filling an array recursively in R language - r

I have a multidimensional array (B_matrix) that I need to fill up with some random values. Since the dimension depends on two parameters K and C that are user defined, I cannot use nested loop to fill the array, so I have decided to fill it up recursively.
The problem with the recursion function (fillUp) is that that even though the array is declared outside the function, the array is set to NULL after the function is run.
B_dim = rep(2,((K+1+C)*2))
B_matrix = array( dim = B_dim, dimnames = NULL)
string = c()
fillUp<-function(level, string ){
if (level>=1){
for(i in c(1,2)){
Recall(level-1, c(string, i))
}
}else{
B_matrix[string] = 1;
}
}
fillUp(length(B_dim), string)
> sum( B_matrix == 1)
[1] NA
I'm new to R, so I'm not sure if the "global" declaration allows fillUp to change the values of the matrix.
Edit:
Note that the line
B_matrix[string] = 1;
is just a test case, and the original idea is to assign some random value that depends of the position of the array element.
Edit2:
Based on what #Bridgeburners hinted, I'm almost there. Replacing B_matrix[string] = 1, by
assign('str', matrix(string,1), envir=.GlobalEnv)
assign('hl', B_half_length, envir=.GlobalEnv)
rul <-runif(1, 0, sum(str[1:hl]))
with( .GlobalEnv,B_matrix[str] <- rul)
I get the error (last line):
Error in eval(expr, envir, enclos) : object 'rul' not found
The problem, I guess, is that I'm working with variables from two different environments at the same time. I don't know how to proceed here.
This option doesn't work either
assign('str',matrix(string,1), envir=.GlobalEnv)
assign('hl', B_half_length, envir=.GlobalEnv)
assign('ru', runif(1, 0, sum(str[1:hl])), envir=.GlobalEnv)
with( .GlobalEnv,B_matrix[str] <- ru)
Note: no visible binding for global variable 'ru'
Edit3:
I've finally solved it:
assign('str',matrix(string,1), envir=.GlobalEnv)
with( .GlobalEnv, B_matrix[str] <- runif(1, 0, sum(str[1:B_half_length])-B_half_length+1) )
where B_half_length is a global variable

Whenever a process is working within a function, it's working in a different environment. The object "B_matrix" is defined in the global environment. Since you're nesting environments (2*(K+C+1) times) you're not impacting the original object. If you simply replace line
B_matrix[string] = 1;
with
assign('str', matrix(string,1), envir=.GlobalEnv)
with(.GlobalEnv,B_matrix[str] <- 1)
your code will work. You simply need to specify which environment your expression is working in. (In the first line you're passing the local value of "string" to a global object named "str".)
Note, also, that indexing an array with a vector doesn't work.
That is, "B_matrix[2,2,2,2,2,2]" is not the same as "B_matrix[c(2,2,2,2,2,2)]".
But it works with a matrix

What you want can be achieved with the following line code once you have initialised you B_matrix array:
B_matrix[] <- runif(length(B_matrix))

