R - "apply" for 2 matrices - r

I am doing an R assignment and I have to write a function that does what dist.xyz does.
dist.xyz(a, b = NULL, all.pairs=FALSE)
a and b are matrices of numbers and the function computes the distances between corresponding rows of
‘a’ and ‘b’.
I tried a for loop (as below) but it takes too long and "apply" only allows us to do operation on 1 matrix at a time.
dis = vector()
for (i in 1:nrow(a)) {
append(dis,sqrt(sum((a[i,] - b[i,]) ^ 2)))
}
Is there some way to "apply" to two matrices?
Thanks in advance

Would be easier if you had example data. But here's my take. This isn't a general solution for '"apply" for 2 matrices'. However, in your case, you only need apply for a single matrix a-b, since the element-wise difference of each row is the first thing you take. Then apply square, sum, and square root to each row to obtain your result.
set.seed(7) # just to ensure reproducible results
rowDist<-function(a,b) {
apply(a-b,1,function(x)sqrt(sum(x^2)))
}
a<-matrix(rnorm(25),5,5)
b<-matrix(rnorm(25),5,5)
rowDist(a,b)
#[1] 2.716251 2.685056 3.699462 2.125998 3.437412

Related

R: How to create a loop for, for a range of data in a function?

I have this parameter:
L_inf <- seq(17,20,by=0.1)
and this function:
fun <- function(x){
L_inf*(1-exp(-B*(x-0)))}
I would to apply this function for a range of value of L_inf.
I tried with loop for, like this:
A <- matrix() # maybe 10 col and 31 row or vice versa
for (i in L_inf){
A[i] <- fun(1:10)
}
Bur R respond: longer object length is not a multiple of shorter object length.
My expected output is a matrix (or data frame, or list maybe) with 10 result (fun(1:10)) for each value of the vector L_inf (lenght=31).
How can to do it?
You are trying to put a vector of 10 elements into one of the matrix cell. You want to assign it to the matrix row instead (you can access the ith row with A[i,]).
But using a for loop in this case is inefficient and it is quite straightforward to use one of the "apply" function. Apply functions typically return a list (which is the most versatile container since there is basically no constraint).
Here sapply is an apply function which tries to Simplify its result to a convenient data structure. In this case, since all results have the same length (10), sapply will simplify the result to a matrix.
Note that I modified your function to make it explicitly depend on L_inf. Otherwise it will not do what you think it should do (see keyword "closures" if you want more info).
L_inf_range <- seq(17,20,by=0.1)
B <- 1
fun <- function(x, L_inf) {
L_inf*(1-exp(-B*(x-0)))
}
sapply(L_inf_range, function(L) fun(1:10, L_inf=L))

Indexing variables in R

I am normally a maple user currently working with R, and I have a problem with correctly indexing variables.
Say I want to define 2 vectors, v1 and v2, and I want to call the nth element in v1. In maple this is easily done:
v[1]:=some vector,
and the nth element is then called by the command
v[1][n].
How can this be done in R? The actual problem is as follows:
I have a sequence M (say of length 10, indexed by k) of simulated negbin variables. For each of these simulated variables I want to construct a vector X of length M[k] with entries given by some formula. So I should end up with 10 different vectors, each of different length. My incorrect code looks like this
sims<-10
M<-rnegbin(sims, eks_2016_kasko*exp(-2.17173), 840.1746)
for(k in 1:sims){
x[k]<-rep(NA,M[k])
X[k]<-rep(NA,M[k])
for(i in 1:M[k]){x[k][i]<-runif(1,min=0,max=1)
if(x[k][i]>=0 & x[i]<=0.1056379){
X[k][i]<-rlnorm(1, 6.228244, 0.3565041)}
else{
X[k][i]<-rlnorm(1, 8.910837, 1.1890874)
}
}
}
The error appears to be that x[k] is not a valid name for a variable. Any way to make this work?
Thanks a lot :)
I've edited your R script slightly to get it working and make it reproducible. To do this I had to assume that eks_2016_kasko was an integer value of 10.
require(MASS)
sims<-10
# Because you R is not zero indexed add one
M<-rnegbin(sims, 10*exp(-2.17173), 840.1746) + 1
# Create a list
x <- list()
X <- list()
for(k in 1:sims){
x[[k]]<-rep(NA,M[k])
X[[k]]<-rep(NA,M[k])
for(i in 1:M[k]){
x[[k]][i]<-runif(1,min=0,max=1)
if(x[[k]][i]>=0 & x[[k]][i]<=0.1056379){
X[[k]][i]<-rlnorm(1, 6.228244, 0.3565041)}
else{
X[[k]][i]<-rlnorm(1, 8.910837, 1.1890874)
}
}
This will work and I think is what you were trying to do, BUT is not great R code. I strongly recommend using the lapply family instead of for loops, learning to use data.table and parallelisation if you need to get things to scale. Additionally if you want to read more about indexing in R and subsetting Hadley Wickham has a comprehensive break down here.
Hope this helps!
Let me start with a few remarks and then show you, how your problem can be solved using R.
In R, there is most of the time no need to use a for loop in order to assign several values to a vector. So, for example, to fill a vector of length 100 with uniformly distributed random variables, you do something like:
set.seed(1234)
x1 <- rep(NA, 100)
for (i in 1:100) {
x1[i] <- runif(1, 0, 1)
}
(set.seed() is used to set the random seed, such that you get the same result each time.) It is much simpler (and also much faster) to do this instead:
x2 <- runif(100, 0, 1)
identical(x1, x2)
## [1] TRUE
As you see, results are identical.
The reason that x[k]<-rep(NA,M[k]) does not work is that indeed x[k] is not a valid variable name in R. [ is used for indexing, so x[k] extracts the element k from a vector x. Since you try to assign a vector of length larger than 1 to a single element, you get an error. What you probably want to use is a list, as you will see in the example below.
So here comes the code that I would use instead of what you proposed in your post. Note that I am not sure that I correctly understood what you intend to do, so I will also describe below what the code does. Let me know if this fits your intentions.
# define M
library(MASS)
eks_2016_kasko <- 486689.1
sims<-10
M<-rnegbin(sims, eks_2016_kasko*exp(-2.17173), 840.1746)
# define the function that calculates X for a single value from M
calculate_X <- function(m) {
x <- runif(m, min=0,max=1)
X <- ifelse(x > 0.1056379, rlnorm(m, 6.228244, 0.3565041),
rlnorm(m, 8.910837, 1.1890874))
}
# apply that function to each element of M
X <- lapply(M, calculate_X)
As you can see, there are no loops in that solution. I'll start to explain at the end:
lapply is used to apply a function (calculate_X) to each element of a list or vector (here it is the vector M). It returns a list. So, you can get, e.g. the third of the vectors with X[[3]] (note that [[ is used to extract elements from a list). And the contents of X[[3]] will be the result of calculate_X(M[3]).
The function calculate_X() does the following: It creates a vector of m uniformly distributed random values (remember that m runs over the elements of M) and stores that in x. Then it creates a vector X that contains log normally distributed random variables. The parameters of the distribution depend on the value x.

