I am exploring using R optim() or optimx() for a (very) nonlinear optimization. Essentially I wrote a function that takes as its inputs:
1) a data.frame with specific column names/types
2) a numeric vector of length 1
3) a numeric vector of length > 1
The function then takes the inputs, performs some calculations and logic tests, then either returns a very negative value if the logic tests are FALSE, or the value of input #2 if the logic tests are TRUE. The goal is to maximize #2 without tripping the logic tests to FALSE.
I tried using optimx() with the following code (the par values correspond to the inputs I refer to above):
optimout <-
optimx:::optimx(
par = c(inputDF, 5000, rep(99,20)),
fn = MyFunction,
maximize = TRUE)
I received the following error message:
Error in optimx.check(par, optcfg$ufn, optcfg$ugr, optcfg$uhess, lower, :
Cannot evaluate function at initial parameters
Ralph
If I understand your problem correctly and you only want to maximize the function based the second input parameter (a numeric vector of length 1), you need to call optimxdifferently, which would make sense given that the data.frame is probably some given input data.
So, try to do the following:
optimout <- optimx(par = c(5000), fn = MyFunction, par1=inputDF,
par2=rep(99,20), maximize = TRUE)
where par1 and par2 are the names of the input variables for your function. Essentially, you are providing optimxwith initial values for the input parameters par1 & par2, which are then not being optimized. Thus, the maximum is searched by only changing the value of your second parameter (a numeric vector of length 1), which you chose to start at 5000.
Related
I have a function below and I have K to be a range of values and want to set lambda =2.
When I run the code I get an error in R- argument k is missing with no default.
What am I doing wrong?
lnmp <- function(mu, k) {
n <- length(k)
-n*mu + log(mu)*sum(k) - sum(lfactorial(k))
}
k<-c(5,12,10,22,45,67,39,28,45,53,45,41,4,lamhda=2)
lnmp(k)
You set your function lnmp with two inputs, mu and k. However, you only passed one. The important thing is that it doesn't matter what the argument you passed is named, it's going to be the first one. If you name it dog, it's still going to be treated as mu inside the function. Since you didn't pass a second argument, k is not defined (it has no value). Therefore, what you have to do is pass both arguments when you call the function (therefore, lnmp(first, second) or lnmp(mu, k)).
I'm trying to calculate crps using the verification package in R. The data appears to read in ok, but I get an error when trying to compute the CRPS itself: "invalid 'times' argument", however all values are real, no negative values and I'm testing for nan/na values and ignoring those. Having searched around I can't find any solution which explains why I'm getting this error. I'm reading the data in from netcdf files into larger arrays, and then computing CRPS for each grid cell in those arrays.
Any help would be greatly appreciated!
The relevant snipped from the code I'm using is:
##for each grid cell, get obs (wbarray) and 25 ensemble members of forecast eps (fcstarray)
for(x in 1:3600){
for(y in 1:1500){
obs=wbarray[x,y]
eps=fcstarray[x,y,1:25]
if(!is.na(obs)){
print(obs)
print(eps)
print("calculating CRPS - real value found")
crpsfcst=(crpsDecomposition(obs,eps)$CRPS)
CRPSfcst[x,y,w]=crpsfcst}}}
(w is specified in an earlier loop)
And the output I get:
obs: 0.3850737
eps: 0.3382506 0.3466184 0.3508921 0.3428135 0.3416993 0.3423528 0.3307764
0.3372431 0.3394377 0.3398165 0.3414395 0.3531360 0.3319155 0.3453161
0.3362813 0.3449474 0.3340050 0.3278898 0.3380596 0.3379150 0.3429202
0.3467927 0.3419354 0.3472489 0.3550797
"calculating CRPS - real value found"
Error in rep(0, nObs * (nMember +1)) : invalid 'times' argument
Calls: crpsDecomposition
Execution halted
If you type crpsDecomposition on your R command prompt you'll get the source code for the function. The first few lines show:
function (obs, eps)
{
nMember = dim(eps)[2]
nObs <- length(obs)
Since your eps data object appears to be (from your output) a one-dimensional vector, the second element of its dimension is going to be NULL, which sets nMember to NULL. Thus nObs*(nMember + 1) gets evaluated to 0. I imagine you simply need to re-examine what form eps should take because it would appear that it needs to be a matrix where each column corresponds to a different "member" (whatever that means in this context).
