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)
Related
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.
I have a list of dataframes (subspec2) which I want to loop through to get the columns with the maximum value from each dataframe, and write these to a new dataframe. I wrote the following loop:
good.data<-data.frame(matrix(nrow=401, ncol=78)) #create empty dataframe
for (i in length(subspec2)) ##subspec2 is the list of dataframes
{
max.name<-names(which.max(apply(subspec2[[i]],MARGIN=2,max))) #find column name with max value
good.data[,i]<-subspec2[[i]][max.name] #write the contents of this column into dataframe
}
This seems to work but only returns values in the last column, nothing else appears to have been saved. Many threads point out the df must be outside the loop, but that is not the problem here.
What am I doing wrong?
Thank you!
I believe you need to change for (i in length(subspec2)) to for (i in 1:length(subspec2)). The former will only do 1 iteration, where i = length(subspec2) whereas the latter iterates over multiple is.
(I am pretty sure that is your issue, but one thing that is great to do is to create a reproducible example so I can run your code to double check, for example I am not exactly sure what subspec2 looks like, and I am not able to run your code as it is, a great resource for this is the reprex package).
I am trying to run a loop to change the names of two columns in my data but the name of these two columns start with a number. For the same work I changed them (they were not starting with a number) by writing as shown below but it is not working.
Here is the code (the loop finishes later) :
#Filtering
for(i in 1:length(names$ID)){
f<-names$ID[[i]]
corrpoints<-sprintf("corrpoints%i",as.numeric(levels(f))[f])
pts=readOGR(dsn="C:/Users/Charlie/Desktop/Stage_permafrost/SIG/Quantification_des_mouvements/Corr_points_disp/Corr_points_ubaye", layer=corrpoints)
pts$Gvalue2004<-pts$2004_red_gr
pts$Gvalue2012<-pts$2012_red_gr
pts$Aspect_mnt<-pts$Aspect_25m_
Any idea on how I could fix this?
Thanks
Your example is not reproducible for us.
Column names CAN start with numbers. Use ticks around the name to access it as below. Whether this is a good idea is an entirely different story.
x <- mtcars
colnames(x)[1] <- '1mpg'
x$`1mpg`
I know that the answer to "how to change names in a list of data frames" has been answered multiple times. However, I'm stuck trying to generate a function that can take any list as an argument and change all of the column names of all of the data frames in the list. I am working with a large number of .csv files, all of which will have the same 3 column names. I'm importing the files in groups as follows:
# Get a group of drying data data files, remove 1st column
files <- list.files('Mang_Run1', pattern = '*.csv', full = TRUE)
mr1 <- lapply(files, read.csv, skip = 1, header = TRUE, colClasses = c("NULL", NA, NA, NA))
I will have 6 such file groups. If I run the following code on a single list, the names of the columns in each data frame within the specified list will be changed correctly.
for (i in seq_along(mr1)) {
names(mr1[[i]]) <- c('Date_Time', 'Temp_F', 'RH')
}
However, if I try to generalize the function (see code below) to take any list as an argument, it does not work correctly.
nameChange <- function(ls) {
for (i in seq_along(ls)) {
names(ls[[i]]) <- c('Date_Time', 'Temp_F', 'RH')
}
return(ls)
}
When I call nameChange on mr1 (list generated from above), it prints the entire contents of the list to the console and does not change the names of the columns in the data frames within the list. I'm clearly missing something fundamental about the inner workings of R here. I've tried the above function with and without return, and have made several modifications to the code, none of which have proven successful. I'd greatly appreciate any help, and would really like to understand the 'why' behind the problem as well. I've had considerable trouble in the past handling functions that take lists as arguments.
Thanks very much in advance for any constructive input.
I think this might be a very simple fix:
First, generalize the function you are using to rename the columns. This only needs to work on one dataframe at a time.
renameFunction<-function(x,someNames){
names(x) <- someNames
return(x)
}
Now we need to define the names we want to change each column name to.
someNames <- c('Date_Time', 'Temp_F', 'RH')
Then we call the new function and apply it to every element of the "mr1" list.
lapply(mr1, renameFunction, someNames)
I may have gotten some of the details wrong with regards to your exact sitiuation, but I've used this method before to solve similar issues. Since you were able to get it to work on the specific case, I'm pretty sure this will generalize readily using lapply
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.