weird error with R when using data.table - r

Question

I'm doing some small calculations and i decided to fill the data inside a data.table since it's much faster than data.frame and rbind
so basically my code is something like that:
df is a data.frame used in the calculation but it's important what does it contain.
l=12000
dti = 1
dt = data.table(ni = 0, nj = 0, regerr = 0)
for (i in seq(1,12000,200)) {
for (j in seq(1, 12000, 200)) {
for (ind in 1:nrow(df)) {
if( i+j >= l/2 ){
df[ind,]$X = df[ind,]$pos * 2
} else {
df[ind,]$X = df[ind,]$pos/l
}
}
for (i in 1:100) { # 100 sample
sample(df$X,nrow(df), replace=FALSE)
fit=lm(X ~ gx, df) #linear regression calculation
regerror=sum(residuals(fit)^2)
print(paste(i,j,regerror))
set(dt,dti,1L,as.double(i))
set(dt,dti,2L,as.double(j))
set(dt,dti,3L,regerror)
dti=dti+1
}
}
}
The code prints the first few rounds of print(paste(i,j,regerror)) and then it quits with this error:
*** caught segfault ***
address 0x3ff00008, cause 'memory not mapped'
Segmentation fault (core dumped)
EDIT
structure(list(ax = c(-0.0242214, 0.19770304, 0.01587302, -0.0374415,
0.05079826, 0.12209738), gx = c(-0.3913043, -0.0242214, -0.4259067,
-0.725, -0.0374415, 0.01587302), pos = c(11222, 13564, 16532,
12543, 12534, 14354)), .Names = c("ax", "gx", "pos"), row.names = c(NA,
-6L), class = "data.frame")
Any ideas are appreciated.

Answer 1

Without meaning to sound rude, I think you may benefit from reading a few R tutorials before going forward. This question is also very likely to be closed as too localized. Also, seg faults are almost always a bug somewhere, but you can avoid a bunch of this headache by understanding what each piece of your code is doing. Since its Friday, lets walk through some of it:
if( i+j >= l/2 ){
data[ind,]$X = df[ind,]$pos * 2
}
else{
data[ind,]$X = df[ind,]$pos/l
}
I'll assume data is meant to be df and go from there. We're inside two loops of i and j that both go from 1 through 20000. They will never sum to less than 1/2 so you will always execute the first statement. Also, if you ever expected the FALSE case to occur, you would need else on the same line as your closing brace:
if (i + j >= 1/2) {
df$X <- df$pos * 2
} else {
df$X <- df$pos
}
R is vectorized so doing the above is the same as looping through every value and multiplying by 2. I also removed the / 1 statement since it doesn't do anything. This whole section can be moved outside of the loop. Since its a constant operation of adding a column X that is double the column pos.
Next, your loop where you do a fit:
for (i in 1:100) { # 100 sample
sample(df$X,nrow(df), replace=FALSE)
fit=lm(X ~ gx, df) #linear regression calculation
regerror=sum(residuals(fit)^2)
print(paste(i,j,regerror))
set(dt,dti,1L,as.double(i))
set(dt,dti,2L,as.double(j))
set(dt,dti,3L,regerror)
dti=dti+1
}
Taking, sample(df$X, nrow(df), replace=FALSE) will only show you the new order. It doesn't actual assign them. Instead df$X <- sample(df$X, nrow(df), replace=FALSE).
Now, It looks like you're going to assign into dt (which is a function much like df and should be avoided as a variable name) at row dti the result of this fit error as well as your indicies? As far as I can tell, nothing depends on i or j. Instead, you're going to perform a randomly ordered fit 60 * 60 * 100 times... If that is what you want to do, by all means go for it! But instead do it in an efficient way:
df$X <- df$pos * 2
fit.fun <- function(n, dat) {
jumble <- sample(nrow(dat))
dat$X <- dat$X[jumble]
sum(residuals(lm(X ~ gx, dat))^2)
}
sapply(1:10, fit.fun, dat=df)