I wrote a for loop to create empty multiple data frames, using a vector of names, but even though it seemed really easy at start I got an error message : Error in ID_names[i] <- data.frame() : replacement has length zero
To be more specific I' ll provide you with a reproducable example:
ID_names <- c("Athens","Rome","Barcelona","London","Paris","Madrid")
for(i in 1:length(ID_names){
ID_names[i] <- data.frame()
}
Do you have any idea why this is wrong? I would like to ask you not only provide a solution, but specify me why this for loop is wrong in order to avoid such kind of mistakes in the future.
You are trying to store a dataframe in one element of a vector (ID_names[i]) which is not possible. You might want to create a list of empty dataframes and assign names to it which can be done using replicate.
ID_names <- c("Athens","Rome","Barcelona","London","Paris","Madrid")
list_data <- setNames(replicate(length(ID_names), data.frame()), ID_names)
However, very rarely such initialisation of empty dataframes will be useful. It ends up creating more confusion down the road. Depending on your actual use case there might be other better ways to handle this.
Related
I am writing this post to ask for some advice for looping code to rename columns by index.
I have a data set that has scale item columns positioned next to each other. Unfortunately, they are oddly named.
I want to re-name each column in this format: SimRac1, SimRac2, SimRac3.... and so on. I know the location of the columns (Columns number 30 to 37). I know these scale items are ordered in such a way that they can be named and numbered in increased order from left to right.
The code I currently have works, but is not efficient. There are other scales, in different locations, that also need to be renamed in a similar fashion. This would result in dozens of code rows.
See below code.
names(Total)[30] <- "SimRac1"
names(Total)[31] <- "SimRac2"
names(Total)[32] <- "SimRac3"
names(Total)[33] <- "SimRac4"
names(Total)[34] <- "SimRac5"
names(Total)[35] <- "SimRac6"
names(Total)[36] <- "SimRac7"
names(Total)[37] <- "SimRac8"
I want to loop this code so that I only have a chunk of code that does the work.
I was thinking perhaps a "for loop" would help.
Hence, the below code
for (i in Total[,30:37]){
names(Total)[i] <- "SimRac(1:8)"
}
This, unfortunately does not work. This chunk of code runs without error, but it doesn't do anything.
Do advice.
In the OP's code, "SimRac(1:8)" is a constant. To have dynamic names, use paste0.
We do not need a loop here. We can use a vectorized function to create the names, then assign the names to a subset of names(Total)
names(Total)[30:37]<-paste0('SimRac', 1:8)
I have a dataset of 80 variables, and I want to loop though a subset of 50 of them and construct returns. I have a list of the names of the variables for which I want to construct returns, and am attempting to use the dplyr command mutate to construct the variables in a loop. Specifically my code is:
for (i in returnvars) {
alldta <- mutate(alldta,paste("r",i,sep="") = (i - lag(i,1))/lag(i,1))}
where returnvars is my list, and alldta is my dataset. When I run this code outside the loop with just one of the `i' values, it works fine. The code for that looks like this:
alldta <- mutate(alldta,rVar = (Var- lag(Var,1))/lag(Var,1))
However, when I run it in the loop (e.g., attempting to do the previous line of code 50 times for 50 different variables), I get the following error:
Error: unexpected '=' in:
"for (i in returnvars) {
alldta <- mutate(alldta,paste("r",i,sep="") ="
I am unsure why this issue is coming up. I have looked into a number of ways to try and do this, and have attempted solutions that use lapply as well, without success.
Any help would be much appreciated! If there is an easy way to do this with one of the apply commands as well, that would be great. I did not provide a dataset because my question is not data specific, I'm simply trying to understand, as a relative R beginner, how to construct many transformed variables at once and add them to my data frame.
EDIT: As per Frank's comment, I updated the code to the following:
for (i in returnvars) {
varname <- paste("r",i,sep="")
alldta <- mutate(alldta,varname = (i - lag(i,1))/lag(i,1))}
This fixes the previous error, but I am still not referencing the variable correctly, so I get the error
Error in "Var" - lag("Var", 1) :
non-numeric argument to binary operator
Which I assume is because R sees my variable name Var as a string, rather than as a variable. How would I correctly reference the variable in my dataset alldta? I tried get(i) and alldta$get(i), both without success.
I'm also still open to (and actively curious about), more R-style ways to do this entire process, as opposed to using a loop.
Using mutate inside a loop might not be a good idea either. I am not sure if mutate makes a copy of the data frame but its generally not a good practice to grow a data frame inside a loop. Instead create a separate data frame with the output and then name the columns based on your logic.
result = do.call(rbind,lapply(returnvars,function(i) {...})
names(result) = paste("r",returnvars,sep="")
After playing around with this more, I discovered (thanks to Frank's suggestion), that the following works:
extended <- alldta # Make a copy of my dataset
for (i in returnvars) {
varname <- paste("r",i,sep="")
extended[[varname]] = (extended[[i]] - lag(extended[[i]],1))/lag(extended[[i]],1)}
This is still not very R-styled in that I am using a loop, but for a task that is only repeating about 50 times, this shouldn't be a large issue.
I am trying to get nested elements from a list. I can extract the elements using: unlist(pull_lists[[i]]$content[[n]]['sha']), however, it seems that I cannot insert them in a nested list. I have extracted a single element of the list in a gist, which creates the reproducible example below. Here is what I have so far:
library("devtools")
pull_lists <- list(source_gist("669dfeccad88cd4348f7"))
sha_list <- list()
for (i in length(pull_lists)){
for (n in length(pull_lists[[i]]$content)){
sha_list[i][n] <- unlist(pull_lists[[i]]$content[[n]]['sha'])
}
}
How can I insert the elements in a nested fashion?
