I need to create 52 resources with capacity 2 in the Simmer simulation package. I am trying to do this by using a while loop that creates these resources for me, instead of creating each resource myself.
The idea is that I have a while loop as given below. In each loop, a resource should be created called Transport_vehicle1, Transport_vehicle2, ..., Transport_vehicle52, with capacity 2.
Now I do not know how to insert the number i in the name of the resource that I am trying to create
i<-1
while (i<=52)
{ env %>%
add_resource("Transport_vehicle"[i],capacity = 2)
i <- i+1
}
Could someone please help me out? Thanks!
You can use the paste method to concatenate the string and the number:
i<-1
while (i<=52)
{ env %>%
add_resource(paste("Transport_vehicle", i),capacity = 2)
i <- i+1
}
If you do not want a space between the string and the number add the sep="" argument
paste("Transport_vehicle", i, sep="")
or use
paste0("Transport_vehicle", i)
Related
i tried updating data in dataframe but its unable to get updating
//Initialize data and dataframe here
user_data=read.csv("train_5.csv")
baskets.df=data.frame(Sequence=character(),
Challenge=character(),
countno=integer(),
stringsAsFactors=FALSE)
/Updating data in dataframe here
for(i in 1:length((user_data)))
{
for(j in i:length(user_data))
{
if(user_data$challenge_sequence[i]==user_data$challenge_sequence[j]&&user_data$challenge[i]==user_data$challenge[j])
{
writedata(user_data$challenge_sequence[i],user_data$challenge[i])
}
}
}
writedata=function( seqnn,challng)
{
#print(seqnn)
#print(challng)
newRow <- data.frame(Sequence=seqnn,Challenge=challng,countno=1)
baskets.df=rbind(baskets.df,newRow)
}
//view data here
View(baskets.df)
I've modified your code to what I believe will work. You haven't provided sample data, so I can't verify that it works the way you want. I'm basing my attempt here on a couple of common novice mistakes that I'll do my best to explain.
Your writedata function was written to be a little loose with it's scope. When you create a new function, what happens in the function technically happens in its own environment. That is, it tries to look for things defined within the function, and then any new objects it creates are created only within that environment. R also has this neat (and sometimes tricky) feature where, if it can't find an object in an environment, it will try to look up to the parent environment.
The impact this has on your writedata function is that when R looks for baskets.df in the function and can't find it, R then turns to the Global Environment, finds baskets.df there, and then uses it in rbind. However, the result of rbind gets saved to a baskets.df in the function environment, and does not update the object of the same name in the global environment.
To address this, I added an argument to writedata that is simply named data. We can then use this argument to pass a data frame to the function's environment and do everything locally. By not making any assignment at the end, we implicitly tell the function to return it's result.
Then, in your loop, instead of simply calling writedata, we assign it's result back to baskets.df to replace the previous result.
for(i in 1:length((user_data)))
{
for(j in i:length(user_data))
{
if(user_data$challenge_sequence[i] == user_data$challenge_sequence[j] &&
user_data$challenge[i] == user_data$challenge[j])
{
baskets.df <- writedata(baskets.df,
user_data$challenge_sequence[i],
user_data$challenge[i])
}
}
}
writedata=function(data, seqnn,challng)
{
#print(seqnn)
#print(challng)
newRow <- data.frame(Sequence = seqnn,
Challenge = challng,
countno = 1)
rbind(data, newRow)
}
I'm not sure what you're programming background is, but your loops will be very slow in R because it's an interpreted language. To get around this, many functions are vectorized (which simply means that you give them more than one data point, and they do the looping inside compiled code where the loops are fast).
With that in mind, here's what I believe will be a much faster implementation of your code
user_data=read.csv("train_5.csv")
# challenge_indices will be a matrix with TRUE at every place "challenge" and "challenge_sequence" is the same
challenge_indices <- outer(user_data$challenge_sequence, user_data$challenge_sequence, "==") &
outer(user_data$challenge, user_data$challenge, "==")
# since you don't want duplicates, get rid of them
challenge_indices[upper.tri(challenge_indices, diag = TRUE)] <- FALSE
# now let's get the indices of interest
index_list <- which(challenge_indices,arr.ind = TRUE)
# now we make the resulting data set all at once
# this is much faster, because it does not require copying the data frame many times - which would be required if you created a new row every time.
baskets.df <- with(user_data, data.frame(
Sequence = challenge_sequence[index_list[,"row"]],
challenge = challenge[index_list[,"row"]]
)
You can suggest me any sort of answers, not necessarily need using conditional statements and Loops.
