I'm converting a rather complicated set of code from Matlab to R. I have zero experience in Matlab and am a functioning novice in R.
I have a segment of code which reads (in matlab):
dSii=(sum(tao.*Sik,1))'-(sum(m'))'.*Sii-beta.*Sii./N.*(Iii+sum(Iik)');
Which I've simplified and will focus on the first segment (if I can solve the first segment I'm confident I can perform the rest):
J = (sum(A.*B,1))' - ...
tao (or A) and Sik (or B) are matrices. So my assumption is I'm performing matrix multiplication here (A * B)and summing the resultant column. The '1' is what is throwing me off in that statement. In R, that 1 would likely indicate we're talking about a sum of rows as opposed to columns(indicated by 2). But I can't find any supporting documentation for that kind of Matlab statement.
I was thinking of using a statement like this (but of course, too many '1's and ',')
J<- (apply(A*B, 1), 1, sum)
Thanks for all your help. I searched for other examples here and elsewhere and couldn't find an answer. I'm willing to work for it but this is akin to me studying French (which I don't know) to translate in Spanish (which I'm moderate in) while interpreting the whole process in English. :D
Because of the different conventions in R and Matlab, the idiosyncrasies have to be learned for each (just like your language analogy!). The Matlab command sum(A.*B,1) means multiply A and B element-wise, so they must be the same shape, and then sum along dimension 1, i.e. add each row together to get the column sums. Dimension 1 is the default so, sum(A.*B) would do the same thing as sum(A.*B,1). Because R treats * as element-wise for matrix multiplication, the following Matlab and R codes will produce the same column of numbers in J:
Matlab:
A=[[1,2,3];[4,5,6];[7,8,9]];
B=[[10,11,12];[13,14,15];[16,17,18]];
J=sum(A.*B,1)'; %the ' means to transpose the column sums to be a 3x1 matrix
R:
A<-matrix(c(1,2,3,4,5,6,7,8,9),3,byrow=T)
B<-matrix(c(10,11,12,13,14,15,16,17,18),3,byrow=T)
J<-matrix(colSums(A*B)) # no transpose needed here: nrow(J)==3
Related
I am trying to use solve() in R to find a solution for a 10x10 matrix. Specifically, I am looking for x in Ax=b where b is a ten dimensional 0 vector. When I input solve(A, rep(0,10)), R returns the trivial solution, namely rep(0,10). I also checked -- det(A) is indeed not equal to 0 and thus not singular.
So how can I stop R from returning this result?
Premultiplying both sides of the equation by the inverse of A gives x=A^{-1}b, i.e. on the right hand side we have a zero vector because b is a zero vector. So, that is the only solution.
My prof introduced a concept that required use of a vector, which he represented as follows (imagine there is only one pair of brackets below, tall enough to encapsulate both terms; I don't have the rep to paste an image and don't know how to format this otherwise):
v =
[-1/2]
[1/2 ]
One of my personal weaknesses is a lack of familiarity with mathematical notation. Is there an accepted way of interpreting this kind notation? Does it vary by discipline, or is this something generalizable that I really should know? Is there something intrinsic about this notation that would lead one to interpret it differently than if it were written v = [-1/4, 1/4]?
Thanks for the help!
A vector is a one-dimensional matrix, but it is a matrix nonetheless. Writing it out horizontally instead of vertically or vice versa changes the dimensionality of the matrix, changing its meaning among the rest of the equations.
Very often you will "transform" a vector by multiplying them by a matrix. For instance, to rotate a vector, you have to multiply it by the rotation matrix, etc. If your vectors are codified in columns, a multiplication by a matrix M will act from the left, M * v, because of the way the multiplication works (every row of M by the column vector v.) Alternatively, if your vectors are codified as rows (v = [-1/4, 1/4]) the multiplication will act from the right: v * M, again, because of the "row by column" definition of the multiplication of "matrices".
So, it is up to you to represent vectors as rows or columns provided your convention is consistent with the way you multiply them by matrices.
I have a problem to solve in either Matlab or R (preferably in R).
Imagine I have a vector A with 10 elements.
I have also a vector B with 30 elements, of which 10 have value 'x'.
