In Matlab, Is it possible to do simple operations between subsequent elements of an array without using a for loop? Something like diff(). For example, I have this vector:
A = [2 4 8 16 32]
and I want each element to be divided by its predecessor:
ans = [2 2 2 2]
How can I do it without going through all elements (without using loops)?
You can use the fact that division in Matlab work on both scalars and matrices, if you use the ./ operator rather than /
>> A = [2 4 8 16 32];
>> A(2:end) ./ A(1:end-1)
ans =
2 2 2 2
Regarding your question about doing dot() between vectors stored in the rows of a matrix. There is an additional argument to dot() that tells it whether your vectors are stored in columns (the default) or rows;
>> x = rand(3);
>> y = rand(3); # random vectors
>> dot(x,y) # dot product of column vectors
ans =
0.5504 0.5561 0.5615
>> dot(x,y,2) # dot product of row vectors
ans =
0.3170
1.0938
0.2572
Most functions in Matlab are vectorized so that they can work on scalars, vectors and matrices, but you sometimes have the read the documentation (e.g. type help dot) to work out how to use them.
Related
I have a situation analogous to the following
z = magic(3) % Data matrix
y = [1 2 2]' % Column indices
So,
z =
8 1 6
3 5 7
4 9 2
y represents the column index I want for each row. It's saying I should take row 1 column 1, row 2 column 2, and row 3 column 2. The correct output is therefore 8 5 9.
I worked out I can get the correct output with the following
x = 1:3;
for i = 1:3
result(i) = z(x(i),y(i));
end
However, is it possible to do this without looping?
Two other possible ways I can suggest is to use sub2ind to find the linear indices that you can use to sample the matrix directly:
z = magic(3);
y = [1 2 2];
ind = sub2ind(size(z), 1:size(z,1), y);
result = z(ind);
We get:
>> result
result =
8 5 9
Another way is to use sparse to create a sparse matrix which you can turn into a logical matrix and then sample from the matrix with this logical matrix.
s = sparse(1:size(z,1), y, 1, size(z,1), size(z,2)) == 1; % Turn into logical
result = z(s);
We also get:
>> result
result =
8
5
9
Be advised that this only works provided that each row index linearly increases from 1 up to the end of the rows. This conveniently allows you to read the elements in the right order taking advantage of the column-major readout that MATLAB is based on. Also note that the output is also a column vector as opposed to a row vector.
The link posted by Adriaan is a great read for the next steps in accessing elements in a vectorized way: Linear indexing, logical indexing, and all that.
there are many ways to do this, one interesting way is to directly work out the indexes you want:
v = 0:size(y,2)-1; %generates a number from 0 to the size of your y vector -1
ind = y+v*size(z,2); %generates the indices you are looking for in each row
zinv = z';
zinv(ind)
>> ans =
8 5 9
Does any one know how I can convert a row vector [6 8 2] into a column vector without using a builtin command. I would like to do it without for loop.A ny idea please. Some one asked me is it a home work I say no, it's part of my work. I am trying to convert MATLAB code to vhdl using hdl coder but hdl coder seems not supporting transpose function.
Some options:
R = 1:10; %// A row vector
%// using built-in transpose
C = R'; %'// be warned this finds the complex conjugate
C = R.'; %'// Just swaps the rows and columns
C = transpose(R);
%// Flattening with the colon operator
C = R(:); %// usually the best option as it also convert columns to columns...
%// Using reshape
C = reshape(R,[],1);
%// Using permute
C = permute(R, [2,1]);
%// Via pre-allocation
C = zeros(numel(R),1);
C(1:end) = R(1:end);
%// Or explicitly using a for loop (note that you really should pre-allocate using zeros for this method as well
C = zeros(numel(R),1); %// technically optional but has a major performance impact
for k = 1:numel(R)
C(k,1) = R(k); %// If you preallocated then C(k)=R(k) will work too
end
%// A silly matrix multiplication method
C = diag(ones(numel(R),1)*R)
You can use the (:) trick
t = [1 2 3];
t(:)
ans =
1
2
3
UPDATE: you should use this method only in the following case: you have a vector (not matrix) and want to make sure it is a column-vector. This method is useful, when you don't know what type (column, row) of vector a variable has.