Making matrix in R

I want to make matrices without using loops such as for , while.
So I tried assigned k and put k in function which makes matrices.
powlist= function(base,startnum,endnum) (base)^(startnum:endnum)
m_maker= function(base) matrix(c(powlist(base,0,19)),4,5)
k= 2:10
a= m_maker((k-1)/k)
But function returns only one matrix.
I think function should return 9 matrices.
Please let me know how should I change this code.
I want to make each matrices that first one is matrix m_maker(1/2) and
second one m_maker(2/3) so on.
When I put k=2 and k=3 each time, it returns what I want.
What I want is way to return 9 matrices at one to go.
You're looking for lapply, like
res <- lapply((k-1)/k, m_maker)
However, you really should use an array for something like this.
ares <- abind(res, along=3)

R: how do you apply a function to a vector and get a vector of different length?

I have a list of means, for example
avgs = c(1,2,3)
and a function:
simulate <- function (avg)
{ rnorm(n=10,m=avg,sd=1) }
What is the best way to get a vector of 30 values, rather than a multidimensional array from
sapply(avgs,simulate)?
In your case, just take advantage of the fact that rnorm is vectorized and thus will accept entire vectors as arguments:
rnorm(30, avgs, 1)
You can also remove dimensions from your matrix with c:
c(sapply(avgs, simulate))
but this approach is slower and less direct.

How to calculate Euclidean distance (and save only summaries) for large data frames

I've written a short 'for' loop to find the minimum euclidean distance between each row in a dataframe and all the other rows (and to record which row is closest). In theory this avoids the errors associated with trying to calculate distance measures for very large matrices. However, while not that much is being saved in memory, it is very very slow for large matrices (my use case of ~150K rows is still running).
I'm wondering whether anyone can advise or point me in the right direction in terms of vectorising my function, using apply or similar. Apologies for what may seem a simple question, but I'm still struggling to think in a vectorised way.
Thanks in advance (and for your patience).
require(proxy)
df<-data.frame(matrix(runif(10*10),nrow=10,ncol=10), row.names=paste("site",seq(1:10)))
min.dist<-function(df) {
#df for results
all.min.dist<-data.frame()
#set up for loop
for(k in 1:nrow(df)) {
#calcuate dissimilarity between each row and all other rows
df.dist<-dist(df[k,],df[-k,])
# find minimum distance
min.dist<-min(df.dist)
# get rowname for minimum distance (id of nearest point)
closest.row<-row.names(df)[-k][which.min(df.dist)]
#combine outputs
all.min.dist<-rbind(all.min.dist,data.frame(orig_row=row.names(df)[k],
dist=min.dist, closest_row=closest.row))
}
#return results
return(all.min.dist)
}
#example
min.dist(df)
This should be a good start. It uses fast matrix operations and avoids the growing object construct, both suggested in the comments.
min.dist <- function(df) {
which.closest <- function(k, df) {
d <- colSums((df[, -k] - df[, k]) ^ 2)
m <- which.min(d)
data.frame(orig_row = row.names(df)[k],
dist = sqrt(d[m]),
closest_row = row.names(df)[-k][m])
}
do.call(rbind, lapply(1:nrow(df), which.closest, t(as.matrix(df))))
}
If this is still too slow, as a suggested improvement, you could compute the distances for k points at a time instead of a single one. The size of k will need to be a compromise between speed and memory usage.
Edit: Also read https://stackoverflow.com/a/16670220/1201032
Usually, built in functions are faster that coding it yourself (because coded in Fortran or C/C++ and optimized).
It seems that the function dist {stats} answers your question spot on:
Description
This function computes and returns the distance matrix computed by using the specified distance measure to compute the distances between the rows of a data matrix.

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