When I download the content, I get a much more complicated structure than you do. For me, it's not pull_lists[[i]]$content, it's pull_lists[[i]]$value$content[[1 or 2]]$parents$sha. The reason nothing is populating is because there is nothing there to populate (ie, n = 0).
I've had to deal with similar data structures before. What I found was that it's much easier to search the naming structure after unlisting rather than to figure out the correct sequence of subsets.
Here's an example:
sha_locations <- grep("sha$",names(unlist(pull_list[[1]])))
unlist(pull_list[[1]])[sha_locations]
Cleaning the for loop a bit, this would look like:
sha_list <- lapply(
pull_list,
function(x) unlist(x)[grep("sha$",names(unlist(x)))]
)
Since there are multiple SHAs, and the question only asks for the SHAs at specific positions, you need to extract those SHAs:
sha_list <- sha_list[[1]][attr(sha_list[[1]], "names")=="value.content.sha"]
I have a data frame loaded using the CSV Library in R, like
mySheet <- read.csv("Table.csv", sep=";")
I now can print a summary on that mySheet object
summary(mySheet)
and it will show me a summary for each column, for example, one column named Diagnose has the unique values RCM, UCM, HCM and it shows the number of occurences of each of these values.
I now filter by a diagnose, like
subSheet <- mySheet[mySheet$Diagnose=='UCM',]
which seems to be working, when I just type subSheet in the console it will print only the rows where the value has been matched with 'UCM'
However, if I do a summary on that subSheet, like
summary(subSheet)
it still 'knows' about the other two possibilities RCM and HCM and prints those having a value of 0. However, I expected that the new created object will NOT know about the possible values of the original mySheet I initially loaded.
Is there any way to get rid of those other possible values after filtering? I also tried subset but this one just seems to be some kind of shortcut to '[' for the interactive mode... I also tried DROP=TRUE as option, but this one didn't change the game.
Totally mind squeezing :D Any help is highly appreciated!
What you are dealing with here are factors from reading the csv file. You can get subSheet to forget the missing factors with
subSheet$Diagnose <- droplevels(subSheet$Diagnose)
or
subSheet$Diagnose <- subSheet$Diagnose[ , drop=TRUE]
just before you do summary(subSheet).
Personally I dislike factors, as they cause me too many problems, and I only convert strings to factors when I really need to. So I would have started with something like
mySheet <- read.csv("Table.csv", sep=";", stringsAsFactors=FALSE)
I am pretty new to R and have a couple of questions about a loop I am attemping to execute. I will try explain myself as best as possible reguarding what I wish the loop to do.
for(i in (1988:1999,2000:2006)){
yearerrors=NULL
binding=do.call("rbind.fill",x[grep(names(x), pattern ="1988.* 4._ data=")])
cmeans=lapply(binding[,2:ncol(binding)],mean)
datcmeans=as.data.frame(cmeans)
finvec=datcmeans[1,]
kk=0
result=RMSE2(yields[(kk+1):(kk+ncol(binding))],finvec)
kk=kk+ncol(binding)
yearerrors=c(result)
}
yearerrors
First I wish for the loop to iterate over file names of data.
Specifically over the years 1988-2006 in the place where 1988 is
placed right now in the binding statement. x is a list of data files
inputted into R and the 1988 is part of the file name. So, I have
file names starting with 1988,1989,...,2006.
yields is a numeric vector and I would like to input the indices of
the vector into the function RMSE2 as indicated in the loop. For
example, over the first iteration I wish for the indices 1 to the
number of columns in binding to be used. Then for the next iteration
I want the first index to be 1 more than what the previous iteration
ended with and continue to a number equal to the number of columns in the next binding
statement. I just don't know if what I have written will accomplish
this.
Finally, I wish to store each of these results in the vector
yearerrors and then access this vector afterwards.
Thanks so much in advance!
OK, there's a heck of a lot of guesswork here because the structure of your data is extremely unclear, I have no idea what the RMSE2 function is (and you've given no detail). Based on your question the other day, I'm going to assume that your data is in .csv files. I'm going to have a stab at your problem.
I would start by building the combined dataframe while reading the files in, not doing one then the other. Like so:
#Set your working directory to the folder containing the .csv files
#I'm assuming they're all in the form "YEAR.something.csv" based on your pattern matching
filenames <- list.files(".", pattern="*.csv") #if you only want to match a specific year then add it to the pattern match
years <- gsub("([0-9]+).*", "\\1", filenames)
df <- mdply(filenames, read.csv)
df$year <- as.numeric(years[df$X1]) #Adds the year
#Your column mean dataframe didn't work for me
cmeans <- as.data.frame(t(colMeans(df[,2:ncol(df)])))
It then gets difficult to know what you're trying to achieve. Since your datcmeans is a one row data.frame, datcmeans[1,] doesn't change anything. So if a one row from a dataframe (or a numeric vector) is an argument required for your RMSE2 function, you can just pass it datcmeans (cmeans in my example).
Your code from then is pretty much indecipherable to me. Without know what yields looks like, or how RMSE2 works, it's pretty much impossible to help more.
If you're going to do a loop here, I'll say that setting kk=kk+ncol(binding) at the end of the first iteration is not going to help you, since you've set kk=0, kk is not going to be equal to ncol(binding), which is, I'm guessing, not what you want. Here's my guess at what you need here (assuming looping is required).
yearerrors=vector("numeric", ncol(df)) #Create empty vector ahead of loop
for(i in 1:ncol(df)) {
yearerrors[i] <- RMSE2(yields[i:ncol(df)], finvec)
}
yearerrors
I honestly can't imagine a function that would work like this, but it seems the most logical adaption of your code.