Check this out
t = [1 2 3]'; %// column vector
t(:)
ans =
1
2
3
However
A=magic(3);
A(:)
ans =
8
3
4
1
5
9
6
7
2
What I intend to do is very simple but yet I haven't found a proper way to do it. I have a function handle which depends on two variables, for example:
f = #(i,j) i+j
(mine is quite more complicated, though)
What I'd like to do is to create a matrix M such that
M(i,j) = f(i,j)
Of course I could use a nested loop but I'm trying to avoid those. I've already managed to do this in Maple in a quite simple way:
f:=(i,j)->i+j;
M:=Matrix(N,f);
(Where N is the dimension of the matrix) But I need to use MATLAB for this. For now I'm sticking to the nested loops but I'd really appreciate your help!
Use bsxfun:
>> [ii jj] = ndgrid(1:4 ,1:5); %// change i and j limits as needed
>> M = bsxfun(f, ii, jj)
M =
2 3 4 5 6
3 4 5 6 7
4 5 6 7 8
5 6 7 8 9
If your function f satisfies the following condition:
C = fun(A,B) accepts arrays A and B of arbitrary, but equal size and returns output of the same size. Each element in the output array C is the result of an operation on the corresponding elements of A and B only. fun must also support scalar expansion, such that if A or B is a scalar, C is the result of applying the scalar to every element in the other input array.
you can dispose of ndgrid. Just add a transpose (.') to the first (i) vector:
>> M = bsxfun(f, (1:4).', 1:5)
Function handles can accept matrices as inputs. Simply pass a square matrix of size N where the values corresponds to the row number for i, and a square matrix of size N where the values correspond to the column number for j.
N = 5;
f = #(i,j) i+j;
M = f(meshgrid(1:N+1), meshgrid(1:N+1)')
I've noticed various cases in Matlab and octave where functions accept both matrices and vectors, but doesn't do the same thing with vectors as it does with matrices.
This can be frustrating because when you input a matrix with a variable number of rows/columns, it could be interpreted as a vector and do something you don't expect when the height/width is 1 making for difficult debugging and weird conditional edge cases.
I'll list a few I've found, but I'm curious what others people have run into
(Note: I'm only looking for cases where code accepts matrices as valid input. Anything that raises an exception when a non-vector matrix is given as an argument doesn't count)
1) "diag" can be used to mean diagonal of a matrix or turn a vector into a diagonal matrix
Since the former is generally only used for square matrices this isn't so egregious in matlab, but in Octave it can be particularly painful when Octave interperets a vector beginning with a nonzero element and everything else zeros as a "diagonal matrix" ie
t=eye(3);
size(diag(t(:,3))) == [3,3]
size(diag(t(:,2))) == [3,3]
size(diag(t(:,1))) == [1,1]
2) Indexing into a row-vector with logicals returns a row-vector
Indexing into anything else with logicals returns a column vector
a = 1:3;
b = true(1,3);
size(a(b)) == [1, 3]
a = [a; a];
b = [b; b];
size(a(b)) == [6, 1]
3) Indexing into a vector v with an index vector i returns a vector of the same (row/col) type as v. But if either v or i is a matrix, the return value has the same size as i.
a = 1:3;
b = a';
size(a(b)) == [1, 3]
b = [b,b];
size(a(b)) == [3, 2]
4) max, min, sum etc. operate on the columns of a matrix M individiually unless M is 1xn in which case they operate on M as a single row-vector
a = 1:3
size(max(a)) == [1, 1]
a = [a;a]
size(max(a)) == [1, 3]
max is particularly bad since it can't even take a dimension as an argument (unlike sum)
What other such cases should I watch out for when writing octave/matlab code?
Each language has its own concepts. An important point of this language is to very often think of matrices as an array of vectors, each column an entry. Things will start to make sense then. If you don't want that behavior, use matrix(:) as the argument to those functions which will pass a single vector, rather than a matrix. For example:
octave> a = magic (5);
octave> max (a)
ans =
23 24 25 21 22
octave> max (a(:))
ans = 25
1) This is not true with at least Octave 3.6.4. I'm not 100% sure but may be related related to this bug which has already been fixed.