Related

Using the try and try catch function on .WAV files

Trying to cut a bunch of audio (.WAV) files into smaller samples in R. For this example, I'm using a loop to cut out 1 minute samples at 140 minutes.
For some files, the recording ends before 140 minutes due to an error in the recording device. When this occurs, an error appears -- and the loop stops. I'm trying to make it so the loop continues by using the try or tryCatch function however keep getting errors.
The code is as follows:
for(i in 1:length(AR_CD288)){
CUT_AR288_5 <- try({readWave(AR_CD288[i], from = 140, to = 141, units = "minutes")})
FILE.OUT_AR288_5<- sub("\\.wav$", "_140.wav", AR_CD288)
OUT.PATH_AR288_5 <- file.path("New files", basename(FILE.OUT_AR288_5))
writeWave(CUT_AR288_5, extensible=FALSE, filename = OUT.PATH_AR288_5[i])
}
I get the following two errors from the code:
Error in readBin(con, int, n = N, size = bytes, signed = (bytes != 1), :
invalid 'n' argument
Error in writeWave(CUT_AR288_5, extensible = FALSE, filename = OUT.PATH_AR288_5[i]) :
'object' needs to be of class 'Wave' or 'WaveMC
The loop still saves some samples into the "New files" directory, however, once the loop reaches a file <140 minutes, the loop stops.
I am very stuck! Any help would be greatly appreciated.
Cheers.
When I use try, I always do one (or both) of:
check the return value to see if it inherits "try-error", indicating that the command failed; or
add try(., silent = TRUE), indicating that I don't care if it succeeded (but this implies that I will not use its return value, either).
Try this:
for (i in seq_along(AR_CD288)) {
CUT_AR288_5 <- try({
readWave(AR_CD288[i], from = 140, to = 141, units = "minutes")
}, silent = TRUE)
if (!inherits(CUT_AR288_5, "try_error")) {
FILE.OUT_AR288_5 <- sub("\\.wav$", "_140.wav", AR_CD288)
OUT.PATH_AR288_5 <- file.path("New files", basename(FILE.OUT_AR288_5))
writeWave(CUT_AR288_5, extensible = FALSE, filename = OUT.PATH_AR288_5[i])
}
}
Three notes:
I changed 1:length(.) to seq_along(.); the latter is more resilient in an automated use when it is feasible that the vector might be length 0. For example, if AR_CD288 can ever be length 2, intuitively we expect 1:length(AR_CD288) to return nothing so that the for loop will not run; unfortunately, it resolves to 1:0 which returns a vector of length 2, which will often fail (based on whatever code is operating in the loop). The use of seq_along(.) will always return a vector of length 0 with an empty input, which is what we need. (Alternatively and equivalent, seq_len(length(AR_CD288)), though that's really what seq_along is intended to do.)
If you do not add silent=TRUE (or explicitly add silent=FALSE), then you will get an error message indicating that the command failed. Unfortunately, the error message may not indicate which i failed, so you may be left in the dark as far as fixing or removing the errant file. You may prefer to add an else to the if (inherits(.,"try-error")) clause so that you can provide a clearer error, such as
if (inherits(CUT_AR288_5, "try_error")) {
warning("'readWave' failed on ", sQuote(AR_CD288[i]), call. = FALSE)
} else {
FILE.OUT_AR288_5 <- sub("\\.wav$", "_140.wav", AR_CD288)
# ...
}
(noting that I put the "things worked" code in the else clause here ... I find it odd to do if (!...) {} else {}, seems like a double-negation :-).
The choice to wrap one function or the whole block depends on your needs: I tend to prefer to know exactly where things fail, so the will-possibly-fail functions are often individually wrapped with try so that I can react (or log/message) accordingly. If you don't need that resolution of error-detection, then you can certainly wrap the whole code-block in a sense:
for (i in seq_along(AR_CD288)) {
ret <- try({
CUT_AR288_5 <- readWave(AR_CD288[i], from = 140, to = 141, units = "minutes")
FILE.OUT_AR288_5 <- sub("\\.wav$", "_140.wav", AR_CD288)
OUT.PATH_AR288_5 <- file.path("New files", basename(FILE.OUT_AR288_5))
writeWave(CUT_AR288_5, extensible = FALSE, filename = OUT.PATH_AR288_5[i])
}, silent = TRUE)
if (inherits(ret, "try-error")) {
# do or log something
}
}

Unexpected symbol error in R that doesn't match my code

I am coding in R-studio and have a function called saveResults(). It takes:
sce - a Single Cell Experiment object.
opt - a list with five things
clusterLabels - simple dataframe with two columns
The important thing is that I receive an error stating:
Error: unexpected symbol in:
"saveResults(sce = sce, opt = opt, clusteInputs()
zhengMix"
which doesn't agree at all with the parameters I pass into the function. You can see this on the last line of the code block below: I pass in proper parameters, but I receive an error that says I have passed in clusteInputs(), and zhengMix instead of clusterLabels. I don't have a function called clusteInputs(), and zhengMix was several lines above.
# Save the clustering data
InstallAndLoadPackagesForSC3Clustering()
opt <- GetOptionInputs()
zhengMix <- FetchzhengMix(opt)
sce <- CreateSingleCellExperiment(zhengMix)
clusterLabels <- getClusterLabels(sce)
opt <- createNewDirectoriesToSaveData(opt)
saveResults <- function(sce, opt, clusterLabels){
print("Beginning process of saving results...")
maxClusters = ncol(clusterLabels)/2+1
for (n in 2:maxClusters){
savePCAasPDF(sce, opt, numOfClusters = n, clusterLabels)
saveClusterLabelsAsRDS(clusterLabels, numOfClusters = n, opt)
}
saveSilhouetteScores(sce, opt)
print("Done.")
}
saveResults(sce = sce, opt = opt, clusterLabels = clusterLabels)
Does anyone have an idea what is going on? I'm pretty stuck on this.
This isn't the best solution, but I fixed my own problem by removing the code out of the function and running it there caused no issues.

