Related
I am using R to parse a list of strings in the form:
original_string <- "variable_name=variable_value"
First, I extract the variable name and value from the original string and convert the value to numeric class.
parameter_value <- as.numeric("variable_value")
parameter_name <- "variable_name"
Then, I would like to assign the value to a variable with the same name as the parameter_name string.
variable_name <- parameter_value
What is/are the function(s) for doing this?
assign is what you are looking for.
assign("x", 5)
x
[1] 5
but buyer beware.
See R FAQ 7.21
http://cran.r-project.org/doc/FAQ/R-FAQ.html#How-can-I-turn-a-string-into-a-variable_003f
You can use do.call:
do.call("<-",list(parameter_name, parameter_value))
There is another simple solution found there:
http://www.r-bloggers.com/converting-a-string-to-a-variable-name-on-the-fly-and-vice-versa-in-r/
To convert a string to a variable:
x <- 42
eval(parse(text = "x"))
[1] 42
And the opposite:
x <- 42
deparse(substitute(x))
[1] "x"
The function you are looking for is get():
assign ("abc",5)
get("abc")
Confirming that the memory address is identical:
getabc <- get("abc")
pryr::address(abc) == pryr::address(getabc)
# [1] TRUE
Reference: R FAQ 7.21 How can I turn a string into a variable?
Use x=as.name("string"). You can use then use x to refer to the variable with name string.
I don't know, if it answers your question correctly.
strsplit to parse your input and, as Greg mentioned, assign to assign the variables.
original_string <- c("x=123", "y=456")
pairs <- strsplit(original_string, "=")
lapply(pairs, function(x) assign(x[1], as.numeric(x[2]), envir = globalenv()))
ls()
assign is good, but I have not found a function for referring back to the variable you've created in an automated script. (as.name seems to work the opposite way). More experienced coders will doubtless have a better solution, but this solution works and is slightly humorous perhaps, in that it gets R to write code for itself to execute.
Say I have just assigned value 5 to x (var.name <- "x"; assign(var.name, 5)) and I want to change the value to 6. If I am writing a script and don't know in advance what the variable name (var.name) will be (which seems to be the point of the assign function), I can't simply put x <- 6 because var.name might have been "y". So I do:
var.name <- "x"
#some other code...
assign(var.name, 5)
#some more code...
#write a script file (1 line in this case) that works with whatever variable name
write(paste0(var.name, " <- 6"), "tmp.R")
#source that script file
source("tmp.R")
#remove the script file for tidiness
file.remove("tmp.R")
x will be changed to 6, and if the variable name was anything other than "x", that variable will similarly have been changed to 6.
I was working with this a few days ago, and noticed that sometimes you will need to use the get() function to print the results of your variable.
ie :
varnames = c('jan', 'feb', 'march')
file_names = list_files('path to multiple csv files saved on drive')
assign(varnames[1], read.csv(file_names[1]) # This will assign the variable
From there, if you try to print the variable varnames[1], it returns 'jan'.
To work around this, you need to do
print(get(varnames[1]))
If you want to convert string to variable inside body of function, but you want to have variable global:
test <- function() {
do.call("<<-",list("vartest","xxx"))
}
test()
vartest
[1] "xxx"
Maybe I didn't understand your problem right, because of the simplicity of your example. To my understanding, you have a series of instructions stored in character vectors, and those instructions are very close to being properly formatted, except that you'd like to cast the right member to numeric.
If my understanding is right, I would like to propose a slightly different approach, that does not rely on splitting your original string, but directly evaluates your instruction (with a little improvement).
original_string <- "variable_name=\"10\"" # Your original instruction, but with an actual numeric on the right, stored as character.
library(magrittr) # Or library(tidyverse), but it seems a bit overkilled if the point is just to import pipe-stream operator
eval(parse(text=paste(eval(original_string), "%>% as.numeric")))
print(variable_name)
#[1] 10
Basically, what we are doing is that we 'improve' your instruction variable_name="10" so that it becomes variable_name="10" %>% as.numeric, which is an equivalent of variable_name=as.numeric("10") with magrittr pipe-stream syntax. Then we evaluate this expression within current environment.