My apologies if this is kind of vague question as I am new to R. While experimenting with R I found one weird behavior. When I create a function like:
myfunction <- function(a,b){
print(a,b)
}
and call it like:
myfunction(b = 10, a = 20)
it returns with result 20, but if I simply call it without function via assigning it directly to variables like:
a <- 20
b <- 10
print(a, b)
I get an error:
Error in print.default(a, b) : invalid 'digits' argument
Furthermore I have read that printing multiple variables in the same line can be accomplished via:
sprintf("%i %i",a, b)
So here is it a bug that it is appearing in function call with result as the first argument?
It might be revealing some underlying differences in how parameters are being handled in different scenarios but I don't think it's a bug.
If your intention is to print both values, consider changing:
print(a,b)
To something like:
print(paste(a,b))
From ?print.default:
# S3 method for default
print(x, digits = NULL, quote = TRUE,
na.print = NULL, print.gap = NULL, right = FALSE,
max = NULL, useSource = TRUE, …)
x the object to be printed.
digits a non-null value for digits specifies the minimum number of
significant digits to be printed in values. The default, NULL, uses
getOption("digits"). (For the interpretation for complex numbers see
signif.) Non-integer values will be rounded down, and only values
greater than or equal to 1 and no greater than 22 are accepted.
...
So R is expecting everything that you actually want to print to be contained in the first variable (x).
Based on your results and some of the comments, apparently in some cases the second variable is being accepted as a valid digits parameter value and in other cases it is not.
While this is a little odd, the more important point is that print(a,b) is not a syntactically correct way to print multiple values.
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.
I have a some true and predicted labels
truth <- factor(c("+","+","-","+","+","-","-","-","-","-"))
pred <- factor(c("+","+","-","-","+","+","-","-","+","-"))
and I would like to build the confusion matrix.
I have a function that works on unary elements
f <- function(x,y){ sum(y==pred[truth == x])}
however, when I apply it to the outer product, to build the matrix, R seems unhappy.
outer(levels(truth), levels(truth), f)
Error in outer(levels(x), levels(x), f) :
dims [product 4] do not match the length of object [1]
What is the recommended strategy for this in R ?
I can always go through higher order stuff, but that seems clumsy.
I sometimes fail to understand where outer goes wrong, too. For this task I would have used the table function:
> table(truth,pred) # arguably a lot less clumsy than your effort.
pred
truth - +
- 4 2
+ 1 3
In this case, you are test whether a multivalued vector is "==" to a scalar.
outer assumes that the function passed to FUN can take vector arguments and work properly with them. If m and n are the lengths of the two vectors passed to outer, it will first create two vectors of length m*n such that every combination of inputs occurs, and pass these as the two new vectors to FUN. To this, outer expects, that FUN will return another vector of length m*n
The function described in your example doesn't really do this. In fact, it doesn't handle vectors correctly at all.
One way is to define another function that can handle vector inputs properly, or alternatively, if your program actually requires a simple matching, you could use table() as in #DWin 's answer
If you're redefining your function, outer is expecting a function that will be run for inputs:
f(c("+","+","-","-"), c("+","-","+","-"))
and per your example, ought to return,
c(3,1,2,4)
There is also the small matter of decoding the actual meaning of the error:
Again, if m and n are the lengths of the two vectors passed to outer, it will first create a vector of length m*n, and then reshapes it using (basically)
dim(output) = c(m,n)
This is the line that gives an error, because outer is trying to shape the output into a 2x2 matrix (total 2*2 = 4 items) while the function f, assuming no vectorization, has given only 1 output. Hence,
Error in outer(levels(x), levels(x), f) :
dims [product 4] do not match the length of object [1]