I'd like to perform this function on a matrix 100 times. How can I do this?
v = 1
m <- matrix(0,10,10)
rad <- function(x) {
idx <- sample(length(x), size=1)
flip = sample(0:1,1,rep=T)
if(flip == 1) {
x[idx] <- x[idx] + v
} else if(flip == 0) {
x[idx] <- x[idx] - v
return(x)
}
}
This is what I have so far but doesn't work.
for (i in 1:100) {
rad(m)
}
I also tried this, which seemed to work, but gave me an output of like 5226 rows for some reason. The output should just be a 10X10 matrix with changed values depending on the conditions of the function.
reps <- unlist(lapply(seq_len(100), function(x) rad(m)))
Ok I think I got it.
The return statement in your function is only inside a branch of an if statement, so it returns a matrix with a probability of ~50% while in the other cases it does not return anything; you should change the code function into this:
rad <- function(x) {
idx <- sample(length(x), size=1)
flip = sample(0:1,1,rep=T)
if(flip == 1) {
x[idx] <- x[idx] + v
} else if(flip == 0) {
x[idx] <- x[idx] - v
}
return(x)
}
Then you can do:
for (i in 1:n) {
m <- rad(m)
}
Note that this is semantically equal to:
for (i in 1:n) {
tmp <- rad(m) # return a modified verion of m (m is not changed yet)
# and put it into tmp
m <- tmp # set m equal to tmp, then in the next iteration we will
# start from a modified m
}
When you run rad(m) is not do changes on m.
Why?
It do a local copy of m matrix and work on it in the function. When function end it disappear.
Then you need to save what function return.
As #digEmAll write the right code is:
for (i in 1:100) {
m <- rad(m)
}
You don't need a loop here. The whole operation can be vectorized.
v <- 1
m <- matrix(0,10,10)
n <- 100 # number of random replacements
idx <- sample(length(m), n, replace = TRUE) # indices
flip <- sample(c(-1, 1), n, replace = TRUE) # subtract or add
newVal <- aggregate(v * flip ~ idx, FUN = sum) # calculate new values for indices
m[newVal[[1]]] <- m[newVal[[1]]] + newVal[[2]] # add new values
Related
I am a R beginner and I tried to make a median function from scratch.
Here is my code:
mymedian <- function(x) {
len <- length(x)
sorted <- sort(x)
if (len %% 2 == 0) {
med1 <- sorted[len / 2]
med2 <- sorted[(len + 1) %/% 2]
result <- sorted[med1 + med2 / 2]
return(result)
} else {
result <- sorted[(len + 1)/2]
return(result)
}
}
Im getting "NA" output. I couldn't find where the problem is.
Main issue is you're trying to index your sorted vector with a non-integer (e.g., 168.5). Compare your function to this:
mymedian <- function(x){
len <- length(x)
sorted <-sort(x)
if(len%%2==0){
i <- len/2
med1<-sorted[i]
med2 <- sorted[i+1]
result <- sum(med1,med2)/2
return(result)
}else{
result<-sorted[(len+1)/2]
return(result)
}
}
I have a large loop that will take too long (~100 days). I'm hoping to speed it up with the snow library, but I'm not great with apply statements. This is only part of the loop, but if I can figure this part out, the rest should be straightforward. I'm ok with a bunch of apply statements or loops, but one apply statement using a function to get object 'p' would be ideal.