I have the data set with several ids and and three alert or groups.
here is the image for the concept:
and here is the actual Data set for one ID: Click me
Concept is:
I have three alert: Relearn - Rebuild - Replace.
and after relearn: rebuild or replace can come but relearn cannot come
and after rebuild: replace can come but relearn cannot come
after replace: relearn and rebuild cannot come. if there is any replace only that can come
I have attached the image and Dataset for better clear understanding and Here is my try:
temp1 = NULL
temp2 = NULL
sql50 = NULL
for(i in 1:nrow(BrokenPins)) #First Loop
{
sql50 = sqldf(paste("select * from rule_data where key = '",BrokenPins[i,1],"'",sep = ""))
for(j in 1:nrow(sql50))
{ #Second Loop
while (head(sql50$Flag,1) == sql50$Flag[j] )
{
temp1 = sql50[j,]
temp2 = rbind(temp2,temp1)
print("Send")
if(j == 1 || sql50$Flag[j] == sql50$Flag[j-1])
{
j = j+1
}
else(sql50$Flag[j] > sql50$Flag[j-1])
{
break
}
}
}
}
First loop will go through each id and second loop will give me all the alert for that id.
so in the image i have added send and dont send. it wont be in actual table. that basically means send means copy it to new dataframe like i am doing above rbind in the code and dont send means dont copy it. This Above piece of code will run but only take the first and end it.
Finally, Based On above Data set Click me: that is for one ID (key), Flag (1 - Relearn, 2-Rebuild,3-replace). so based on this dataset. my output should have Row 1, 2 and 7 because First Relearn[Flag 1] came then Rebuild[Flag 2] then again relearn[Flag 1]cannot come, only rebuild [Flag 2] and replace[Flag 2] can.
can you help me solve this concept?
One thing I notice is that you use else and also provided a condition; you should only use else when you want every case that is not included in the condition of your if statement. Basically, instead of using else(sql50$Flag[j] > sql50$Flag[j-1]) you should be using else if(sql50$Flag[j] > sql50$Flag[j-1]).
Been going around for hours with this. My 1st question online on R. Trying to creat a function that contains a loop. The function takes a vector that the user submits like in pollutantmean(4:6) and then it loads a bunch of csv files (in the directory mentioned) and binds them. What is strange (to me) is that if I assign the variable id and then run the loop without using a function, it works! When I put it inside a function so that the user can supply the id vector then it does nothing. Can someone help ? thank you!!!
pollutantmean<-function(id=1:332)
{
#read files
allfiles<-data.frame()
id<-str_pad(id,3,pad = "0")
direct<-"/Users/ped/Documents/LearningR/"
for (i in id) {
path<-paste(direct,"/",i,".csv",sep="")
file<-read.csv(path)
allfiles<-rbind(allfiles,file)
}
}
Your function is missing a return value. (#Roland)
pollutantmean<-function(id=1:332) {
#read files
allfiles<-data.frame()
id<-str_pad(id,3,pad = "0")
direct<-"/Users/ped/Documents/LearningR/"
for (i in id) {
path<-paste(direct,"/",i,".csv",sep="")
file<-read.csv(path)
allfiles<-rbind(allfiles,file)
}
return(allfiles)
}
Edit:
Your mistake was that you did not specify in your function what you want to get out from the function. In R, you create objects inside of function (you could imagine it as different environment) and then specify which object you want it to return.
With my comment about accepting my answer, I meant this: (...To mark an answer as accepted, click on the check mark beside the answer to toggle it from greyed out to filled in...).
