How to combine two loops - r

In a tutorial on for() Loops came across the following exercise:
Exercise 4.4. Write a function to perform matrix-vector multiplication. It should take a matrix A and a vector b as arguments, and return the vector Ab. Use two loops to do this, rather than %*% or any vectorization.
Lets say I use a specific matrix A(dim:3,4) and vector b(length(3)).
> # Ex 4.4
> out<-c(1,1,1)
> Ab<-function(A,b) {
+ for(i in 1:dim(A)[1]) {
+
+ out[i]=sum(A[i,]*b)
+ }
+ out
+ }
> a = c(1,1,1)
> A
[,1] [,2] [,3]
[1,] 1 4 7
[2,] 2 5 8
[3,] 3 6 10
> a
[1] 1 1 1
> Ab(A,a)
[1] 12 15 19
This works for a very specific case, i.e. matrix with 3 rows and vector of length 3, but leaves much to be desired, i don't know what a good solution to this exercise would be but the question says 'use two loops'. Suggestions will be much appreciated.
thx

You are hiding the inner loop with A[i,]*b which is doing vectorized multiplication (ie. a hidden loop). So, if you expand that out explicitly you will have the two required loops.
Ab<-function(A,b) {
if (dim(A)[2] != NROW(b)) stop("wrong dimensions")
out <- matrix(, nrow(A), 1)
for(i in 1:dim(A)[1]) {
s <- 0
for (j in 1:dim(A)[2]) s <- s + A[i,j] * b[j]
out[i] <- s
}
out
}

Related

How to perform matrix operations similar to multiplication fast?

I have two matrices.
A<-matrix(c(1,0,2,3),2,2)
B<-matrix(c(0,1,4,2),2,2)
Instead of multiplication (A%*%B) and having results like:
C[1,1]<-A[1,1]*B[1,1]+ A[1,2]*B[2,1]
C[1,2]<-A[1,1]*B[1,2]+ A[1,2]*B[2,2]
C[2,1]<-A[2,1]*B[1,1]+ A[2,2]*B[2,1]
C[2,2]<-A[2,1]*B[1,2]+ A[2,2]*B[2,2]
How can I have a modified version of multiplication and get results like:
C[1,1]<-min(A[1,1],B[1,1])+ min(A[1,2],B[2,1])
C[1,2]<-min(A[1,1],B[1,2])+ min(A[1,2],B[2,2])
C[2,1]<-min(A[2,1],B[1,1])+ min(A[2,2],B[2,1])
C[2,2]<-min(A[2,1],B[1,2])+ min(A[2,2],B[2,2])
?
I know that I can do it with rotation, but am looking for a faster solution.
result <- matrix(nrow= 2, ncol= 2)
for(i in 1:2){
minMat <-t(apply(B,2,function(x) pmin(x, A[i,])))
result[i,]<-rowSums(minMat)
}
A piece of the solution could be to use a function as follow (from one of the comments above):
## Defining the function
sum.min.row <- function(i, A, B) {
minMat <-t(apply(B,2,function(x) pmin(x, A[i,])))
rowSums(minMat)
}
## Applying it to the whole matrix
t(sapply(1:nrow(A), sum.min.row, A, B))
# [,1] [,2]
# [1,] 1 3
# [2,] 1 2
This is still not optimal though...

R programming:How to use loop on variables labelled in a consecutive manner?