2) If you index with boolean values, it will considered to be a mask and treated as such. If you index with non-boolean values, then it's treated as the indexes for the values. This makes perfect sense to me.
3) This is not true. The returned has always the same size of the index, independent if it's a matrix or vector. The only exception is that if the index is a vector, the output will be a single row. The idea is that indexing with a single vector/matrix returns something of the same size:
octave> a = 4:7
a =
4 5 6 7
octave> a([1 1])
ans =
4 4
octave> a([1 3])
ans =
4 6
octave> a([1 3; 3 1])
ans =
4 6
6 4
4) max does take dimension as argument at least in Octave. From the 3.6.4 help text of max:
For a vector argument, return the maximum value. For a matrix
argument, return the maximum value from each column, as a row vector,
or over the dimension DIM if defined, in which case Y should be set to
the empty matrix (it's ignored otherwise).
The rest applies like I said on the intro. If you supply a matrix, it will think of each column as a dataset.
1) As pointed out by the other user, this is not true with at Octave >= 3.6.4.
In case 2) the rule is for vectors, return always the same shape of vector, for anything else return a column vector, consider:
>> a = reshape (1:3, 1,1,3)
a(:,:,1) =
1.0000e+000
a(:,:,2) =
2.0000e+000
a(:,:,3) =
3.0000e+000
>> b = true(1,3)
b =
1×3 logical array
1 1 1
>> a(b)
ans(:,:,1) =
1.0000e+000
ans(:,:,2) =
2.0000e+000
ans(:,:,3) =
3.0000e+000
>> a = [a;a]
a(:,:,1) =
1.0000e+000
1.0000e+000
a(:,:,2) =
2.0000e+000
2.0000e+000
a(:,:,3) =
3.0000e+000
3.0000e+000
>> b = [b;b]
b =
2×3 logical array
1 1 1
1 1 1
>> a(b)
ans =
1.0000e+000
1.0000e+000
2.0000e+000
2.0000e+000
3.0000e+000
3.0000e+000
You can see that this makes sense since vectors have a clear 'direction' but other shaped matrices do not when you remove elements. EDIT: actually I just checked and Octave doesn't seem work this way exactly, but probably should.
3) This is consistent with 2). Essentially if you supply a list of indices the direction of the indexed vector is preserved. If you supply indices with a shape like a matrix, the new information is the index matrix shape is used. This is more flexible, since you can always do a(b(:)) to preserve the shape of a if you so wish. You may say it is not consistent, but remember indexing with logicals may reduce the number of elements to be returned, so they cannot be reshaped in this way.
4) As pointed out in a comment, you can specify dimension for max/min to operate on: min(rand(3),[],1) or max(rand(3),[],2), but in this case there are 'legacy' issues with these functions which data back to when they were first created and now are very difficult to change without upsetting people.
I have two expressions in MATLAB that represent a 365x24 matrix. The first expression has 10, 365x24 matrices and is therefore
PV_power_output(:,:,K)
and the second expression which is again 365x24 but with three possible matrices therefore is
WT_energy_supply(:,:,M);ode here
Now, I am looking to create a third matrix that adds the elements in the same position above and thus form a 365x24 matrix. However I want a set of matrix with all possible combinations of the two expressions shown above (therefore this matrix must be 365x24x30.
How do I go about this?
What about the bsxfun function in MATLAB?
Expand the original matrices (which for clarity I name a and b) with repmat and then just add them, bsxfun is not needed.
repmat(a,[1 1 size(b,3)]) + repmat(b,[1 1 size(a,3)]))
Update
>> size(a)
ans =
364 24 10
>> size(b)
ans =
364 24 3
>> c=repmat(a,[1 1 size(b,3)])+repmat(b,[1 1 size(a,3)]);
>> size(c)
ans =
364 24 30
It looks fine to me. Of course you'll have to replace my variables a and b with your variables PV_power_output and WT_energy_supply.