I have 50 files to read in R and I created this loop to help me. I would like to know if it is possible to do something like this in R.
How can I write it properly in R?
library(foreign)
for(i in 1:50 ){
tpi <- read.dbf('toto_%i%')
}
Help please.
We can do this using lapply
lst <- lapply(1:50, function(i) read.dbf(paste0("toto_", i)))
You want to use the function paste. As written your loop will overwrite tpi everytime it increments, so you will want to use a list to store the data.
toto = list()
for(i in 1:50)
{
toto[i] = read.dbf(paste0("toto_", i))
}
A shortcut using lapply gets the same results:
toto = lapply(1:50, function(x) read.dbf(paste0("toto_", x)))
Related
This code chunk creates a 10 objects based of length of alpha.
alpha <- seq(.1,1,by=.1)
for (i in 1:length(alpha)){
assign(paste0("list_ts_ses_tune", i),NULL)
}
How do I put each function into the new list_ts_ses_tune1 ... null objects I've created? Each function puts in a list, and works if I set list_ts_ses_tune1 <- lapply ...
for (i in 1:length(alpha))
{
list_ts_ses_tune[i] <- lapply(list_ts, function(x)
forecast::forecast(ses(x,h=24,alpha=alpha[i])))
list_ts_ses_tune[i] <- lapply(list_ts_ses_tune[i], "[", c("mean"))
}
Maybe this is a better way to do this? I need each individual output in a list of values.
Edit:
for (i in 1:length(alpha))
{
list_ts_ses_tune[[i]] <- lapply(list_ts[1:(length(list_ts)/2)],
function(x)
forecast::forecast(ses(x,h=24,alpha=alpha[i])))
list_ts_ses_tune[[i]] <- lapply(list_ts_ses_tune[[i]], "[", c("mean"))
}
We can use mget to return all the objects into a list
mget(ls(pattern = '^list_ts_ses_tune\\d+'))
Also, the NULL list can be created more easily instead of 10 objects in the global environment
list_ts_ses_tune <- vector('list', length(alpha))
Now, we can just use the OP's code
for (i in 1:length(alpha))
{
list_ts_ses_tune[[i]] <- lapply(list_ts, function(x)
forecast::forecast(ses(x,h=24,alpha=alpha[i])))
}
If we want to create a single data.frame
for(i in seq_along(alpha)) {
list_ts_ses_tune[[i]] <- data.frame(Mean = do.call(rbind, lapply(list_ts, function(x)
forecast::forecast(ses(x,h=24,alpha=alpha[i]))$mean)))
}
You could simply accomplish everything by doing:
library(forecast)
list_ts_ses_tune <- Map(function(x)
lapply(alpha, function(y)forecast(ses(x,h=24,alpha=y))['mean']), list_ts)
I have problems storing user defined functions in R list when they are put on it in a for loop.
I have to define some segment-specific functions based on some parameters, so I create functions and put them on a list looping through segments with for-loop. The problem is I get same function everywhere on a result list.
The code looks like this:
n <- 100
segmenty <- 1:n
segment_functions <- list()
for (i in segmenty){
segment_functions[[i]] <- function(){return(i)}
}
When i run the code what I get is the same function (last created in the loop) for all indexes:
## for all k
segment_functions[[k]]()
[1] 100
There is no problem when I put the functions on list manually e.g.
segment_functions[[1]] <- function(){return(1)}
segment_functions[[2]] <- function(){return(2)}
segment_functions[[3]] <- function(){return(3)}
works just fine.
I honsetly have no idea what's wrong. Could you help?
You need to use the force function to ensure that the evaluation of i is done during the assignment into the list:
n <- 100
segmenty <- 1:n
segment_functions <- list()
f <- function(i) { force(i); function() return(i) }
for (i in segmenty){
segment_functions[[i]] <- f(i)
}
I'd use lapply and capture i in a clousre of the wrapper:
segment_functions <- lapply(1:100, function(i) function() i)
Please tell me how can I save my loop results into the table(data.frame) or even .csv file. I can't handle with that issue by myself.
for (i in 1:10000){
x<-pois(1,40)
sum<-round(digits = 2, sum(rlnorm(x, log(10), log(5))))
}
If you want to use a for loop, create an empty vector and then iterate through each position.
mySums <- numeric(10000)
for (i in 1:10000){
x <- rpois(1,40)
mySums[i] <- round(digits = 2, sum(rlnorm(x, log(10), log(5))))
}
It is straightforward to then turn this into a dataframe or any other format you require.
