I have script main.R, where I create inv_cov_mat variable. I later load metrics.R and use it to calculate function value (I use it as kind of inter-script function closure). I get error "object 'inv_cov_mat' not found". My code:
main.R:
knn <- function(...)
{
# some code
source("./source/metrics.R")
if (metric == "mahalanobis")
inv_cov_mat <- solve(cov(training_set))
# other code
# calculate distance in given metric between current vector and every row vector from training set matrix
distances <- apply(training_set, 1, metric, vec2=curr_vec) # error
metrics.R:
mahalanobis <- function(vec1, vec2)
{
diff <- vec1 - vec2
sqrt(t(diff) %*% inv_cov_mat %*% diff)
}
I've found simple, even if not elegant answer: use inv_cov_mat as global variable, not inside knn function. Then other scripts can see it.
It's not entirely clear what you want, but if I understand you correctly---you have a character string identifying the metric you want to use, and a function with the same name. So you should be able to use get to retrieve the function based on the name.
metric == "mahalanobis"
metric.fun = get(metric)
distances <- apply(training_set, 1, metric.fun, vec2=curr_vec)
That said, there are probably better ways to organize your code that would avoid this problem entirely, e.g. create a named list of functions for accessing metrics.
EDIT regarding the issue of inv_cov_mat, either pass it as an argument to your metric function or use get inside that function to access variables from the parent environment using the envir argument. Passing the variable as an argument to your metric function is definitely the better and cleaner approach.
Related
testing<-function(formula=NULL,data=NULL){
if(with(data,formula)==T){
print('YESSSS')
}
}
A<-matrix(1:16,4,4)
colnames(A)<-c('x','y','z','gg')
A<-as.data.frame(A)
testing(data=A,formula=(2*x+y==Z))
Error in eval(expr, envir, enclos) : object 'x' not found
##or I can put formula=(x=1)
##reason that I use formula is because my dataset had different location and I would want
##to 'subset' my data into different set
This is the main flow of my code. I had done some search and seems to be no one ask this kind of stupid question or it is not possible to pass a formula in a if statement. Thank you in advance
if you just want subset of your data.frame create a character object representing the formula like this:
formula="2*x+y==z"
testing<-function(data,formula){with(data = data,expr = eval(parse(text = formula)))}
subset(A,testing(A,formula=formula))
#x y z gg
#2 2 6 10 14
You can change the formula as per your need.
If we need to evaluate it, one option is eval(parse
testing<-function(formula=NULL,data=NULL){
data <- deparse(substitute(data))
if(any(eval(parse(text=paste("with(", data, ",",
deparse(substitute(formula)), ")")))))
print("YESSS")
}
testing(data=A,formula=(2*x+y==z))
#[1] "YESSS"
When you call a function in R it evaluates its arguments first before executing the function.
For example, prod(2+2, 3) is first turned into prod(4, 3) before the function prod() is even called.
Thus, in your code, R starts by trying to solve (2*x+y==Z). It fails because there is no x object outside of the function code. So, it not even begin running testing().
To use your function correctly you should make it clear to R that it is not supposed to calculate (2*x+y==Z). Instead it should pass this information as is. You could do that using the functions expression() and eval().
testing<-function(formula=NULL,data=NULL){
if(with(data,eval(formula==T)){
print('YESSSS')
}
}
A<-matrix(1:16,4,4)
colnames(A)<-c('x','y','z','gg')
A<-as.data.frame(A)
testing(data=A,formula=expression(2*x+y==Z))
However, you will notice that there other problems with your code.
For Z is different than z. Notice that the in colnames you use z and in the formula Z.
The if() only works for when there is a single value of true or false. In your case, you will have one value for each row in A. When this happens, if() will only check if the first row fits the criteria.
If your purpose is subsetting, it is much more easier to do:
A.subset <- subset(A, 2*A$x+A$y == A$z)
After a discussion with my colleague,
here is a kind of solution
testing<-function(cx,cy,px,py,z,data=NULL){
list<-NULL
for(m in 1:nrow(data)){
if(cx*data$x[m]^px+cy*data$y[m]^py+data$z==0){
print(m)}
}
}
but this can deal with polynomial only and with a lot of arguments in the function. I am think of a way to reduce it as a general equation.or maybe this is the most easiest equation.
I am a beginner in R and need to use it to estimate a MLE. I am using the mle2 function. However since i have many independent variables, I need to pass many parameter values when calling mle2. For example
library(bbmle)
x <- mle2(probit, start=list(b0=1,b1=1,b2=1,c0=1,c1=1,d0=1,d1=1),method="BFGS")
Instead, I would like to create a vector theta of length 7 and pass that when i call mle2. Something like
x <- mle2(probit, start=theta,method="BFGS")
Exactly this does not work. Do I need to change how I define the function probit accordingly to use a vector as argument?
I went through the existing threads related to passing vectors as arguments and they suggest using do.call, but can that be used to call mle2? If so, how?
many thanks!
