How can I multiply columns of a matrix and obtain a column vector.
Example:
A =
1 1 4
3 2 2
2 1 1
4 1 1
Expected output:
C =
4
12
2
4
Any ideas without for?
You can simply use the inbuilt prod function as prod(A,2) or prod(A')'.
For example:
>>
A = [
1 1 4
3 2 2
2 1 1
4 1 1
];
Now:
>> prod(A,2)
ans =
4
12
2
4
For more details, try help prod.
Once again, google and the Matlab documentation are your friend.
You are looking for the function prod:
prod(A,2)
A simple loop could do this for you too if you don't know the size of the matrix in hand. Just build a new vector by row-wise multiplication of elements in A.
Related
Suppose that I have a matrix , let's call it A, as follows:
1 2 3 4 5 1 2 3 4 5
0 2 4 6 8 1 3 5 7 9
And I want to reshape it into a matrix like this:
1 2 3 4 5
0 2 4 6 8
1 2 3 4 5
1 3 5 7 9
So, basically, what I want to be done is that MATLAB first reads a block of size (2,5) and then splits the remaining matrix to the next row and then repeats this so on so forth until we get something like in my example.
I tried to do this using MATLAB's reshape command in several ways but I failed. Any help is appreciated. In case that it matters, my original data is larger. It's (2,1080). Thanks.
I don't believe you can do this in a single command, but perhaps someone will correct me. If speed isn't a huge concern a for loop should work fine.
Alternatively you can get your results by reshaping each row of A and then placing the results into every other row of a new matrix. This will also work with your larger data.
A = [1 2 3 4 5 1 2 3 4 5
0 2 4 6 8 1 3 5 7 9];
An = zeros(numel(A)/5, 5); % Set up new, empty matrix
An(1:2:end,:) = reshape(A(1,:), 5, [])'; % Write the first row of A to every other row of An
An(2:2:end,:) = reshape(A(2,:), 5, [])' % Write second row of A to remaining rows
An =
1 2 3 4 5
0 2 4 6 8
1 2 3 4 5
1 3 5 7 9
You may need to read more about indexing in the Matlab's documentation.
For your example, it is easy to do the following
A=[1 2 3 4 5 1 2 3 4 5; 0 2 4 6 8 1 3 5 7 9]
a1=A(:,1:5); % extract all rows, and columns from 1 to 5
a2=A(:,6:end); % extract all rows, and columns from 6 to end
B=[a1;a2] % construct a new matrix.
It is not difficult to build some sort of loops to extract the rest.
Here's a way you can do it in one line using the reshape and permute commands:
B = reshape(permute(reshape(A,2,5,[]), [1,3,2]), [], 5);
The reshape(A,2,5,[]) command reshapes your A matrix into a three-dimensional tensor of dimension 2 x 5 x nblocks, where nblocks is the number of blocks in A in the horizontal direction. The permute command then swaps the 2nd and 3rd dimensions of this 3D tensor, so that it becomes a 2 x nblocks x 5 tensor. The final reshape command then transforms the 3D tensor into a matrix of dimension (2*nblocks) x 5.
Looking at the results at each stage may give you a better idea of what's happening.
I ran into an operation I cannot seem to achieve via vectorization.
Let's say I want to find the matrix of the application defined by
h: X -> cross(V,X)
where V is a predetermined vector (both X and V are 3-by-1 vectors).
In Matlab, I would do something like
M= cross(repmat(V,1,3),eye(3,3))
to get this matrix. For instance, V=[1;2;3] yields
M =
0 -3 2
3 0 -1
-2 1 0
Let's now suppose that I have a 3-by-N matrix
V=[V_1,V_2...V_N]
with each column defining its own cross-product operation. For N=2, here's a naive try to find the two cross-product matrices that V's columns define
V=[1,2,3;4,5,6]'
M=cross(repmat(V,1,3),repmat(eye(3,3),1,2))
results in
V =
1 4
2 5
3 6
M =
0 -6 2 0 -3 5
3 0 -1 6 0 -4
-2 4 0 -5 1 0
while I was expecting
M =
0 -3 2 0 -6 5
3 0 -1 6 0 -4
-2 1 0 -5 4 0
2 columns are inverted.
Is there a way to achieve this without for loops?
Thanks!
First, make sure you read the documentation of cross very carefully when dealing with matrices:
It says:
C = cross(A,B,DIM), where A and B are N-D arrays, returns the cross
product of vectors in the dimension DIM of A and B. A and B must
have the same size, and both SIZE(A,DIM) and SIZE(B,DIM) must be 3.
Bear in mind that if you don't specify DIM, it's automatically assumed to be 1, so you're operating along the columns. In your first case, you specified both the inputs A and B to be 3 x 3 matrices. Therefore, the output will be the cross product of each column independently due to the assumption that DIM=1. As such, you expect that the i'th column of the output contains the cross product of the i'th column of A and the i'th column of B and the number of rows is expected to be 3 and the number of columns needs to match between A and B.
You're getting what you expect because the first input A has [1;2;3] duplicated correctly over the columns three times. From your second piece of code, what you're expecting for V as the first input (A) looks like this:
V =
1 1 1 4 4 4
2 2 2 5 5 5
3 3 3 6 6 6
However, when you do repmat, you are in fact alternating between each column. In fact, you are getting this:
V =
1 4 1 4 1 4
2 5 2 5 2 5
3 6 3 6 3 6
repmat tile matrices together and you specified that you wanted to tile V horizontally three times. That's obviously not correct. This explains why the columns are swapped because the second, fourth and sixth columns of V actually should appear at the last three columns instead. As such, the ordering of your input columns is the reason why the output appears swapped.