Hope that helps someone who'd wander around here 8 years later ;-)
Other than assign, one other way to assign value to string named object is to access .GlobalEnv directly.
# Equivalent
assign('abc',3)
.GlobalEnv$'abc' = 3
Accessing .GlobalEnv gives some flexibility, and my use case was assigning values to a string-named list. For example,
.GlobalEnv$'x' = list()
.GlobalEnv$'x'[[2]] = 5 # works
var = 'x'
.GlobalEnv[[glue::glue('{var}')]][[2]] = 5 # programmatic names from glue()
I have to run 10's of different permutations with same structure but different base names for the output. to avoid having to keep replacing the whole character names within each formula, I was hoping to great a variable then use paste function to assign the variable to the name of the output..
Example:
var<-"Patient1"
(paste0("cells_", var, sep="") <- WhichCells(object=test, expression = test > 0, idents=c("patient1","patient2"))
The expected output would be a variable called "cells_Patient1"
Then for subsequent runs, I would just copy and paste these 2 lines and change var <-"Patient1" to var <-"Patient2"
[please note that I am oversimplifying the above step of WhichCells as it entails ~10 steps and would rather not have to replace "Patient1" by "Patient2" using Search and Replaced
Unfortunately, I am unable to crate the variable "cells_Patient1" using the above command. I am getting the following error:
Error in variable(paste0("cells_", var, sep = "")) <-
WhichCells(object = test, : target of assignment expands to
non-language object
Browsing stackoverflow, I couldn't find a solution. My understanding of the error is that R can't assign an object to a variable that is not a constant. Is there a way to bypass this?
1) Use assign like this:
var <- "Patient1"
assign(paste0("cells_", var), 3)
cells_Patient1
## [1] 3
2) environment This also works.
e <- .GlobalEnv
e[[ paste0("cells_", var) ]] <- 3
cells_Patient1
3) list or it might be better to make these variables into a list:
cells <- list()
cells[[ var ]] <- 3
cells[[ "Patient1" ]]
## [1] 3
Then we could easily iterate over all such variables. Replace sqrt with any suitable function.
lapply(cells, sqrt)
## $Patient1
## [1] 1.732051
I am using R to parse a list of strings in the form:
original_string <- "variable_name=variable_value"
First, I extract the variable name and value from the original string and convert the value to numeric class.
parameter_value <- as.numeric("variable_value")
parameter_name <- "variable_name"
Then, I would like to assign the value to a variable with the same name as the parameter_name string.
variable_name <- parameter_value
What is/are the function(s) for doing this?
assign is what you are looking for.
assign("x", 5)
x
[1] 5
but buyer beware.
See R FAQ 7.21
http://cran.r-project.org/doc/FAQ/R-FAQ.html#How-can-I-turn-a-string-into-a-variable_003f
You can use do.call:
do.call("<-",list(parameter_name, parameter_value))
There is another simple solution found there:
http://www.r-bloggers.com/converting-a-string-to-a-variable-name-on-the-fly-and-vice-versa-in-r/
To convert a string to a variable:
x <- 42
eval(parse(text = "x"))
[1] 42
And the opposite:
x <- 42
deparse(substitute(x))
[1] "x"
The function you are looking for is get():
assign ("abc",5)
get("abc")
Confirming that the memory address is identical:
getabc <- get("abc")
pryr::address(abc) == pryr::address(getabc)
# [1] TRUE
Reference: R FAQ 7.21 How can I turn a string into a variable?
Use x=as.name("string"). You can use then use x to refer to the variable with name string.
