I want to use V1 to scale something in the 2nd coordinates, but I don't know how to calculate V1', the first coordinates use Transform T to the 2nd coordinate.
what should I do?
thanks.
If I understand your question properly,
V1' = V * T-1
where T-1 is inverse matrix to transform T
About multiply order - there are both "left multiply" and "right multiply" conventions. Vectors for them are row and column. And transformation matrices are transposed. For example, DirectX and OpenGL use different conventions. I don't know what one is used in cocos.
Related
I have two tensors of rank 3 each, in other words two 3D matrix. I want to take dot product of these two matrix. I am confused to continue with this problem. Help me out with formula to do so.
A 3-way tensor (or equivalently 3D array or 3-order array) need not necessarily be of rank-3; Here, "rank of a tensor" means the minimum number of rank-1 tensors (i.e. outer product of vectors; For N-way tensor, it's the outer product of N vectors) needed to get your original tensor. This is explained in the below figure of so-called CP decomposition.
In the above figure, the original tensor(x) can be written as a sum of R rank-1 tensors, where R is a positive integer. In CP decomposition, we aim to find a minimum R that yields our original tensor X. And this minimum R is called the rank of our original tensor.
For a 3-way tensor, it is the minimum number of (a1,a2,a3...aR; b1,b2,b3...bR; c1,c2,c3...cR) vectors (where each of the vectors is n dimensional) required to obtain the original tensor. The tensor can be written as the outer product of these vectors as:
In terms of element-wise, we can write the 3-way tensor as:
Now, with that background, to answer your specific question, to take the dot product (also called tensor inner product), both tensors must be of same shape (for e.g. 3x2x5 and 3x2x5), then the inner product is defined as the sum of the element-wise product of their values.
where the script X and Y are the same-shape tensors.
P.S.: The tilde in the above formulae should not be interpreted as an approximation.
The vector inner product sum the elementwise products. The tensor inner product follows the same idea. Match the elements, multiply them, and add them all .
Assume X is a 5x5 matrix (which represents pixel luminosity from a picture):
I would like to select the element that fit within a given simple geometrical shape (e.g. square, circle, ovale) superposed to this matrix. In this example I would like to select the elements from the matrix that fall (even slightly) within the ovale.
Ultimately those elements position would be return into a vector logical vector elementInOvale on which I could perform a simple operation such as X[elementInOvale] <- 0
I have the feeling that this is a common problem which has already been solved, I just don't know how to formulate or where to find information about it, and to do with R.
I've got a big problem.
I've got a large raster (rows=180, columns=480, number of cells=86400)
At first I binarized it (so that there are only 1's and 0's) and then I labelled the clusters.(Cells that are 1 and connected to each other got the same label.)
Now I need to calculate all the distances between the cells, that are NOT 0.
There are quiet a lot and that's my big problem.
I did this to get the coordinates of the cells I'm interested in (get the positions (i.e. cell numbers) of the cells, that are not 0):
V=getValues(label)
Vu=c(1:max(V))
pos=which(V %in% Vu)
XY=xyFromCell(label,pos)
This works very well. So XY is a matrix, which contains all the coordinates (of cells that are not 0). But now I'm struggling. I need to calculate the distances between ALL of these coordinates. Then I have to put each one of them in one of 43 bins of distances. It's kind of like this (just an example):
0<x<0.2 bin 1
0.2<x<0.4 bin2
When I use this:
pD=pointDistance(XY,lonlat=FALSE)
R says it's not possible to allocate vector of this size. It's getting too large.
Then I thought I could do this (create an empty data frame df or something like that and let the function pointDistance run over every single value of XY):
for (i in 1:nrow(XY))
{pD=PointDistance(XY,XY[i,],lonlat=FALSE)
pDbin=as.matrix(table(cut(pD,breaks=seq(0,8.6,by=0.2),Labels=1:43)))
df=cbind(df,pDbin)
df=apply(df,1,FUN=function(x) sum(x))}
It is working when I try this with e.g. the first 50 values of XY.
But when I use that for the whole XY matrix it's taking too much time.(Sometimes this XY matrix contains 10000 xy-coordinates)
Does anyone have an idea how to do it faster?
I don't know if this will works fast or not. I recommend you try this:
Let say you have dataframe with value 0 or 1 in each cell. To find coordinates all you have to do is write the below code:
cord_matrix <- which(dataframe == 1, arr.ind = TRUE)
Now, you get the coordinate matrix with row index and column index.
To find the euclidean distance use dist() function. Go through it. It will look like this:
dist_vector <- dist(cord_matrix)
It will return lower triangular matrix. can be transformed into vector/symmetric matrix. Now all you have to do is calculating bins according to your requirement.
Let me know if this works within the specific memory space.
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.
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).