As such, you need to re-order V so that the first three vectors are [1;2;3], followed by the next three vectors as [4;5;6] after. Therefore, you can generate your original V matrix first, then create a new matrix such that the odd column comes first in a group of three, followed by the even column in a group of three after:
>> V = [1,2,3;4,5,6].';
>> V = V(:, [1 1 1 2 2 2])
V =
1 1 1 4 4 4
2 2 2 5 5 5
3 3 3 6 6 6
Now use V with cross and maintain the same second input:
>> M = cross(V, repmat(eye(3), 1, 2))
M =
0 -3 2 0 -6 5
3 0 -1 6 0 -4
-2 1 0 -5 4 0
Looks good to me!
Supose there is a Matrix
A =
1 3 2 4
4 2 5 8
6 1 4 9
and I have a Vector containing the "class" of each column of this matrix for example
v = [1 , 1 , 2 , 3]
How can I sum the columns of the matrix to a new matrix as column vectors each to the column of their class? In this example columns 1 and 2 of A would added to the first column of the new matrix, column 2 to the 3 to the 2nd, column 4 the the 3rd.
Like
SUM =
4 2 4
6 5 8
7 4 9
Is this possible without loops?
One of the perfect scenarios to combine the powers of accumarray and bsxfun -
%// Since we are to accumulate columns, first step would be to transpose A
At = A.' %//'
%// Create a vector of linear IDs for use with ACCUMARRAY later on
idx = bsxfun(#plus,v(:),[0:size(A,1)-1]*max(v))
%// Use ACCUMARRAY to accumulate rows from At, i.e. columns from A based on the IDs
out = reshape(accumarray(idx(:),At(:)),[],size(A,1)).'
Sample run -
A =
1 3 2 4 6 0
4 2 5 8 9 2
6 1 4 9 8 9
v =
1 1 2 3 3 2
out =
4 2 10
6 7 17
7 13 17
An alternative with accumarray in 2D. Generate a grid with the vector v and then apply accumarray:
A = A.';
v = [1 1 2 3];
[X, Y] = ndgrid(v,1:size(A,2));
Here X and Y look like this:
X =
1 1 1
1 1 1
2 2 2
3 3 3
Y =
1 2 3
1 2 3
1 2 3
1 2 3
Then apply accumarray:
B=accumarray([X(:) Y(:)],A(:)),
SUM = B.'
SUM =
4 2 4
6 5 8
7 4 9
As you see, using [X(:) Y(:)] create the following array:
ans =
1 1
1 1
2 1
3 1
1 2
1 2
2 2
3 2
1 3
1 3
2 3
3 3
in which the vector v containing the "class" is replicated 3 times since there are 3 unique classes that are to be summed up together.
EDIT:
As pointed out by knedlsepp you can get rid of the transpose to A and B like so:
[X2, Y2] = ndgrid(1:size(A,1),v);
B = accumarray([X2(:) Y2(:)],A(:))
which ends up doing the same. I find it a bit more easier to visualize with the transposes but that gives the same result.
How about a one-liner?
result = full(sparse(repmat(v,size(A,1),1), repmat((1:size(A,1)).',1,size(A,2)), A));
Don't optimize prematurely!
The for loop performs fine for your problem:
out = zeros(size(A,1), max(v));
for i = 1:numel(v)
out(:,v(i)) = out(:,v(i)) + A(:,i);
end
BTW: With fine, I mean: fast, fast, fast!
I can't figure out how to create a vector based on condition on more than one other vectors. I have three vectors and I need values of one vector if values on other vectors comply to condition.
As an example below I would like to choose values from vector a if values on vector b==2 and values on vector c==0 obviously I expect [2 4]
a = [1 2 3 4 5 6 7 8 9 10];
b = [1 2 1 2 1 2 1 2 1 2];
c = [0 0 0 0 0 1 1 1 1 1]
I thought something like:
d = a(b==2) & a(c==0)
but I have d = 1 1 1 1 1 not sure why.
It seems to be basic problem but I can find solution for it.
In your case you can consider using a(b==2 & c==0)
Use ismember to find the matching indices along the rows after concatenating b and c and then index to a.
Code
a(ismember([b;c]',[2 0],'rows'))
Output
ans =
2
4
You may use bsxfun too for the same result -
a(all(bsxfun(#eq,[b;c],[2 0]'),1))
Or you may just tweak your method to get the correct result -
a(b==2 & c==0)
I have a two column matrix of the following form:
1. 1 1
2. 1 1
3. 1 2
4. 1 2
5. 2 2
6. 2 2
7. 3 2
8. 3 2
9. 3 3
10. 4 3
11. 4 4
I would like to sample a single number from the first column using say randsample().
Let's say the results is 2.
What I would like to know is which ROW was the sample taken from? (in this case it could have been sampled both from row 5 or row 6)
Is this possible?
It's easy with find and ==:
>> A = [
1 1
1 1
1 2
1 2
2 2
2 2
3 2
3 2
3 3
4 3
4 4];
>> R = randsample(4,1)
>> find(A(:,1) == R)
R =
4
ans =
10
11
Or, as indicated by igor milla,
>> I = randi(11)
>> A(I, :)
I =
9
ans =
3 3
If you just need to sample one value, the solution as given by #igor milla is fine. But if you want to use the options given by randsample then I would recommend you to sample the column numbers rather than the sample directly.
A = rand(11,2); %suppose this is your matrix
k = 1; %This is the size of your desired sample
mysampleid = randsample(size(A,1),k)
mysample = A(mysampleid,:)
Now mysampleid contains the numbers of the columns, and mysample contains the rows that you sampled.
If you just want to sample the first column you can use A(mysampleid,1) instead.