Now, I want to replace all the 'x' in B by the corresponding values taken from A, in the order that is established in A. Once a value in A is taken, the next one is ready to be used when the next 'x' in B is found.
Note that the sizes of A and B are different, it's the number of 'x' cells that coincides with the size of A.
I have tried different ways to do it. Any suggestion on how to program this?
As long as the number of x entries in B matches the length of A, this will do what you want:
B[B=='x'] <- A
(It should be clear that this is the R solution.)
MATLAB Solution
In MATLAB it's quite simple, use logical indexing:
B(B == 'x') = A;
How can I turn a regular matrix into a matrix full-ranked in R? Is there an available method for that?
I have a matrix that may have linearly dependent columns and I need to
pass it to a function that requires its argument to be a matrix with
full rank. Since linearly dependent columns are not of interest
anyway, I am looking for a function that removes such columns until
the matrix is full rank. There may be several solutions of course, but
any one of them should be fine.
Right now I am just constructing the matrix column by column and only
add a column if its the resulting matrix is still fullrank, but it
feels like there should be a better way to do this.
Another approach is to minimize |y - Ax|2 + c |x|2,
by tacking an identity matrix on to A and zeros to y.
The parameter c (a.k.a. λ)
trades off fitting y - Ax, and keeping |x| small.
Then run a second fit with the r largest components of x,
r = rank(A) (or any number you please).
I have 2 vectors with 11 dimentions.
a <- c(-0.012813841, -0.024518383, -0.002765056, 0.079496744, 0.063928973,
0.476156960, 0.122111977, 0.322930189, 0.400701256, 0.454048860,
0.525526219)
b <- c(0.64175768, 0.54625694, 0.40728261, 0.24819750, 0.09406221,
0.16681692, -0.04211932, -0.07130129, -0.08182200, -0.08266852,
-0.07215885)
cosine_sim <- cosine(a,b)
which returns:
-0.05397935
I used cosine() from lsa package.
for some values i am getting negative cosine_sim like the given one. I am not sure how the similarity can be negative. It should be between 0 and 1.
Can anyone explain what is going on here.
The nice thing about R is that you can often dig into the functions and see for yourself what is going on. If you type cosine (without any parentheses, arguments, etc.) then R prints out the body of the function. Poking through it (which takes some practice), you can see that there is a bunch of machinery for computing the pairwise similarities of the columns of the matrix (i.e., the bit wrapped in the if (is.matrix(x) && is.null(y)) condition, but the key line of the function is
crossprod(x, y)/sqrt(crossprod(x) * crossprod(y))
Let's pull this out and apply it to your example:
> crossprod(a,b)/sqrt(crossprod(a)*crossprod(b))
[,1]
[1,] -0.05397935
> crossprod(a)
[,1]
[1,] 1
> crossprod(b)
[,1]
[1,] 1
So, you're using vectors that are already normalized, so you just have crossprod to look at. In your case this is equivalent to
> sum(a*b)
[1] -0.05397935
(for real matrix operations, crossprod is much more efficient than constructing the equivalent operation by hand).
As #Jack Maney's answer says, the dot product of two vectors (which is length(a)*length(b)*cos(a,b)) can be negative ...
For what it's worth, I suspect that the cosine function in lsa might be more easily/efficiently implemented for matrix arguments as as.dist(crossprod(x)) ...
edit: in comments on a now-deleted answer below, I suggested that the square of the cosine-distance measure might be appropriate if one wants a similarity measure on [0,1] -- this would be analogous to using the coefficient of determination (r^2) rather than the correlation coefficient (r) -- but that it might also be worth going back and thinking more carefully about the purpose/meaning of the similarity measures to be used ...
The cosine function returns
crossprod(a, b)/sqrt(crossprod(a) * crossprod(b))
In this case, both the terms in the denominator are 1, but crossprod(a, b) is -0.05.
The cosine function can take on negative values.
While cosine of two vectors can take any value between -1 and +1, cosine similarity (in dicument retreival) used to take values from the [0,1] interval. The reason is simple: in the WordxDocument matrix there are no negative values, so the maximum angle of two vectors is 90 degrees, for wich the cosine is 0.