I don't know, if it answers your question correctly.
strsplit to parse your input and, as Greg mentioned, assign to assign the variables.
original_string <- c("x=123", "y=456")
pairs <- strsplit(original_string, "=")
lapply(pairs, function(x) assign(x[1], as.numeric(x[2]), envir = globalenv()))
ls()
assign is good, but I have not found a function for referring back to the variable you've created in an automated script. (as.name seems to work the opposite way). More experienced coders will doubtless have a better solution, but this solution works and is slightly humorous perhaps, in that it gets R to write code for itself to execute.
Say I have just assigned value 5 to x (var.name <- "x"; assign(var.name, 5)) and I want to change the value to 6. If I am writing a script and don't know in advance what the variable name (var.name) will be (which seems to be the point of the assign function), I can't simply put x <- 6 because var.name might have been "y". So I do:
var.name <- "x"
#some other code...
assign(var.name, 5)
#some more code...
#write a script file (1 line in this case) that works with whatever variable name
write(paste0(var.name, " <- 6"), "tmp.R")
#source that script file
source("tmp.R")
#remove the script file for tidiness
file.remove("tmp.R")
x will be changed to 6, and if the variable name was anything other than "x", that variable will similarly have been changed to 6.
I was working with this a few days ago, and noticed that sometimes you will need to use the get() function to print the results of your variable.
ie :
varnames = c('jan', 'feb', 'march')
file_names = list_files('path to multiple csv files saved on drive')
assign(varnames[1], read.csv(file_names[1]) # This will assign the variable
From there, if you try to print the variable varnames[1], it returns 'jan'.
To work around this, you need to do
print(get(varnames[1]))
If you want to convert string to variable inside body of function, but you want to have variable global:
test <- function() {
do.call("<<-",list("vartest","xxx"))
}
test()
vartest
[1] "xxx"
Maybe I didn't understand your problem right, because of the simplicity of your example. To my understanding, you have a series of instructions stored in character vectors, and those instructions are very close to being properly formatted, except that you'd like to cast the right member to numeric.
If my understanding is right, I would like to propose a slightly different approach, that does not rely on splitting your original string, but directly evaluates your instruction (with a little improvement).
original_string <- "variable_name=\"10\"" # Your original instruction, but with an actual numeric on the right, stored as character.
library(magrittr) # Or library(tidyverse), but it seems a bit overkilled if the point is just to import pipe-stream operator
eval(parse(text=paste(eval(original_string), "%>% as.numeric")))
print(variable_name)
#[1] 10
Basically, what we are doing is that we 'improve' your instruction variable_name="10" so that it becomes variable_name="10" %>% as.numeric, which is an equivalent of variable_name=as.numeric("10") with magrittr pipe-stream syntax. Then we evaluate this expression within current environment.
Hope that helps someone who'd wander around here 8 years later ;-)
Other than assign, one other way to assign value to string named object is to access .GlobalEnv directly.
# Equivalent
assign('abc',3)
.GlobalEnv$'abc' = 3
Accessing .GlobalEnv gives some flexibility, and my use case was assigning values to a string-named list. For example,
.GlobalEnv$'x' = list()
.GlobalEnv$'x'[[2]] = 5 # works
var = 'x'
.GlobalEnv[[glue::glue('{var}')]][[2]] = 5 # programmatic names from glue()
I am a beginner so I'd appreciate any thoughts, and I understand that this question might be too basic for some of you.
Also, this question is not about the difference between <- and =, but about the way they get evaluated when they are part of the function argument. I read this thread, Assignment operators in R: '=' and '<-' and several others, but I couldn't understand the difference.
Here's the first line of code:
My objective is to get rid of variables in the environment. From reading the above thread, I would believe that <- would exist in the user workspace, so there shouldn't be any issue with deleting all variables.
Here is my code and two questions:
Question 1
First off, this code doesn't work.
rm(ls()) #throws an error
I believe this happens because ls() returns a character vector, and rm() expects an object name. Am I correct? If so, I would appreciate if someone could guide me how to get object names from character array.