Why does this happen when a user-defined R function does not return a value?

In the function shown below, there is no return. However, after executing it, I can confirm that the value entered d normally.
There is no return. Any suggestions in this regard will be appreciated.
Code
#installed plotly, dplyr
accumulate_by <- function(dat, var) {
var <- lazyeval::f_eval(var, dat)
lvls <- plotly:::getLevels(var)
dats <- lapply(seq_along(lvls), function(x) {
cbind(dat[var %in% lvls[seq(1, x)], ], frame = lvls[[x]])
})
dplyr::bind_rows(dats)
}
d <- txhousing %>%
filter(year > 2005, city %in% c("Abilene", "Bay Area")) %>%
accumulate_by(~date)
In the function, the last assignment is creating 'dats' which is returned with bind_rows(dats) We don't need an explicit return statement. Suppose, if there are two objects to be returned, we can place it in a list
In some languages like python, for memory efficiency, generators are used which will yield instead of creating the whole output in memory i.e. Consider two functions in python
def get_square(n):
result = []
for x in range(n):
result.append(x**2)
return result
When we run it
get_square(4)
#[0, 1, 4, 9]
The same function can be written as a generator. Instead of returning anything,
def get_square(n):
for x in range(n):
yield(x**2)
Running the function
get_square(4)
#<generator object get_square at 0x0000015240C2F9E8>
By casting with list, we get the same output
list(get_square(4))
#[0, 1, 4, 9]
There is always a return :) You just don't have to be explicit about it.
All R expressions return something. Including control structures and user-defined functions. (Control-structures are just functions, by the way, so you can just remember that everything is a value or a function call, and everything evaluates to a value).
For functions, the return value is the last expression evaluated in the execution of the function. So, for
f <- function(x) 2 + x
when you call f(3) you will invoke the function + with two parameters, 2 and x. These evaluate to 2 and 3, respectively, so `+`(2, 3) evaluates to 5, and that is the result of f(3).
When you call the return function -- and remember, this is a function -- you just leave the control-flow of a function early. So,
f <- function(x) {
if (x < 0) return(0)
x + 2
}
works as follows: When you call f, it will call the if function to figure out what to do in the first statement. The if function will evaluate x < 0 (which means calling the function < with parameters x and 0). If x < 0 is true, if will evaluate return(0). If it is false, it will evaluate its else part (which, because if has a special syntax when it comes to functions, isn't shown, but is NULL). If x < 0 is not true, f will evaluate x + 2 and return that. If x < 0 is true, however, the if function will evaluate return(0). This is a call to the function return, with parameter 0, and that call will terminate the execution of f and make the result 0.
Be careful with return. It is a function so
f <- function(x) {
if (x < 0) return;
x + 2
}
is perfectly valid R code, but it will not return when x < 0. The if call will just evaluate to the function return but not call it.
The return function is also a little special in that it can return from the parent call of control structures. Strictly speaking, return isn't evaluated in the frame of f in the examples above, but from inside the if calls. It just handles this special so it can return from f.
With non-standard evaluation this isn't always the case.
With this function
f <- function(df) {
with(df, if (any(x < 0)) return("foo") else return("bar"))
"baz"
}
you might think that
f(data.frame(x = rnorm(10)))
should return either "foo" or "bar". After all, we return in either case in the if statement. However, the if statement is evaluated inside with and it doesn't work that way. The function will return baz.
For non-local returns like that, you need to use callCC, and then it gets more technical (as if this wasn't technical enough).
If you can, try to avoid return completely and rely on functions returning the last expression they evaluate.
Update
Just to follow up on the comment below about loops. When you call a loop, you will most likely call one of the built-in primitive functions. And, yes, they return NULL. But you can write your own, and they will follow the rule that they return the last expression they evaluate. You can, for example, implement for in terms of while like this:
`for` <- function(itr_var, seq, body) {
itr_var <- as.character(substitute(itr_var))
body <- substitute(body)
e <- parent.frame()
j <- 1
while (j < length(seq)) {
assign(x = itr_var, value = seq[[j]], envir = e)
eval(body, envir = e)
j <- j + 1
}
"foo"
}
This function, will definitely return "foo", so this
for(i in 1:5) { print(i) }
evalutes to "foo". If you want it to return NULL, you have to be explicit about it (or just let the return value be the result of the while loop -- if that is the primitive while it returns NULL).
The point I want to make is that functions return the last expression they evaluate has to do with how the functions are defined, not how you call them. The loops use non-standard evaluation, so the last expression in the loop body you provide them might be the last value they evaluate and might not. For the primitive loops, it is not.
Except for their special syntax, there is nothing magical about loops. They follow the rules all functions follow. With non-standard evaluation it can get a bit tricky to work out from a function call what the last expression they will evaluate might be, because the function body looks like it is what the function evaluates. It is, to a degree, if the function is sensible, but the loop body is not the function body. It is a parameter. If it wasn't for the special syntax, and you had to provide loop bodies as normal parameters, there might be less confusion.

