What does declaring function at the start of a line do? - r

I encountered this code:
res <- lapply(strsplit(s, "\n")[[1]],
(function (str) paste(rev(strsplit(str, "")[[1]]), collapse = "")))
The secodnd line reverses each of the splitted strings at the first line.
How does it do that? Namely, what does calling 'function' at the start do?

Calling lapply takes and performs some function on each list element. It takes the form lapply(list_data, some_function). So, for instance, if I have a list of integers and want to find out how many integers are in each list element, I would run:
list_data <- list(list1 = 1:5,
list2 = 6:10,
list3 = 11:30)
lapply(list_data, length)
The function here is length, which is a function that is inherent in R. Some functions aren't defined in R, say if I want define my own formula for each value in the list, I could define my own function. Calling a function allows users to define a function that is not already in R or an R library. Like so:
lapply(list_data, function(x) x^2+4-x^3)
The function here is x^2+4-x^3, which is not defined in R programming itself.
So in your example, your data is strsplit(s, "\n")[[1]] and it is taking that data and applying the function paste(rev(strsplit(str, "")[[1]]), collapse = "")) to each element in the data.
Note that in my example, I put function(x) - your example puts function(str) - what's in the parentheses doesn't matter and is user defined. For example lapply(list_data, function(str) str^2+4-str^3) will return the same thing as lapply(list_data, function(x) x^2+4-x^3)
Please note that broad "learning" style questions like this are not exactly what this site is for, and this question will likely get removed and/or receive some negative feedback. Since you are new to this site and to R, I'm providing this answer but I would not be surprised if the question is removed. Just trying to help both you and the SO community!

Related

Is there literature on how to set up own square bracket functions in R?

I'm struggling writing functions for my own class. Hadley Wickham suggests to implement functions for square bracket functions like [, [<- etc. But how is this done?
After the comment from r.user.05apr, I manage to write
`[.test` <- function(x, y){
substr(x, start = y[1], stop = y[length(y)])
}
foo <- "hello world"
class(foo) <- "test"
foo[2:5] #what correctly returns "ello"
I did find sites on how they are called, but no explanations and examples on how they are defined. For example for the mentioned [<- function I guess, three arguments are needed, the object to manipulate, the index showing which elements of the objects shall be replaced and finally a replacement. I managed to get the result wanted by
`[<-.test` <- function(obj, index, value){
tmp <- unlist(strsplit(obj, ""))
tmp[index] <- value
return(paste(tmp, collapse = ""))
}
foo <- "hello world"
class(foo) <- "test"
foo[c(2, 5)] <- "X"
but only after figuring out that the third element has to be called value and nothing else. So I'm looking for a good, easy to understand (this criteria is not met by Writing R Extensions; more like Creating R Packages by Friedrich Leisch) piece of literature covering the methods mentioned by Wickham.

Preserving data structure when returning values from function in R

I currently have a basic script written in R, which has two functions embedded within another:
FunctionA <- Function() {
results_from_B <- FunctionB()
results_from_C <- FunctionC()
}
Function B generates some data which is then analysed in Function C.
If I stop the code within function A, I can see the structure of results_from_C - this appears under 'values' and I can refer to different elements using the syntax results_from_C$column_name1.
I achieved this within Function C by specifying the returned values using:
return(list(column_name_1 = value1, column_name_2 = value2)
However, I cannot work out how I can return these same values (in the same structure) from Function A - everything I try returns a list which is formatted as 'Data' rather than 'Values' and cannot be indexed using the syntax results_from_A$column_name1.
Can anyone help me to understand what I need to do in order to extract results from Function C outside of Function A?
Thanks in advance
I don't understand what you mean by formatted as 'Data' rather than 'Values'.
There's nothing wrong with the setup you describe, I every now and then use functions inside functions, it's perfectly OK.
(Note that R is case sensitive, it's function not Function.)
FunctionA <- function() {
FunctionB <- function() 1:2*pi
FunctionC <- function(x)
list(column_name_1 = x[1], column_name_2 = x[2])
results_from_B <- FunctionB()
results_from_C <- FunctionC(results_from_B)
results_from_C
}
result <- FunctionA()
result
$column_name_1
[1] 3.141593
$column_name_2
[1] 6.283185
result$column_name_1
[1] 3.141593
Is this it? If not, please clarify your question.