Question 2
I googled this topic and found that this code below deletes all variables.
rm(list = ls())
While this does help me, I am unsure why = is used instead of <-. If I run the following code, I get an error Error in rm(list <- ls()) : ... must contain names or character strings
rm(list <- ls())
Why is this? Can someone please guide me? I'd appreciate any help/guidance.
I read this thread, Assignment operators in R: '=' and '<-' and several others, but I couldn't understand the difference.
No wonder, since the answers there are actually quite confusing, and some are outright wrong. Since that’s the case, let’s first establish the difference between them before diving into your actual question (which, it turns out, is mostly unrelated):
<- is an assignment operator
In R, <- is an operator that performs assignment from right to left, in the current scope. That’s it.
= is either an assignment operator or a distinct syntactic token
=, by contrast, has several meanings: its semantics change depending on the syntactic context it is used in:
If = is used inside a parameter list, immediately to the right of a parameter name, then its meaning is: “associate the value on the right with the parameter name on the left”.
Otherwise (i.e. in all other situations), = is also an operator, and by default has the same meaning as <-: i.e. it performs assignment in the current scope.
As a consequence of this, the operators <- and = can be used interchangeably1. However, = has an additional syntactic role in an argument list of a function definition or a function call. In this context it’s not an operator and cannot be replaced by <-.
So all these statements are equivalent:
x <- 1
x = 1
x[5] <- 1
x[5] = 1
(x <- 1)
(x = 1)
f((x <- 5))
f((x = 5))
Note the extra parentheses in the last example: if we omitted these, then f(x = 5) would be interpreted as a parameter association rather than an assignment.
With that out of the way, let’s turn to your first question:
When calling rm(ls()), you are passing ls() to rm as the ... parameter. Ronak’s answer explains this in more detail.
Your second question should be answered by my explanation above: <- and = behave differently in this context because the syntactic usage dictates that rm(list = ls()) associates ls() with the named parameter list, whereas <- is (as always) an assignment operator. The result of that assignment is then once again passed as the ... parameter.
1 Unless somebody changed their meaning: operators, like all other functions in R, can be overwritten with new definitions.
To expand on my comment slightly, consider this example:
> foo <- function(a,b) b+1
> foo(1,b <- 2) # Works
[1] 3
> ls()
[1] "b" "foo"
> foo(b <- 3) # Doesn't work
Error in foo(b <- 3) : argument "b" is missing, with no default
The ... argument has some special stuff going on that restricts things a little further in the OP's case, but this illustrates the issue with how R is parsing the function arguments.
Specifically, when R looks for named arguments, it looks specifically for arg = val, with an equals sign. Otherwise, it is parsing the arguments positionally. So when you omit the first argument, a, and just do b <- 1, it thinks the expression b <- 1 is what you are passing for the argument a.
If you check ?rm
rm(..., list = character(),pos = -1,envir = as.environment(pos), inherits = FALSE)
where ,
... - the objects to be removed, as names (unquoted) or character strings (quoted).
and
list - a character vector naming objects to be removed.
So, if you do
a <- 5
and then
rm(a)
it will remove the a from the global environment.
Further , if there are multiple objects you want to remove,
a <- 5
b <- 10
rm(a, b)
This can also be written as
rm(... = a, b)
where we are specifying that the ... part in syntax takes the arguments a and b
Similarly, when we want to specify the list part of the syntax, it has to be given by
rm(list = ls())
doing list <- ls() will store all the variables from ls() in the variable named list
list <- ls()
list
#[1] "a" "b" "list"
I hope this is helpful.