Consider even an lapply and do.call which would not need return being last line of function:
pollutantmean <- function(id=1:332) {
id <- str_pad(id,3,pad = "0")
direct_files <- paste0("/Users/ped/Documents/LearningR/", id, ".csv")
# READ FILES INTO LIST AND ROW BIND
allfiles <- do.call(rbind, lapply(direct_files, read.csv))
}
ok, I got it. I was expecting the files that are built to be actually created and show up in the environment of R. But for some reason they don't. But R still does all the calculations. Thanks lot for the replies!!!!
pollutantmean<-function(directory,pollutant,id)
{
#read files
allfiles<-data.frame()
id2<-str_pad(id,3,pad = "0")
direct<-paste("/Users/pedroalbuquerque/Documents/Learning R/",directory,sep="")
for (i in id2) {
path<-paste(direct,"/",i,".csv",sep="")
file<-read.csv(path)
allfiles<-rbind(allfiles,file)
}
#averaging polutants
mean(allfiles[,pollutant],na.rm = TRUE)
}
pollutantmean("specdata","nitrate",23:35)
I got warnings when running this code.
For example, when I put
tm1<- summary(tmfit)[c(4,8,9)],
I can get the result, but I need to run this code for each $i$.
Why do I get this error?
Is there any way to do this instead of via a for loop?
Specifically, I have many regressants ($y$) with the same two regressors ($x$'s).
How I can get these results of regression analysis(to make some comparisons)?
dreg=read.csv("dayreg.csv")
fundr=read.csv("fundreturnday.csv")
num=ncol(fundr)
exr=dreg[,2]
tm=dreg[,4]
for(i in 2:num)
{
tmfit=lm(fundr[,i]~exr+tm)
tm1[i]<- summary(tmfit)[c(4,8,9)]
}
Any help is highly appreciated
Try storing your result into a list instead of a vector.
dreg=read.csv("dayreg.csv")
fundr=read.csv("fundreturnday.csv")
num=ncol(fundr)
exr=dreg[,2]
tm = list()
for(i in 2:num)
{
tmfit=lm(fundr[,i]~exr+tm)
tm1[[i]]<- summary(tmfit)[c(4,8,9)]
}
You can look at an element in the list like so
tm1[[2]]
I am having a very hard time trying to subtract two vectors in my OpenBUGS model. The last line of the code below keeps giving the error "expected right parenthesis error":
model {
for ( i in 1:N) {
for(j in 1:q) {
vv[i,j] ~ dnorm(vz[i,j],tau.eta[j])
}
vz[i,1:q] ~ dmnorm(media.z[i,], K.delta[,])
for(j in 1:q) {
mean.z[i,j] <- inprod(K[i,] , vbeta[j,])
}
K[i,1] <- 1.0
for(j in 1:N) {
K[i,j+1] <- sum(ve[,i] - ve[,j])
}
}
If I change that line to K[i,j+1] <- sum(ve[,i]) - sum(ve[,j]), then the model works fine, but that is not what I want to do. I would like to subtract element-wise.
I searched SO for OpenBUGS, but there are only a few unrelated topics:
OpenBUGS - Variable is not defined
OpenBUGS: missing value in Bernoulli distribution
In Stats Stack Exchange there is this post which is close, but I still could not make how to implement this in my model:
https://stats.stackexchange.com/questions/20653/vector-multiplication-in-bugs-and-jags/20739#20739
I understand I have to write a for loop, but this thing is sure giving me a big headache. :)
I tried changing that line to:
for(k in 1:p) { temp [k] <- ve[k,i] - ve[k,j] }
K[i,j+1] <- sum(temp[])
where 'p' is the number of rows in each 've'. Now I keep getting the error "multiple definitions of node temp[1]".
I could definitely use some help. It will be much appreciated.
Best regards to all and thanks in advance!
PS: I wanted to add the tag "OpenBUGS" to this question but unfortunately I couldn't because it would be a new tag and I do not have enough reputation. I added "winbugs" instead.
The "multiple definitions" error is because temp[k] is redefined over and over again within a loop over i and another loop over j - you can only define it once. To get around that, use i and j subscripts like
for(k in 1:p) { temp[k,i,j] <- ve[k,i] - ve[k,j] }
K[i,j+1] <- sum(temp[,i,j])
Though if that compiles and runs, I'd check the results to make sure that's exactly what you want mathematically.