I'm trying to figure out, how I can run a loop on some variables that have a consecutive label.
I want to do matrix.2-Matrix.1 and store it in matrix x.1, then Matrix.3-matrix.2 and store it in matrix x.2. There are 300 matrices(Matrix.1,Matrix.2,...Matrix.300) but for this example, I would like to just work on matrix 1,2 and 3.
I first tried an approach that involved the list function, but it didn't work, and then I thought about using a MACRO just like in SAS (the % symbol). But the Macro approach seemed not to work in R.
My code is below:
(The list approach)
> Matrix.1=matrix(c(1:6),nrow=2,ncol=3,byrow=TRUE)
> Matrix.2=matrix(c(1,8,9,17,15,2),nrow=2,ncol=3,byrow=TRUE)
> Matrix.3=matrix(c(0,1,2,3,6,0),nrow=2,ncol=3,byrow=TRUE)
> x.1=matrix(rep(0,6),nrow=2,ncol=3,byrow=TRUE)
> x.2=matrix(rep(0,6),nrow=2,ncol=3,byrow=TRUE)
> m=list(Matrix.1=Matrix.1,Matrix.2=Matrix.2,Matrix.3=Matrix.3)
> x=list(x.1=x.1,x.2=x.2)
> m[1]
$Matrix.1
[,1] [,2] [,3]
[1,] 1 2 3
[2,] 4 5 6
> m[2]
$Matrix.2
[,1] [,2] [,3]
[1,] 1 8 9
[2,] 17 15 2
> m[3]
$Matrix.3
[,1] [,2] [,3]
[1,] 0 1 2
[2,] 3 6 0
> x[1]
$x.1
[,1] [,2] [,3]
[1,] 0 0 0
[2,] 0 0 0
> x[2]
$x.2
[,1] [,2] [,3]
[1,] 0 0 0
[2,] 0 0 0
> for (i in 1:2){
+ x[i]=m[i+1]-m[i]
+ print(x[i])
+ }
Error in m[i + 1] - m[i] : non-numeric argument to binary operator
>
How can I make operations on list?
> #Other approach inspired from SAS
> for (i in i:2){
+ x.i=Matrix.i+1-Matrix.i
+ print(R.i)
+ }
Error: object 'Matrix.i' not found
This second approach isn't even doable in R.
What is the best way of dealing loops involving consecutively labelled variables?
Since m and x are both lists, you need to use m[[1]] and x[[1]] to extract its elements.
for (i in 1:2){
x[[i]] <- m[[i+1]]-m[[i]]
print(x[[i]])
}
On the other hand, if you have 300 matrices (Matrix.1, Matrix.2, ... Matrix.300), you could use get and assign to deal with the numerical labels. Here I first assign values to 300 matrices with names Matrix.1 through Matrix.300. Then I use get function to extract these matrices and generate list x.
for (i in 1:300) {
assign(paste("Matrix.", i, sep = ""), matrix(rnorm(9), 3, 3))
}
x <- list()
for (i in 2:300) {
x[[i-1]] <- get(paste("Matrix.", i, sep = "")) - get(paste("Matrix.", i-1, sep = ""))
}
It is the preferred method in R to use the apply family of functions to loop through objects. For lists, you can use lapply which returns a list, or sapply which returns the most simplified object it can without losing any information. With these functions, you output is stored in the same order as the input, which makes comparisons or additional steps much easier.
myProcessedList <- lapply(x, FUN=<some function>)
This is a lot simpler and more straightforward than using assign and get and is worth the investment to learn. SO has many useful examples.

R: How to do this matrix operation without loops or more efficient?

