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How to multiply two rows or columns?
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Closed 6 years ago.
Why cannot multiply two matrices of same dimensions (1x3) ? :
>> a = [1 1 1]
a =
1 1 1
>> b = [1 1 1]
b =
1 1 1
>> a * b
error: operator *: nonconformant arguments (op1 is 1x3, op2
is 1x3)
* is the matrix multiplication operator so if A is an n-by-m matrix and B is an x-by-y matrix for
A*B
to be valid you need the inner dimension to match in size thus m must equal x. Or said another way, the number of columns of A (in your case 3) must equal the number of columns of B (in your case 1).
Some solutions:
You wanted the inner product:
A*B.' % result is a scalar (1-by-1)
or the outer product:
A.'*B % result is a 3-by-3 matrix
Or else you wanted the element-wise multiplication (i.e. multiply each element with it's corresponding element, assuming the two matrices have the identical size) which is the .* operator:
A.*B % result is a 1-by-3
Matrix dimensions should match: the columns of the second matrix must be the same number rows of the first matrix. For your case, you can multiply a*b^T or a^T*b, depending on what you're trying to achieve. There is also Hadamard product (elementwise multiplication) but it's not considered as matrix multiplication.
Related
I am following https://taninamdar.files.wordpress.com/2013/11/submatrices3.pdf to find total number of sub matrix of a matrix.But am stuck how to find how many sub matrix of a given size is present in a matrix.
Also 0<=A<=M and 0<=B<=N.
where AxB(submatrix size) and MxN(matrix size).
I didn't go through the pdf (math and I aren't friends), however simple logic is enough here. Simply, try to reduce the dimension: How many vectors of length m can you put in a vector of length n ?
Answer: n-m+1. To convince you, just go through the cases. Say n = 5 and m = 5. You've got one possibility. With n = 5 and m = 4, you've got two (second vector starts at index 0 or index 1). With n = 5 and m = 3, you've got three (vector can start at index 0, 1 or 2). And for n = 5 and m = 1, you've got 5, seems logic.
So, in order to apply that to a matrix, you have to add a dimension. How do you do that ? Multiplication. How many vectors of length a can you put inside a vector of length n ? n-a+1. How many vectors of length b can you put inside a vector of length m ? m-b+1.
So, how many matrices of size A*B can you put in a matrix of length N*M ? (N-A+1)*(M-B+1).
So, I didn't handle the case where one of the dimension is 0. It depends on how you consider this case.
Say I have a IxJ matrix of values,
V= [1,4;2,5;3,6];
and a IxR matrix X of indexes,
X = [1 2 1 ; 1 2 2 ; 2 1 2];
I want to get a matrix Vx that is IxR such that for each row i, I want to read R times a (potentially) different column of V, which are given by the numbers in each corresponding column in X.
Vx(i,r) = V(i,X(i,r)).
For instance in this case it would be
Vx = [1,4,1;2,5,5;6,3,6];
Any help to do this fast, (without any looping) is much appreciated!
So what you want to achieve is using vectorization to achieve speed. This is one of the major strength of MATLAB. What you want is a matrix (index in the following code) whose elements are linear indexes that will be used to pick out value from the source matrix(V in your case). The first two lines of codes are doing exactly the same thing as sub2ind, turning subscripts to linear indexes. I'm coding this way so the logic of index conversion is clear.
[m,n] = ndgrid(1:size(X,1),1:size(X,2));
index = m + (X-1)*size(X,1);
Vx = V(index);
You can use bsxfun for an efficient solution -
N = size(V,1)
Vx = V(bsxfun(#plus,[1:N]',(X-1)*N))
Sample run -
>> V
V =
1 4
2 5
3 6
>> X
X =
1 2 1
1 2 2
2 1 2
>> N = size(V,1);
Vx = V(bsxfun(#plus,[1:N]',(X-1)*N))
Vx =
1 4 1
2 5 5
6 3 6
Another method would be to use repmat combined with sub2ind. sub2ind takes in row and column locations and the output are column-major linear indices that you can use to vectorize access into a matrix. Specifically, you want to build a 2D matrix of row indices and column indices which is the same size as X where the column indices are exactly specified as X but the row indices are the same for each row that we're concerned with. Concretely, the first row of this matrix will be all 1s, the next row all 2s, etc. To build this row matrix, first generate a column vector that goes from 1 up to as many rows as there are X and replicate this for as many columns as there are in X. With this new matrix and X, use sub2ind to generate column-major linear indices to finally index V to produce the matrix Vx:
subs = repmat((1:size(X,1)).', [1 size(X,2)]); %'
ind = sub2ind(size(X), subs, X);
Vx = V(ind);
a =
1
2
3
b =
1 2 3
a.*b
ans =
1 2 3
2 4 6
3 6 9
I used the .* operator to multiply a row vector and a column vector in Octave to see the results. I dont understand how the answer is obtained.