Edit: This is assuming you meant to use rpois() or something similar.
Assuming that you want to call rpois() (and not pois()):
mySums <- replicate(10000, round(digits=2, sum(rlnorm(rpois(1,40), log(10), log(5)))))
yes indeed, I meant rpois function. I solved it using the next formula:
Results=matrix(0,10000,1)
for ( i in 1:10000)
{
x<-rpois(1,40)
Results[i,1]<-round(digits = 2, sum(rlnorm(x,log(20), log(4))))
}
i've following problem:
I use the for-loop within R to get specific data from a matrix.
my code is as follows.
for(i in 1:100){
T <- as.Date(as.mondate (STARTLISTING)+i)
DELIST <- (subset(datensatz_Start_End.frame, TIME <= T))[,1]
write.table(DELIST, file = paste("tab", i, ".csv"), sep="," )
print(DELIST)
}
Using print, R delivers the data.
Using write.table, R delivers the data into different files.
My aim is to aggregate the results from the for-loop within one matrix. (each row for 'i')
But unfortunately I can not make it.
sorry, i'm a real noob within R.
for(i in 1:100)
{
T <- as.Date(as.mondate (STARTLISTING)+i)
DELIST <- (subset(datensatz_Start_End.frame, TIME <= T))[,1]
assign(paste('b',i,sep=''),DELIST)
}
this delivers 100 objects, which contain my results.
But what i need is one matrix/dataframe with 100 columns or one list.
Any ideas?
Hey!
Hence I'm not allowed to edit my own answers, here my (simple) solution as follows:
DELIST <- vector("list",100)
for(i in 1:100)
{
T <- as.Date(as.mondate (STARTLISTING)+i)
DELIST[[i]] <- as.character((subset(datensatz_Start_End.frame, TIME <= T))[,1])
}
DELIST[[99]] ## it is possible to requist the relevant companies for every 'i'
Thx to everyone!
George
If you want a list you can use lapply instead of loop
LL <- lapply(1:100,
function(i) {
T <- as.Date(as.mondate (STARTLISTING)+i)
DELIST <- (subset(datensatz_Start_End.frame, TIME <= T))[,1]
assign(paste('b',i,sep=''),DELIST)
}
)
After that you can rbind results together using do.call
result <- do.call(rbind, LL)
Or if you are confident that columns of all elements of LL are going to be of same, then you can use more efficient rbindlist from package data.table
result <- rbindlist(LL)
check out rbind function. You can start with empty DELIST.DF and append each row to it inside the loop -
DELIST.DF <- NULL
for(i in 1:100){
T <- as.Date(as.mondate (STARTLISTING)+i)
DELIST <- (subset(datensatz_Start_End.frame, TIME <= T))[,1]
DELIST.DF <- rbind(DELIST.DF, DELIST)
write.table(DELIST, file = paste("tab", i, ".csv"), sep="," )
print(DELIST)
}
I'm breaking my head on how to write multiple files from each row of the input matrix, after some calculations. The code that I'm using now looks like this:
akl <- function(dii) {
ddi <- as.matrix(dii)
m <- rowMeans(ddi)
M <- mean(m) # mean(ddi) == mean(m)
r <- sweep(ddi, 1, m)
b <- sweep(r, 2, m)
return(b + M)
}
require(plyr)
akl.list <- llply(1:nrow(aa), function(i) {
akl(dist(aa[i, ]))
})
The akl.list that I create is too large for large input matrix and I cannot store it in the RAM. My idea was to write on files each matrix that I obtain in the llply loop. Is there an easy way to do that?
thank you!!
gibbi
you can use do_ply since you want just the loop feature
d_ply(aa, 1,function(row){
a <- akl(dist(row))
write.table(a) ## you save in a file here
},.progress='text' ## to show progress (optional)
)