More 'feels like it should be' simple stuff which seems to be eluding me today. Thanks in advance for assistance.
Within a loop, that's within a function, I'm trying to add a column, and name it based on a formula.
I can bind a column & its name is taken from the bound object: data<-cbind(data,bothdata)
I can bind a column & manually name the bound object: data<-cbind(data,newname=bothdata)
I can bind a column which is the product of an equation & manually name the bound object: data<-cbind(data,newname2=bothdata-1)
Or another way: data <- transform(data, newColumn = bothdata-1)
What I can't do is have the name be the product of a formula. My actual formula-derived example name is paste("E_wgt",rev(which(rev(Esteps) == q))-1,"%") & equation for column: baddata - q.
A simpler one: data<-cbind(data,paste("magic",100,"beans")=bothdata-1). This fails because cbind isn't expecting the = even though it's fine in previous examples. Same fail for transform.
My first thought was assign but while I've used this successfully for creating forumla-named objects, I can't see how to get it to work for formula-named columns.
If I use an intermediary step to put the naming formula in an object container then use that, e.g.:
name <- paste("magic",100,"beans")
data<-cbind(data,name=bothdata-1)
the column name is "name" not "magic100beans". If I assign the equation result to an formula-named object:
assign(paste("magic",100,"beans"),bothdata-1)
Then try to cbind that via get:
data<-cbind(data,get(paste("magic",100,"beans")))
The column is called "get(paste("magic",100,"beans"))". Boo! Any thoughts anyone? It occurs to me that I can do cbind then separately colnames(data)[ncol(data)] <- paste("magic",100,"beans")) which I guess I'll settle for for now, but would still be interested to find if there was a direct way.
Thanks.
Chances are that cbind is overkill for your use case. In almost every instance, you can simply mutate the underlying data frame using data$newname2 <- data$bothdata - 1.
In the case where the name of the column is dynamic, you can just refer to it using the [[ operator -- data[["newcol"]] <- data$newname + 1. See ?'[' and ?'[.data.frame' for other tips and usages.
EDIT: Incorporated #Marek's suggestion for [["newcol"]] instead of [, "newcol"]
It may help you to know that data$col1 is the same than data[,"col1"] which is the same than data[,x] if x is "col1". This is how I usually access/set columns programmatically.
So this should work:
name <- paste("magic",100,"beans")
data[,name] <- obsdata-1
Note that you don't have to use the temporary variable name. This is equivalent to:
data$magic100beans <- obsdata-1
Itself equivalent, for a data.frame, to:
data<-cbind(data, magic100beans=bothdata-1)
Just so you know, you could also set the names afterwards:
old_names <- names(data)
name <- paste("magic",100,"beans")
data <- cbind(data, bothdata-1)
data <- setNames(data, c(old_names, name))
# or
names(data) <- c(old_names, name)
Related
I am learning R, and I am trying to understand the indexing properties. I cannot seem to understand why the following code to change a column name does not work:
state.all <- as.data.frame(state.x77)
head(state.all)
state.all$States <- rownames(state.all)
rownames(state.all) <- NULL
# why the following row does not work?
names(state.all["States"]) <- "Test"
colnames(state.all)
While this works:
state.all <- as.data.frame(state.x77)
head(state.all)
state.all$States <- rownames(state.all)
rownames(state.all) <- NULL
# This work
names(state.all)[which(colnames(state.all)=="States")] <- "Test"
colnames(state.all)
Shouldn't the function be able to overwrite the name of the column also in the first example? Is it something to do with the local vs. global environment?
Thanks in advance!
What you're trying to do is replacing the name of column number 9.
the expression which(colnames(state.all)=="States") results in the index if the column named "States" (if there is any) and then takes this index and replaces the value in the names vector.
the expression state.all["States"] just returns the values of this column so of course nothing will happen.
I suggest something like colnames(state.all)[which(colnames(state.all)=="States")] <- "Test".
I would like to know how to make a reference to a data frame and variable generic, please. Say I have a data frame named 's' and a variable in that data frame named 'Y'.
Regular R code:
look = s$Y
What I would like to do:
data = s
variable = Y
look = data$variable (which functions the same as look = s$Y)
Any thoughts? The reason I would like to do this is that I have s$Y throughout my code, and later I may want to change s for t (or Y for some other variable), and don't want to have to go through all of my code manually replacing s$Y with t$Y where I need it changed.
Thanks!