If I understand correctly, your goal is to reduce boilerplate by finding a way to pass a single vector of parameter values. You could achieve this by defining a utility function which converts a single vector of un-named entries into a named list suitable for your model and for mle2:
## vector of un-named entries
theta <- c(1,1,1,1,1,1,1)
## utility function
toList <- function(th) structure(as.list(th), names=c("b0","b1","b2","c0","c1","d0","d1"))
## check
identical(toList(theta), list(b0=1,b1=1,b2=1,c0=1,c1=1,d0=1,d1=1))
## now this should work
x <- mle2(probit, start=toList(theta), method="BFGS")
I am trying to use the package Deriv, to compute symbolic derivatives of a function depending on one or two variables and a vector of parameters. However, i always obtain the error:
Error in FUN(X[[i]], ...) : Could not retrieve body of '[()'
I have tried:
test_fun <- function(x,y,par){x\*par[1]+y\*par[2]}
Deriv(test_fun,"x",par=c(2,2))
which yields the above error. So does
par <- c(2,2)
test_fun <- function(x,y,par){x\*par[1]+y\*par[2]}
Deriv(test_fun,"x")
Obviously
test_fun <- function(x,y,par){x\*2+y\*2}
Deriv(test_fun,"x")
works as intended, but is not what I want.
Reading the Documentation for the Deriv-package, it seems that directly passing additional arguments to the function is not supported. Is there any other way to achieve the desired result?
Updated answer (9/16):
There's two options below for passing the parameters through Deriv. Also note that Derive will retain curly braces in the original function, so either define your functions without them or call eval() afterwards.
Option 1. Define within the call to Deriv
test_fun <- function(x,y,par){x*par[1]+y*par[2]}
eval(Deriv(test_fun(par=c(2,2)),'x'))
# [1] 2
Option 2. Define the parameter values in as a function.
test_fun <- function(x,y,par){x*par[1]+y*par[2]}
tpar <- function() c(2,2)
eval(Deriv(test_fun(par=tpar()), "x"))
# [1] 2
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'm attempting to use the bnlearn package to calculate conditional probabilities, and I'm running into a problem when the "cpquery" function is used within a loop. I've created an example, shown below, using data included with the package. When using the cpquery function in a loop, a variable created in the loop ("evi" in the example) is not recognized by the function. I receive the error:
Error in parse(text = evi) : object 'evi' not found
The creation steps of "evi" are based on examples provided by the author.
Any help you could provide would be great. I'm desperate to find a way that I can apply the cpquery function for a large number of observations.
library(bnlearn)
data(learning.test)
fitted = bn.fit(hc(learning.test), learning.test)
bn.function <- function(network, evidence_data) {
a <- NULL
b <- nrow(evidence_data)
for (i in 1:b) {
evi <- paste("(", names(evidence_data), "=='",
sapply(evidence_data[i,], as.character), "')",
sep = "", collapse = " & ")
a[i] <- cpquery(network, (C=='c'), eval(parse(text=evi)))
}
return(a)
}
test <- bn.function(fitted, learning.test)
Thanks in advance!
I don't know if this is due to a bugfix or just because I tried another approach - anyways, looping works if you iteratively build up the evidence list outside of the cpquery-function.
An example for an iteration through a list called evidenceData with all-positive evidences:
for(i in names(evidenceData)){
loopEvidenceList <- list()
loopEvidenceList[[i]] <- "TRUE"
a =cpquery(fitted = bayesNet, event = queryNode == "TRUE",
evidence = loopEvidenceList, method = "lw", n = 100000)
print(a)
}
Depending on the way your evidence is availible, you might need more sophisticated preparation of the "loopEvidenceList" but once you got that prepared, it works fine.
To avoid the scoping problem, you can postpone the call to eval and do it inside the cpquery function. If you directly pass evi (the character variable) to cpquery and then parse it inside the definition, the chain of environments gets shifted and cpquery will have access to evi.
You can use m.cpquery <- edit(cpquery) to fork your own version of the function and insert the following line at its beginning:
evidence = parse(text = evidence)
and then save your new function.
So the heading of m.cpquery will look like:
> m.cpquery
function (fitted, event, evidence, cluster = NULL, method = "ls",
..., debug = FALSE)
{
evidence = parse(text = evidence)
check.fit(fitted)
check.logical(debug)
...
Now you can use m.cpquery in your own function like before, except we'll pass the plain character variable to it:
a[i] <- m.cpquery(network, (C=='c'), evi)
Note that in the first line of m.cpquery, we only parsed the evidence character variable and didn't call eval on it. cpquery is a front-end to conditional.probability.query (see here) and we're relying on conditional.probability.query's subsequent call to eval.
I should say that this is a rather ugly workaround. And it only works if you are using logic sampling (method='ls'). But if you want to use likelihood weighting, the check.mutilated.evidence function will raise an error. I haven't checked if injecting an eval expression before it gets called would result in a mayhem of subsequent errors leading to hell.
I feel like the problem is you are using the same variable in evidence as well as event. Learning.test contains the values of "C" variable. then we are trying to predict C as the event. Maybe using a subset of the original dataset excluding C will do the trick