If my function doesn't work on every object, how do I skip those objects?

I am trying to write a function and apply it to a list. Inside my function is a function written by some one else. If I make my list very easy, everything will work fine. But if I use all the real data I have, there are some bad objects and the outside function doesn't work and my whole function won't go through.
What do I type to say "If the outside function doesn't work, skip that object and move to the next one in the list."? With or without NA, doesn't matter.
I cannot figure out how to write a reproducible example that would result in a list of dataframes, which is what happens inside this function. I'm willing to take any help to improve this question.
My function is something like this:
do_this<- function(x){
outside_function(x))%>% #this returns a dataframe for each object
filter()%>%
select()%>%
summarise_each(funs(mean(., na.rm = TRUE))) #by the end the df is down to one row
}
This is how I apply the function to the list to come up with my final dataframe.
df<-bind_rows(lapply(my_list, do_this))
An example:
myfun <- function(x) {if (x == 1) {stop("bad")} else x}
throws error on input of 1:
lapply(1:4, myfun) # stops from error
Just wrap it in try (as long as you don't need more complex error handling):
L <- lapply(1:4, function(x) try(myfun(x)))
And then you can use Filter to get rid of the "bad" cases:
Filter(function(x) !inherits(x, "try-error"), L)
Although you may want to just make your wrapper function more robust, or return NULL (or some other appropriate value) under the condition that makes the inner function fail.

R: passing by parameter to function and using apply instead of nested loop and recursive indexing failed

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.

How do I call a namespaced function without evaluating the parameters you are giving it?

In R, the idiomatic way to call another function without evaluating the parameters you give it is apparently as follows:
Call <- match.call(expand.dots = TRUE)
# Modify parameters here as needed and set unneeded ones to NULL.
Call[[1L]] <- as.name("name.of.function.to.be.called.here")
eval.parent(Call)
However, when I put a namespaced name (e.g. utils::write.csv) in the as.name() call, I get an error:
"could not find function "utils::write.csv"
What is the proper way of using this R idiom to call a namespaced function?
Here is a solution using do.call(), which both constructs and evaluates the function call.
Like the approach you started with, this one uses the fact that R calls are lists in which: (a) the first element is the name of a function; and (b) all following elements are arguments to that function.
j <- function(x, file) {
Call <- match.call(expand.dots = TRUE)
arglist <- as.list(Call)[-1]
do.call(utils::write.csv, arglist)
}
dat <- data.frame(x=1:10, y=rnorm(10))
j(dat, file="outfilename.csv")
EDIT: FWIW, here's an example from plot.formula in base R, which uses a construct similar to the one above:
{
m <- match.call(expand.dots = FALSE)
eframe <- parent.frame()
. . .
. . .
m <- as.list(m)
m[[1L]] <- stats::model.frame.default
m <- as.call(c(m, list(na.action = NULL)))
mf <- eval(m, eframe)
. . .
. . .
}
The function uses the do.call() construct later on. Going a bit deeper into the weeds, my reading is that in the snippet shown here, it instead uses several steps mostly because of the need to add na.action=NULL to the list of arguments.
In any case, it looks like the do.call() options is as close to canonical as could be desired.
As #Josh O'Brien answered, do.call is much more straight forward to use.
The first argument to do.call can be either a function name or an actual function.
The function name can NOT contain the namespace qualifier. The :: part is actually a function that takes the names on both sides and find the corresponding function, so it must be evaluated separately to work.
So, with do.call, you need something like:
# ...Stuff from Josh's answer goes here
# And then:
do.call(utils::write.csv, arglist)
And with eval:
Call <- match.call(expand.dots = TRUE)
# Modify parameters here as needed and set unneeded ones to NULL.
Call[[1L]] <- utils::write.csv
eval.parent(Call)
Note the lack of quotes around the function name. That evaluates to the function closure.
Another way of getting the function from a namespace-qualified name:
eval(parse(text="utils::write.csv"))
Again, the :: function is called that correctly finds the function.
Another more manual way is to extract the namespace name & function name and then do the lookup yourself:
x <- strsplit("utils::write.csv", "::")[[1]]
get(x[2], asNamespace(x[1]))

Resources