This is the reason that the $-operator is considered poor-practice inside function definitions, i.e. it "locks you in" to a particular spelling of a column name. You are not going to do this, however:
variable = Y
Rather you are going to do this:
variable = "Y"
And that is because the first version would have caused the R-interpreter to go out and try to identify a value for the symbol Y someplace in what is known as its "search path" which is roughly speaking all that functions and values that have been called and are still being processed since code was started. In the case of the second version "Y" is its own value and no further searching is needed. With that fundamental confusion corrected you would now do this
look <- data[[ variable ]] # although using 'data' as a name is another "poor-practice"
Whereupon R will look for a value of variable and find it in the global environment, returning the character "Y" and delivering a column named "Y" from the dataset s. Column names are not considered first-class objects in R, whereas named dataframes are. The "names" of columns are not true R names (even though they are called colnames).. The $-operator is just shorthand for "[[" with a character value. Here's a full transcript to test this:
> s <- data.frame(Y=1:10, X=LETTERS[1:10]); data = s
>
> variable <- "Y"
>
> look1 <- data$Y; look2 <- data[["Y"]]
> identical(look1, look2)
[1] TRUE
The confusion that this "non-standard evaluation" (NSE) shorthand feature of R has caused new users appears to be one of the motivations for the creation of first the ggplot aes function and later the evolution of the package-dplyr and the tidyverse-bundle-of-packages. Those packages allow the use of non-quoted names or tokens to refer to column identities.
In addition to #42-'s answer, you can dynamically reference columns like this:
colName <- "something"
myDataFrame[,colname]
Edit: Since you also asked about dynamically referencing data.frames #Rich Scriven suggested making a function that takes the data.frame as an argument, which is one working solution. You can also just load the data you need at the top of your script, which is easy to change on the fly if you need:
fileName <- "file1.csv"
data <- read.table(fileName, header = TRUE, stringsAsFactors = FALSE)
As per -42 above, the best choice seems to be the packages referenced. Using a function is close but doesn't seem to allow 'data' and 'variable' to be generic in 'data$variable'.
Thanks everyone!
I am implementing k-means in R.
In a loop, I am initiating several vectors that will be used to store values that belong to a particular cluster, as seen here:
for(i in 1:k){
assign(paste("cluster",i,sep=""),vector())
}
I then want to add to a particular "cluster" vector, depending on the value I get for the variable getIndex. So if getIndex is equal to 2 I want to add the variable minimumDistance to the vector called cluster2. This is what I am attempting to do:
minimumDistance <- min(distanceList)
getIndex <- match(minimumDistance,distanceList)
clusterName <- paste("cluster",getIndex,sep="")
name <- c(name, minimumDistance)
But obviously the above code does not work because in order to append to a vector that I'm naming I need to use assign as I do when I instantiate the vectors. But I do not know how to use assign, when using paste, when also appending to a vector.
I cannot use the index such as vector[i] because I don't know what index of that particular vector I want to add to.
I need to use the vector <- c(vector,newItem) format but I do not know how to do this in R. Or if there is any other option I would greatly, greatly appreciate it. If I were using Python I would simply use paste and then use append but I can't do that in R. Thank you in advance for your help!
You can do something like this:
out <- list()
for (i in 1:nclust) {
# assign some data (in this case a list) to a cluster
assign(paste0("N_", i), list(...))
# here I put all the clusters data in a list
# but you could use a similar statement to do further data manipulation
# ie if you've used a common syntax (here "N_" <index>) to refer to your elements
# you can use get to retrieve them using the same syntax
out[[i]] <- get(paste0("N_", i))
}
If you want a more complete code example, this link sounds like a similar problem emclustr::em_clust_mvn
I have a data.frame mapping which contains path and map.
I also have another data.frame DATA which contains the raw path and value.
EDIT: Path might have two components or more: e.g. "A>C" or "A>C>B"
set.seed(24);
DATA <- data.frame(
path=paste0(sample(LETTERS[1:3], 25, replace=TRUE), ">", sample(LETTERS[1:3], 25, replace=TRUE)),
value=rnorm(25)
)
mapping <- data.frame(path=c("A","B","C"), map=c("X","Y","Z"))
lapply(mapping, function (x) {
for (i in 1:nrow(DATA)) {
DATA$path[i] <- gsub(as.character(x["path"]),as.character(x["map"]),as.character(DATA$path[i]))
}
})
I'm trying to replace the path in DATA with the map value in mapping but this doesn't seem to be working for me.
"A>C" will be converted to "X>Z".
I understand that for loops are not good in R, but I can't think of another way to code it. Data size I'm working with is 6m row in DATA and 16k rows in mapping.
Clarification on Data: While the path consists of alphabets (ABC) now, the real path are actually domain names. Number of steps in a path is also not fixed at 2 and can be any number.
