I want to repeat a column vector that have 300rows about 241times and to concatonate it. The data is downloadable in this link.
https://1drv.ms/u/s!AiZLoqatH-p7rD0og-RufSi6fljB
I tried the following code.
read.csv("stack_overflow.csv")
fund_name = d[,1]
fund_name_panel=c()
for (i in 1:300{x1=rep(fund_name[i], 241) fund_name_Panel=append(x1,fund_name_panel)}
Result: unfortunately, My code repeats only the very last row of the data. How can i repeat each of the 300rows rather than the very last?
Any hint is appreciated.
From your description of the problem you are committing a very simple error a lot of people make when first learning for loops. First since you are making a new variable (fund_name_panel) you need to create an empty vector the length of the vector you will use in the for loop.
fund_name_panel <- numeric(length(fund_name))
Use nrow() instead of length() if fund_name is a data.frame and not a vector.
Secondly, you will need to specify the row (i) in both the now new vector (fund_name_panel) and the vector of you are referencing in the for loop (fund_name) see code below.
fund_name_panel <- numeric(length(fund_name))
for(i in 1:length(fund_name)){
x[i]=y[i]
}
Related
I am currently looking for a way to simplify searching through a column within a dataframe for a vector of values and replacing each of of those values with another value (also contained within a separate vector). I can run a for loop for this, but it must be possible within the apply family, I'm just not seeing it yet. Very new to using the apply family and could use help.
So far, I've been able to have it replace all instances of the first value in my vector with the new first value in the new vector, it just isn't iterating past the first level. I hope this makes sense. Here is the code I have:
#standardize tank location
old_tank_list <- c("7.C.4","7.C.5","7.C.6","7.C.7","7.C.8","7.C.9","7.C.10","7.C.11")
new_tank_list <- c("7.B.3-4","7.C.3-4","7.C.1-2","7.C.5-6","7.C.7-8","7.C.9-10","7.E.9-10","7.C.11-12")
sapply(df_growth$Tank,function(y) gsub(old_tank_list,std_tank_list,y))
Tank is the name of the column I am trying to replace all of these values within. I haven't assigned it back yet, because I want to test the functionality first. Thanks for any help you can offer.
Hopefully, this image will help. The photo on the left is the column before my function is applied. The column on the right is after. Basically, I just want to batch change text values.
Before and After
library(dplyr)
df %>%
mutate(Tank = recode(Tank, !!!setNames(new_tank_list, old_tank_list)))
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 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 have 75 matrices that I want to search through. The matrices are named a1r1, a1r2, a1r3, a1r4, a1r5, a2r1,...a15r5, and I have a list with all 75 of those names in it; each matrix has the same number of rows and columns. Inside some nested for loops, I also have a line of code that, for the first matrix looks like this:
total <- (a1r1[row,i]) + (a1r1[row,j]) + (a1r1[row,k])
(i, j, k, and row are all variables that I am looping over.) I would like to automate this line so that the for loops would fully execute using the first matrix in the list, then fully execute using the second matrix and so on. How can I do this?
(I'm an experienced programmer, but new to R, so I'm willing to be told I shouldn't use a list of the matrix names, etc. I realize too that there's probably a better way in R than for loops, but I was hoping for sort of quick and dirty at my current level of R expertise.)
Thanks in advance for the help.
Here The R way to do this :
lapply(ls(pattern='a[0-9]r[0-9]'),
function(nn) {
x <- get(nn)
sum(x[row,c(i,j,k)])
})
ls will give a list of variable having a certain pattern name
You loop through the resulted list using lapply
get will transform the name to a varaible
use multi indexing with the vectorized sum function
It's not bad practice to build automatically lists of names designating your objects. You can build such lists with paste, rep, and sequences as 0:10, etc. Once you have a list of object names (let's call it mylist), the get function applied on it gives the objects themselves.
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.