Original data
dim(m1) == x x # x >>> 0
dim(m2) == y x # y >>> 0, y > x, y > x-10
dim(mout) == x x
thresh == x-10 #specific to my data, actual number probably unimportant
len(v1) == y #each element is a random integer, min==1, max==thresh
len(v2) == y #each element is a random integer, min==1, max==thresh
Original loop
p <- rep(NA,y)
for (k in 1:y){
mout <- m1 * matrix(m2[k,],x,x)
mout <- mout/sum(mout)
if (v1[k] < thresh + 1){
if(v2[k] < thresh + 1){
p[k] <- out[v1[k],v2[k]]
}
if(v2[k] > thresh){
p[k] <- sum(mout[v1[k],(thresh+1):x])
}
}
#do stuff with object 'p'
}
library(snow)
dostuff <- function(k){
#contents of for-loop
mout <- m1 * matrix(m2[k,],x,x)
mout <- mout/sum(mout)
if (v1[k] < thresh + 1){
if(v2[k] < thresh + 1){
p <- out[v1[k],v2[k]]
}
if(v2[k] > thresh){
p <- sum(mout[v1[k],(thresh+1):x])
}
}
#etc etc
return(list(p,
other_vars))
}
exports = c('m1',
'm2',
'thresh',
'v1',
'x' ,
'v2')
cl = makeSOCKcluster(4)
clusterExport(cl,exports)
loop <- as.array(1:y)
out <- parApply(cl,loop,1,dostuff)
p <- rep(NA,y)
for(k in 1:y){
p[k] <- out[[k]][[1]]
other_vars[k] <- out[[k]][[2]]
}
A have code that creates a random graph in the form of a matrix. Now I would like it to create many, say m, random graphs so the output is m matrices. I am trying to do this with a for loop. This would be my preferred method however I am open to other suggestions (apply family?). Here is my code, where n is the number of nodes/vertices the graph has and beta is the amount of preferential attachment (keep this between 0 and 1.5)
multiplerandomgraphs <- function(n, beta, m) {
for(k in 1:m) {
randomgraph <- function(n, beta) {
binfunction <- function(y) {
L <- length(y)
x <- c(0, cumsum(y))
U <- runif(1, min = 0 , max = sum(y))
for(i in 1:L) {
if(x[i] <= U && x[i+1] > U){
return(i)
}
}
}
mat <- matrix(0,n,n)
mat[1,2] <- 1
mat[2,1] <- 1
for(i in 3:n) {
degvect <- colSums(mat[ , (1:(i-1))])
degvect <- degvect^(beta)
j <- binfunction(degvect)
mat[i,j] <- 1
mat[j,i] <- 1
}
return(mat)
}
}
}
You can define your randomgraph function as randomgraph <- function(i, n, beta) {} with the body the same as your definition, leaves the parameter i as a dummy parameter. And then use apply function as listOfMatrix <- lapply(1:m, randomgraph, n, beta) which return a list of matrix.
Im trying to creat a function in r that will let me sum data from the last three rows into a new column (i.e. if im looking at day 4 = days 3+2+1)
This is what i've worked out so far, however it doesnt work.
S3<- function(x){
res <- numeric(nrow(x))
for (i in 1:nrow(x)){
res[i] <- i
if (i > 3) {
res[i] <- x[i-3,10]
} else {
res[i] <- x[i,10]
}
}
x$PP3 <- res
return(x)
}
Use this:
x$PP3 <- x[,10] + c(0,head(x[,10],-1)) + c(0,0,head(x[,10],-2))
If you want a function:
S3 <- function(x, j=10){
within(x, PP3 <- x[,j] + c(0,head(x[,j],-1)) + c(0,0,head(x[,j],-2)))
}
I am trying to create a function to calculate the Box-Cox transformation in R, where you iterate values of lambda (lambdas) in a formula to maximize L. What I ultimately want is a vector of L, such that for all i in lambda, there is a corresponding L value.
y <- c(256,256,231,101,256,213,241,246,207,143,287,240,262,234,146,255,184,161,252,229,283,132,218,113,194,237,181,262,104)
df <- 28
n=29
lambdas <- seq(-3,3,0.001)
L <- c(rep(NA,length(lambdas)))
for(i in lambdas) {
if(i != 0) {
yprime <- (((y^i)-1)/i)
} else
{ yprime <- log(y)
}
st2 <- var(yprime)
L <- (((-df/2)*(log(st2))) + ((i-1)*(df/n)*(sum(log(y)))))
}
What I typically end up with L as a vector of 1, with the final iteration calculated.
Use seq_along to generate an index for lambdas[] and L[]
for(i in seq_along(lambdas)) {
if(i != 0) {
yprime <- (((y^lambdas[i])-1)/lambdas[i])
} else {
yprime <- log(y)
}
st2 <- var(yprime)
L[i] <- (((-df/2)*(log(st2))) + ((lambdas[i]-1)*(df/n)*(sum(log(y)))))
}
plot(L)