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$variable'.
Thanks everyone!
Related
I am currently trying to create a table from a list of variable names (something I feel should be relatively simple) and I can't for the life of me, figure out how to do it correctly.
I have a data table that I've named 'file' and there are a list of 3 variable names within this file. What I want to do is create a table of each variable and then rbind them together. For further context, these few lines of code will be worked into a much larger function. The list of variable names must be able to accommodate the number of variables the user defines.
I have tried the following:
file<-as.data.table(dt)
variable_list<-list("outcome", "type")
for (variable in variable_list){
var_table<-as.data.table(table(file$variable_list))
na_table<-as.data.table(table(is.na(file$variable)))
}
When I run the above code, R returns empty tables of var_table and na_table. What am I doing wrong?
An option is to loop over the 'variable_list, extract the column, apply tableandrbindwithindo.call`
do.call(rbind, lapply(variable_list, function(nm) table(file[[nm]])))
NOTE: assuming that the levels of the columns are the same
If the levels are not the same, make it same by converting the columns to factor with levels specified
lvls <- na.omit(sort(unique(unlist(file[, unlist(variable_list), with = FALSE]))))
do.call(rbind, lapply(variable_list, function(nm)
table(factor(file[[nm]], levels = lvls))))
Or if we have a data.table, use the data.table methods
rbindlist(lapply(variable_list, function(nm) file[, .N,by = c(nm)]), fill = TRUE)
The problem (at least one of the problems) might be that you are attempting to use the $ operator incorrectly. You cannot substitute text values into the second argument. You can use its syntactic equivalent [[ instead of $, however. So this would be a possible improvement. (I've not tested it since you provided no test material.)
file<-as.data.table(dt)
variable_list<-list("outcome", "type")
for (variable in variable_list){
var_table<-as.data.table(table(file[[variable]])) # clearly not variable_list
na_table<-as.data.table(table(is.na(file[[variable]] )))
}
I'm guessing you might have done something like, ...
var_table <- file[, table(variable ) ]
... since data.table syntax evaluates text values in the environment of the file (which in this case is confusing named "file". It's better not to use such names, since in this case there's also an R function by that name.
This is certainly a simple question but I can't find a solution.
I want to clean my environment by removing some variables I don't need anymore and keep some others.
I unterstand ls() can list them and ls()[[i]] returns the name of the variable, as a string.
So If I want to remove the 10th, let's say it's the variable age , ls()[[10]] will return "age", and I would like to do something like rm(ls()[[10]), but it does not work. I can't figure out to force rm(ls([10])) to be be equivalent to rm(age).
I guess I need to force some evaluation of string "age" to return the variable age but can't find the proper function in R documentation.
Thanks if you can help.
The list argument of rm will help you. It accepts a character vector. Consider:
age <- 1
rm(list = "age") # Same effect as rm(age)
age
#Error: object 'age' not found
So running e.g.
rm(list = ls())
will clear all visible objects in the specified environment.
In your case rm(list = ls()[10]) will do what you want. However, note that ls() always returns a sorted character vector, so the 10th entry can change rather easily. You probably want to do the following
objects_to_remove <- c("age", "another_object") # etc
rm(list = objects_to_remove)
How about the following:
1: Grab the list in the environment,
2: Define the items you want to remove,
3: Filter the list by the items you want to remove
4: Then remove them
list <- ls()
to_remove <- c("Item1", "Item2")
list_to_remove <- list[ list %in% to_remove]
list_to_remove
rm(list=list_to_remove)
DF <- data.frame(CpGId, tframe$t, tframe$p, q)
dimnames(DF)[[2]] <- c("CpGId", "t_value", "p_value", "q_value")
DFhyper <- DF[with(DF, q_value < 0.05 & t_value> 0), ]
DFhyper <- data.frame(DFhyper, row.names = NULL)
DFhyper <- DFhyper [order(p_value), ]
Until fourth line of code, things work fine but then why R gives an error stating p_value object not found?
R executes the bracketed expression first, without paying any attention to how it is going to be used. When you type
DFhyper[order(p_value),]
R will look for p_value in the current scope (probably the global scope), however, as this is bound into the dataframe, it will not be able to find it. You need to do something to tell it where this is located.
Either
DFhyper[order(DFhyper$p_value),]
or
DFhyper[with(DFhyper,order(p_value)),]
(or nearly equivalent, with(DFhyper,DFHyper[order(p_value),])) will work. The first command tells R specifically that you are referencing the column in the data frame, and the second tells R to look in the dataframe for the variable if it can't find it in scope.
Finally, you can just bind the dataframe into the scope as well, executing
attach(DFhyper)
DFhyper[order(p_value),]
The attach command adds the dataframe columns to the current scope. It can be useful for when you have many operations on the dataframe columns, but don't want to keep referencing it. You can then detach it with detach(DFhyper) when you are done.
It needs to be
DFhyper <- DFhyper [order(Dfhyper$p_value), ]
More 'feels like it should be' simple stuff which seems to be eluding me today. Thanks in advance for assistance.
Within a loop, that's within a function, I'm trying to add a column, and name it based on a formula.
I can bind a column & its name is taken from the bound object: data<-cbind(data,bothdata)
I can bind a column & manually name the bound object: data<-cbind(data,newname=bothdata)
I can bind a column which is the product of an equation & manually name the bound object: data<-cbind(data,newname2=bothdata-1)
Or another way: data <- transform(data, newColumn = bothdata-1)
What I can't do is have the name be the product of a formula. My actual formula-derived example name is paste("E_wgt",rev(which(rev(Esteps) == q))-1,"%") & equation for column: baddata - q.
A simpler one: data<-cbind(data,paste("magic",100,"beans")=bothdata-1). This fails because cbind isn't expecting the = even though it's fine in previous examples. Same fail for transform.
My first thought was assign but while I've used this successfully for creating forumla-named objects, I can't see how to get it to work for formula-named columns.
If I use an intermediary step to put the naming formula in an object container then use that, e.g.:
name <- paste("magic",100,"beans")
data<-cbind(data,name=bothdata-1)
the column name is "name" not "magic100beans". If I assign the equation result to an formula-named object:
assign(paste("magic",100,"beans"),bothdata-1)
Then try to cbind that via get:
data<-cbind(data,get(paste("magic",100,"beans")))
The column is called "get(paste("magic",100,"beans"))". Boo! Any thoughts anyone? It occurs to me that I can do cbind then separately colnames(data)[ncol(data)] <- paste("magic",100,"beans")) which I guess I'll settle for for now, but would still be interested to find if there was a direct way.
Thanks.
Chances are that cbind is overkill for your use case. In almost every instance, you can simply mutate the underlying data frame using data$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)
Is there an R type equivalent to the Matlab structure type?
I have a few named vectors and I try to store them in a data frame. Ideally, I would simply access one element of an object and it would return the named vectors (like a structure in Matlab). I feel that using a data frame is not the right thing to do since it can store the values of the named vectors but not the names when they differ from one vector to the other.
More generally, is it possible to store a bunch of different objects in a single one in R?
Edit: As Joran said I think that list does the job.
l = list()
l$vec1 = namedVector1
l$vec2 = namedVector2
...
If I have a list of names
name1 = 'vec1'
name2 = 'vec2'
is there any way for the interpreter to understand that when I use a variable name like name1, I am not referring to the variable name but to its content? I have tried get(name1) but it does not work.
I could still be wrong about what you're trying to do, but I think this is the best you're going to get in terms of accessing each list element by name:
l <- list(a= 1:3,b = 1:10)
> ind <- "a"
> l[[ind]]
[1] 1 2 3
Namely, you're going to have to use [[ explicitly.