I have two lists of lists. humanSplit and ratSplit. humanSplit has element of the form::
> humanSplit[1]
$Fetal_Brain_408_AGTCAA_L001_R1_report.txt
humanGene humanReplicate alignment RNAtype
66 DGKI Fetal_Brain_408_AGTCAA_L001_R1_report.txt 6 reg
68 ARFGEF2 Fetal_Brain_408_AGTCAA_L001_R1_report.txt 5 reg
If you type humanSplit[[1]], it gives the data without name $Fetal_Brain_408_AGTCAA_L001_R1_report.txt
RatSplit is also essentially similar to humanSplit with difference in column order. I want to apply fisher's test to every possible pairing of replicates from humanSplit and ratSplit. Now I defined the following empty vector which I will use to store the informations of my fisher's test
humanReplicate <- vector(mode = 'character', length = 0)
ratReplicate <- vector(mode = 'character', length = 0)
pvalue <- vector(mode = 'numeric', length = 0)
For fisher's test between two replicates of humanSplit and ratSplit, I define the following function. In the function I use `geneList' which is a data.frame made by reading a file and has form:
> head(geneList)
human rat
1 5S_rRNA 5S_rRNA
2 5S_rRNA 5S_rRNA
Now here is the main function, where I use a function getGenetype which I already defined in other part of the code. Also x and y are integers :
fishertest <-function(x,y) {
ratReplicateName <- names(ratSplit[x])
humanReplicateName <- names(humanSplit[y])
## merging above two based on the one-to-one gene mapping as in geneList
## defined above.
mergedHumanData <-merge(geneList,humanSplit[[y]], by.x = "human", by.y = "humanGene")
mergedRatData <- merge(geneList, ratSplit[[x]], by.x = "rat", by.y = "ratGene")
## [here i do other manipulation with using already defined function
## getGenetype that is defined outside of this function and make things
## necessary to define following contingency table]
contingencyTable <- matrix(c(HnRn,HnRy,HyRn,HyRy), nrow = 2)
fisherTest <- fisher.test(contingencyTable)
humanReplicate <- c(humanReplicate,humanReplicateName )
ratReplicate <- c(ratReplicate,ratReplicateName )
pvalue <- c(pvalue , fisherTest$p)
}
After doing all this I do the make matrix eg to use in apply. Here I am basically trying to do something similar to double for loop and then using fisher
eg <- expand.grid(i = 1:length(ratSplit),j = 1:length(humanSplit))
junk = apply(eg, 1, fishertest(eg$i,eg$j))
Now the problem is, when I try to run, it gives the following error when it tries to use function fishertest in apply
Error in humanSplit[[y]] : recursive indexing failed at level 3
Rstudio points out problem in following line:
mergedHumanData <-merge(geneList,humanSplit[[y]], by.x = "human", by.y = "humanGene")
Ultimately, I want to do the following:
result <- data.frame(humanReplicate,ratReplicate, pvalue ,alternative, Conf.int1, Conf.int2, oddratio)
I am struggling with these questions:
In defining fishertest function, how should I pass ratSplit and humanSplit and already defined function getGenetype?
And how I should use apply here?
Any help would be much appreciated.
Up front: read ?apply. Additionally, the first three hits on google when searching for "R apply tutorial" are helpful snippets: one, two, and three.
Errors in fishertest()
The error message itself has nothing to do with apply. The reason it got as far as it did is because the arguments you provided actually resolved. Try to do eg$i by itself, and you'll see that it is returning a vector: the corresponding column in the eg data.frame. You are passing this vector as an index in the i argument. The primary reason your function erred out is because double-bracket indexing ([[) only works with singles, not vectors of length greater than 1. This is a great example of where production/deployed functions would need type-checking to ensure that each argument is a numeric of length 1; often not required for quick code but would have caught this mistake. Had it not been for the [[ limit, your function may have returned incorrect results. (I've been bitten by that many times!)
BTW: your code is also incorrect in its scoped access to pvalue, et al. If you make your function return just the numbers you need and the aggregate it outside of the function, your life will simplify. (pvalue <- c(pvalue, ...) will find pvalue assigned outside the function but will not update it as you want. You are defeating one purpose of writing this into a function. When thinking about writing this function, try to answer only this question: "how do I compare a single rat record with a single human record?" Only after that works correctly and simply without having to overwrite variables in the parent environment should you try to answer the question "how do I apply this function to all pairs and aggregate it?" Try very hard to have your function not change anything outside of its own environment.