This is because Octave (in a notable difference from Matlab) automatically broadcasts.
The * operator in Octave is the matrix multiplication operator. So in your case a*b would output (in Matlab as well)
a*b
ans =
1 2 3
2 4 6
3 6 9
which should be expected. The product of a 3-by-1 matrix with a 1-by-3 matrix would have dimensions 3-by-3 (inner dimensions must match, the result takes the outer dimensions).
However the .* operator is the element-wise multiplication operation. That means that instead of matrix multiplication, this would multiply each corresponding elements of the two inputs independent from the rest of the matrix. So [1,2,3].*[1,2,3] (or a'.*b) results in [1,4,9]. Again this is in Matlab and Octave.
When using element-wise operations, it is important that the dimensions of the inputs exactly match. So [1,2,3].*[1,2] will through an error because the dimensions do not match. In Matlab, your a.*b will through an error as well. HOWEVER in Octave it won't, instead it will automatically broadcast. You can imagine this is as if it takes one of your inputs and replicates it on a singleton dimension (so in a column vector, the second dimension is a singleton dimension because it's size is 1) and then applies the operator element-wise. In your case you have two matrices with singleton dimensions (i.e. a columan vector and a row vector) so it actually broadcasts twice and you effectively (but note that it does not actually expand the matrices in memory and is often far faster than using repmat) get
[1,2,3;1,2,3;1,2,3].*[1,1,1;2,2,2;3,3,3]
which produces the result you see.
In matlab, to achieve the same result you would have to explicitly call the bsxfun function (binary singleton expansion function) like so:
bsxfun(#times, a, b)
I want to convert 127 by 4 matrix to 1 by 4 matrix such that each value in the output row equivalent to average of all values in that particular column.
Simply use the mean function:
A = rand(127,4);
B = mean(A,1); % Average of A along the first dimension
Best,
I am new to Octave. I have two matrices. I have to compare a particular column of a one matrix with the other(my matrix A is containing more than 5 variables, similarly matrix B is containing the same.) and if elements in column one of matrix A is equal to elements in the second matrix B then I have to use the third column of second matrix B to compute certain values.I am doing this with octave by using for loop , but it consumes a lot of time to do the computation for single day , i have to do this for a year . Because size of matrices is very large.Please suggest some alternative way so that I can reduce my time and computation.
Thank you in advance.
Thanks for your quick response -hfs
continuation of the same problem,
Thank u, but this will work only if both elements in both the rows are equal.For example my matrices are like this,
A=[1 2 3;4 5 6;7 8 9;6 9 1]
B=[1 2 4; 4 2 6; 7 5 8;3 8 4]
here column 1 of first element of A is equal to column 1 of first element of B,even the second column hence I can take the third element of B, but for the second element of column 1 is equal in A and B ,but second element of column 2 is different ,here it should search for that element and print the element in the third column,and am doing this with for loop which is very slow because of larger dimension.In mine actual problem I have given for loop as written below:
for k=1:37651
for j=1:26018
if (s(k,1:2)==l(j,1:2))
z=sin((90-s(k,3))*pi/180) , break ,end
end
end
I want an alternative way to do this which should be faster than this.
You should work with complete matrices or vectors whenever possible. You should try commands and inspect intermediate results in the interactive shell to see how they fit together.
A(:,1)
selects the first column of a matrix. You can compare matrices/vectors and the result is a matrix/vector of 0/1 again:
> A(:,1) == B(:,1)
ans =
1
1
0
If you assign the result you can use it again to index into matrices:
I = A(:,1) == B(:,1)
B(I, 3)
This selects the third column of B of those rows where the first column of A and B is equal.
I hope this gets you started.