I'm trying to make this operation matrices, multiplying the first column with 2, 3 and 4, the first hold value, and then multiply the second column with 3 and 4, keep the value of the third and multiply the third column with 4. I want to do this without using a "for" loop, wanted to use functions like sapply or mapply. Does anyone have an idea how to do it?
Example with one line:
a[1,1]*(a[1,2], a[1,3], a[1,4]) = 2 4 4 4
a[1,1] a[1,2]*(a[1,3], a[1,4]) = 2 4 16 16 #keep a[1,1] a[1,2]
a[1,1] a[1,2] a[1,3] a[1,3]*(a[1,4]) = 2 4 16 256 # #keep a[1,1] a[1,2] a[1,3]
Input:
> a<- matrix(2,4,4) # or any else matrix like a<- matrix(c(1,8,10,1,4,1),3,3)
> a
[,1] [,2] [,3] [,4]
[1,] 2 2 2 2
[2,] 2 2 2 2
[3,] 2 2 2 2
[4,] 2 2 2 2
Output:
> a
[,1] [,2] [,3] [,4]
[1,] 2 4 16 256
[2,] 2 4 16 256
[3,] 2 4 16 256
[4,] 2 4 16 256
EDIT: LOOP VERSION
a<- matrix(2,4,4);
ai<-a[,1,drop=F];
b<- matrix(numeric(0),nrow(a),ncol(a)-1);
i<- 1;
for ( i in 1:(ncol(a)-1)){
a<- a[,1]*a[,-1,drop=F];
b[,i]<- a[,1];
}
b<- cbind(ai[,1],b);
b
If I understand correctly, what you are trying to do is, starting with a matrix A with N columns, perform the following steps:
Step 1. Multiply columns 2 through N of A by column 1 of A. Call the resulting matrix A1.
Step 2. Multiply columns 3 through N of A1 by column 2 of A1. Call the resulting matrix A2.
...
Step (N-1). Multiply column N of A(N-2) by column (N-1) of A(N-2). This is the desired result.
If this is indeed what you are trying to do, you need to either write a double for loop (which you want to avoid, as you say) or come up with some iterative method of performing the above steps.
The double for way would look something like this
DoubleFor <- function(m) {
res <- m
for(i in 1:(ncol(res)-1)) {
for(j in (i+1):ncol(res)) {
res[, j] <- res[, i] * res[, j]
}
}
res
}
Using R's vectorized operations, you can avoid the inner for loop
SingleFor <- function(m) {
res <- m
for(i in 1:(ncol(res)-1))
res[, (i+1):ncol(res)] <- res[, i] * res[, (i+1):ncol(res)]
res
}
When it comes to iterating a procedure, you may want to define a recursive function, or use Reduce. The recursive function would be something like
RecursiveFun <- function(m, i = 1) {
if (i == ncol(m)) return(m)
n <- ncol(m)
m[, (i+1):n] <- m[, (i+1):n] * m[, i]
Recall(m, i + 1) # Thanks to #batiste for suggesting using Recall()!
}
while Reduce would use a similar function without the recursion (which is provided by Reduce)
ReduceFun <- function(m) {
Reduce(function(i, m) {
n <- ncol(m)
m[, (i+1):n] <- m[, (i+1):n] * m[, i]
m
}, c((ncol(m)-1):1, list(m)), right = T)
}
These will all produce the same result, e.g. testing on your matrix
a <- matrix(c(1, 8, 10, 1, 4, 1), 3, 3)
DoubleFor(a)
# [,1] [,2] [,3]
# [1,] 1 1 1
# [2,] 8 32 2048
# [3,] 10 10 1000
all(DoubleFor(a) == SingleFor(a) & SingleFor(a) == RecursiveFun(a) &
RecursiveFun(a) == ReduceFun(a))
# [1] TRUE
Just out of curiosity, I did a quick speed comparison, but I don't think any one of the above will be significantly faster than the others for your size of matrices, so I would just go with the one you think is more readable.
a <- matrix(rnorm(1e6), ncol = 1e3)
system.time(DoubleFor(a))
# user system elapsed
# 22.158 0.012 22.220
system.time(SingleFor(a))
# user system elapsed
# 27.349 0.004 27.415
system.time(RecursiveFun(a))
# user system elapsed
# 25.150 1.336 26.534
system.time(ReduceFun(a))
# user system elapsed
# 26.574 0.004 26.626