calling variable name of a dataframe inside functions in R using "$"

I have a dataframe(master) that has some variables which i have stored in the list below:
cont<-list("Quantity","Amt_per_qty","Trans_tax","Total_trans_amt")
catg<-list("Gender","Region_code","SubCategory")
I am trying to create a function where I can access the variables from dataframe and perform some function on them, though x and val in below function seems to resolve, how can I access the variables using the $ sign inside function
univar<-function (x){
for (val in cont){
print (val)
n<-nrow(x$val) }
print (n) }
univar(master)
Its returning NULL, I tried even with n<-nrow(x[,val]), that also don't seem to work.
i) x[val] returns a data.frame
ii) x[,val,drop = TRUE] returns a vector
iii) x[[val]] shall return as a vector. Advantage of this is : it also works with data.tables
n <- nrow(x) or length(x[[val]])
The reason is that the OP created a list, it could be unlisted and then use [
cont <- unlist(cont)
univar<-function(x){
for (val in cont){
print (val)
n<-nrow(x[[val]]) }
print (n) }
univar(master)

How to use S4 object programming in R

What's wrong with my R script? I'm trying to use a vector of user-defined objects (here a vector of "Page" objects) within another user-defined object (here a "Book" object)
setClass("Page",
slots = c(PageNo = "numeric", #scalar
Contents = "character") #vector of strings
)
setClass("Book",
slots = c(Pages = "vector", # Something wrong here? vector of pages ? "Page" or vector" or "list"
Title = "character") #vector of strings
)
setGeneric(name="AddPage", def=function(aBook, pageNo){standardGeneric("AddPage")})
setMethod(f="AddPage", signature="Book",
definition=function(aBook, pageNo)
{
page1 = new("Page")
page1#PageNo = pageNo
aBook#Pages = c(aBook#Pages, page1) # Something wrong here?
}
)
book1 = new("Book")
book1#Title = "Sample Book"
book1
book1#Pages
AddPage(book1, 1)
AddPage(book1, 2)
book1#Pages
Remember that R does not use reference semantics, so AddPage(book1, 1) creates a copy of book1, and updates that. In the method you don't return the updated object, and book1 remains unchanged.
Update the method so that it returns the modified object
setMethod(f="AddPage", signature="Book",
definition=function(aBook, pageNo)
{
page1 = new("Page")
page1#PageNo = pageNo
aBook#Pages = c(aBook#Pages, page1) # Something wrong here?
aBook
}
)
and assign the return value to the old variable
book1 = AddPage(book1, 1)
But this is a very inefficient approach -- the line aBook#Pages = c(aBook#Pages, page1) makes a copy of all existing pages (on the right-hand side, to create a longer vector; this will scale with the square of the number of Pages added to the book) and then copies the entire Book (for the assignment). In addition, creating individual objects is expensive and does not exploit R's 'vectorization'. A first step is to think of the object 'Page' as instead 'Pages', where the object models the columns rather than rows of a data frame. 'Book' then doesn't have vector of Page objects, but a single Pages object. This also implies a different approach to creating your 'book'.

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