPOSE. The .' operator is the ordinary TRANSPOSE operator.
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.