Efficient implementation of summed area table/integral image in R

I am trying to construct a summed area table or integral image given an image matrix. For those of you who dont know what it is, from wikipedia:
A summed area table (also known as an integral image) is a data structure and algorithm for quickly and efficiently generating the sum of values in a rectangular subset of a grid
In other words, its used to sum up values of any rectangular region in the image/matrix in constant time.
I am trying to implement this in R. However, my code seems to take too long to run.
Here is the pseudo code from this link. in is the input matrix or image and intImg is whats returned
for i=0 to w do
sum←0
for j=0 to h do
sum ← sum + in[i, j]
if i = 0 then
intImg[i, j] ← sum
else
intImg[i, j] ← intImg[i − 1, j] + sum
end if
end for
end for
And here is my implementation
w = ncol(im)
h = nrow(im)
intImg = c(NA)
length(intImg) = w*h
for(i in 1:w){ #x
sum = 0;
for(j in 1:h){ #y
ind = ((j-1)*w)+ (i-1) + 1 #index
sum = sum + im[ind]
if(i == 1){
intImg[ind] = sum
}else{
intImg[ind] = intImg[ind-1]+sum
}
}
}
intImg = matrix(intImg, h, w, byrow=T)
Example of input and output matrix:
However, on a 480x640 matrix, this takes ~ 4 seconds. In the paper they describe it to take on the order of milliseconds for those dimensions.
Am I doing something inefficient in my loops or indexing?
I considered writing it in C++ and wrapping it in R, but I am not very familiar with C++.
Thank you
You could try to use apply (isn't faster than your for-loops if you pre-allocating the memory):
areaTable <- function(x) {
return(apply(apply(x, 1, cumsum), 1, cumsum))
}
areaTable(m)
# [,1] [,2] [,3] [,4]
# [1,] 4 5 7 9
# [2,] 4 9 12 17
# [3,] 7 13 16 25
# [4,] 9 16 22 33

R: Vectorize Finite Difference Equations

I'm trying to move some Fortran code to R for finite differences related to chemical kinetics.
Sample Fortran loop:
DOUBLE PRECISION, DIMENSION (2000,2) :: data=0.0
DOUBLE PRECISION :: k1=5.0, k2=20.0, dt=0.0005
DO i=2, 2000
data(i,1) = data(i-1,1) + data(i-1,1)*(-k1)*dt
data(i,2) = data(i-1,2) + ( data(i-1,1)*k1*dt - data(i-1,2)*k2*dt )
...
END DO
The analogous R code:
k1=5
k2=20
dt=0.0005
data=data.frame(cbind(c(500,rep(0,1999)),rep(0,2000)))
a.fun=function(y){
y2=y-k1*y*dt
return(y2)
}
apply(data,2,a.fun)
This overwrites my first value in the dataframe and leaves zeros elsewhere. I'd like to run this vectorized and not using a for loop since they are so slow in R. Also, my function only calculates the first column so far. I can't get the second column working until I get the syntax right on the first.
Its not necessarily true that R is bad at loops. It very much depends on what you are doing. Using k1, k2, dt and data from the question (i.e. the four lines beginning with k1=5) and formulating the problem in terms of an iterated matrix, the loop in the last line below returns nearly instantaneously on my PC:
z <- as.matrix(data)
m <- matrix(c(1-k1*dt, k1*dt, 0, 1-k2*dt), 2)
for(i in 2:nrow(z)) z[i, ] <- m %*% z[i-1, ]
(You could also try storing the vectors in columns of z rather than rows since R stores matrices by column.)
Here is the first bit of the result:
> head(z)
X1 X2
[1,] 500.0000 0.000000
[2,] 498.7500 1.250000
[3,] 497.5031 2.484375
[4,] 496.2594 3.703289
[5,] 495.0187 4.906905
[6,] 493.7812 6.095382
May be this can help.
I think you need to have the initial condition for data[1,2]. I assumed both data[1,1] as 500 and data[1,2 as 0 at the initial condition.
The code goes like this:
> ## Define two vectors x and y
> x <- seq(from=0,length=2000,by=0)
> y <- seq(from=0,length=2000,by=0)
>
> ## Constants
> k1 = 5.0
> dt = 0.0005
> k2 = 20.0
>
> ## Initialize x[1]=500 and y[1]=0
> x[1]=500
> y[1] = 0
>
> for (i in 2:2000){
+ x[i]=x[i-1]+x[i-1]*-k1*dt
+ y[i] = y[i-1]+x[i-1]*k1*dt-y[i-1]*k2*dt
+ }
>
> finaldata <- data.frame(x,y)
> head(finaldata)
x y
1 500.0000 0.000000
2 498.7500 1.250000
3 497.5031 2.484375
4 496.2594 3.703289
5 495.0187 4.906905
6 493.7812 6.095382
I hope this helps.

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