You can use chartr
DATA$path <- chartr('ABC', 'XYZ', DATA$path)
Or if we are using the data from 'mapping'
DATA$path <- chartr(paste(mapping$path, collapse=''),
paste(mapping$map, collapse=''), DATA$path)
Or using gsubfn
library(gsubfn)
pat <- paste0('[', paste(mapping$path, collapse=''),']')
indx <- setNames(as.character(mapping$map), mapping$path)
gsubfn(pat, as.list(indx), as.character(DATA$path))
Or a base R option based on #smci's comment
vapply(strsplit(as.character(DATA$path), '>'), function(x)
paste(indx[x], collapse=">"), character(1L))
Using data.table (1.9.5+), especially advisable b/c of the size of your data.
library(data.table)
setDT(DATA); setDT(mapping)
DATA[,paste0("path",1:2):=tstrsplit(path,split=">")]
setkey(DATA,path1)[mapping,new.path1:=i.map]
setkey(DATA,path2)[mapping,new.path2:=i.map]
DATA[,new.path:=paste0(new.path1,">",new.path2)]
If you want to get rid of the extra columns:
DATA[,paste0(c("","","new.","new."),"path",rep(1:2,2)):=NULL]
If you just want to overwrite path, use path on the LHS of the last line instead of new.path.
This could also be written more concisely:
library(data.table)
setDT(mapping)
setkey(setkey(setDT(DATA)[,paste0("path",1:2):=tstrsplit(path,split=">")
],path1)[mapping,new.path1:=i.map],path2
)[mapping,new.path:=paste0(new.path1,">",i.map)]
I think you're using the wrong apply.
mapply allows you to use two arguments to the function, here the path and the map. Note that in mapply, the argument FUN comes first. You also do not need to do this row by row, you can just do the entire column at once. Finally, in an apply the variables do not get updated as they do in a for loop, so you need to assign them in the .GlobalEnv. You can do this with an explicit call to assign() or using <<- which assigns them in the first place it finds them in the stack. In this case, that will be back in .GlobalEnv.
After defining mapping and DATA as you do above, try this.
head(DATA)
invisible(mapply( function (x,y) {
DATA$path <<- gsub(x,y,DATA$path)
},mapping$path, mapping$map))
head(DATA)
note that the call to invisible suppresses output from mapply.
If you really want to use lapply, you can. But you need to transpose mapping. You can do that but it will be converted to a matrix, so you have to convert it back. Then, you can just use the same tricks with <<- and not using a for loop as above to get this code:
invisible(lapply(as.data.frame(t(mapping)), function (x) {
DATA$path <<- gsub(x[1],x[2],DATA$path)
}))
head(DATA)
Thanks for sharing, I learned a lot answering this question.
I'm having some trouble understanding how R handles subsetting internally and this is causing me some issues while trying to build some functions. Take the following code:
f <- function(directory, variable, number_seq) {
##Create a empty data frame
new_frame <- data.frame()
## Add every data frame in the directory whose name is in the number_seq to new_frame
## the file variable specify the path to the file
for (i in number_seq){
file <- paste("~/", directory, "/",sprintf("%03d", i), ".csv", sep = "")
x <- read.csv(file)
new_frame <- rbind.data.frame(new_frame, x)
}
## calculate and return the mean
mean(new_frame[, variable], na.rm = TRUE)*
}
*While calculating the mean I tried to subset first using the $ sign new_frame$variable and the subset function subset( new_frame, select = variable but it would only return a None value. It only worked when I used new_frame[, variable].
Can anyone explain why the other subseting didn't work? It took me a really long time to figure it out and even though I managed to make it work I still don't know why it didn't work in the other ways and I really wanna look inside the black box so I won't have the same issues in the future.
Thanks for the help.
This behavior has to do with the fact that you are subsetting inside a function.
Both new_frame$variable and subset(new_frame, select = variable) look for a column in the dataframe withe name variable.
On the other hand, using new_frame[, variable] uses the variablename in f(directory, variable, number_seq) to select the column.
The dollar sign ($) can only be used with literal column names. That avoids confusion with
dd<-data.frame(
id=1:4,
var=rnorm(4),
value=runif(4)
)
var <- "value"
dd$var
In this case if $ took variables or column names, which do you expect? The dd$var column or the dd$value column (because var == "value"). That's why the dd[, var] way is different because it only takes character vectors, not expressions referring to column names. You will get dd$value with dd[, var]
I'm not quite sure why you got None with subset() I was unable to replicate that problem.