Errors in apply()
Had your function worked properly despite these errors, you would have received the following error from apply:
apply(eg, 1, fishertest(eg$i, eg$j))
## Error in match.fun(FUN) :
## 'fishertest(eg$i, eg$j)' is not a function, character or symbol
When you call apply in this sense, it it parsing the third argument and, in this example, evaluates it. Since it is simply a call to fishertest(eg$i, eg$j) which is intended to return a data.frame row (inferred from your previous question), it resolves to such, and apply then sees something akin to:
apply(eg, 1, data.frame(...))
Now that you see that apply is being handed a data.frame and not a function.
The third argument (FUN) needs to be a function itself that takes as its first argument a vector containing the elements of the row (1) or column (2) of the matrix/data.frame. As an example, consider the following contrived example:
eg <- data.frame(aa = 1:5, bb = 11:15)
apply(eg, 1, mean)
## [1] 6 7 8 9 10
# similar to your use, will not work; this error comes from mean not getting
# any arguments, your error above is because
apply(eg, 1, mean())
## Error in mean.default() : argument "x" is missing, with no default
Realize that mean is a function itself, not the return value from a function (there is more to it, but this definition works). Because we're iterating over the rows of eg (because of the 1), the first iteration takes the first row and calls mean(c(1, 11)), which returns 6. The equivalent of your code here is mean()(c(1, 11)) will fail for a couple of reasons: (1) because mean requires an argument and is not getting, and (2) regardless, it does not return a function itself (in a "functional programming" paradigm, easy in R but uncommon for most programmers).
In the example here, mean will accept a single argument which is typically a vector of numerics. In your case, your function fishertest requires two arguments (templated by my previous answer to your question), which does not work. You have two options here:
Change your fishertest function to accept a single vector as an argument and parse the index numbers from it. Bothing of the following options do this:
fishertest <- function(v) {
x <- v[1]
y <- v[2]
ratReplicateName <- names(ratSplit[x])
## ...
}
or
fishertest <- function(x, y) {
if (missing(y)) {
y <- x[2]
x <- x[1]
}
ratReplicateName <- names(ratSplit[x])
## ...
}
The second version allows you to continue using the manual form of fishertest(1, 57) while also allowing you to do apply(eg, 1, fishertest) verbatim. Very readable, IMHO. (Better error checking and reporting can be used here, I'm just providing a MWE.)
Write an anonymous function to take the vector and split it up appropriately. This anonymous function could look something like function(ii) fishertest(ii[1], ii[2]). This is typically how it is done for functions that either do not transform as easily as in #1 above, or for functions you cannot or do not want to modify. You can either assign this intermediary function to a variable (which makes it no longer anonymous, figure that) and pass that intermediary to apply, or just pass it directly to apply, ala:
.func <- function(ii) fishertest(ii[1], ii[2])
apply(eg, 1, .func)
## equivalently
apply(eg, 1, function(ii) fishertest(ii[1], ii[2]))
There are two reasons why many people opt to name the function: (1) if the function is used multiple times, better to define once and reuse; (2) it makes the apply line easier to read than if it contained a complex multi-line function definition.
As a side note, there are some gotchas with using apply and family that, if you don't understand, will be confusing. Not the least of which is that when your function returns vectors, the matrix returned from apply will need to be transposed (with t()), after which you'll still need to rbind or otherwise aggregrate.
This is one area where using ddply may provide a more readable solution. There are several tutorials showing it off. For a quick intro, read this; for a more in depth discussion on the bigger picture in which ddply plays a part, read Hadley's Split, Apply, Combine Strategy for Data Analysis paper from JSS.