I am normally a maple user currently working with R, and I have a problem with correctly indexing variables.
Say I want to define 2 vectors, v1 and v2, and I want to call the nth element in v1. In maple this is easily done:
v[1]:=some vector,
and the nth element is then called by the command
v[1][n].
How can this be done in R? The actual problem is as follows:
I have a sequence M (say of length 10, indexed by k) of simulated negbin variables. For each of these simulated variables I want to construct a vector X of length M[k] with entries given by some formula. So I should end up with 10 different vectors, each of different length. My incorrect code looks like this
sims<-10
M<-rnegbin(sims, eks_2016_kasko*exp(-2.17173), 840.1746)
for(k in 1:sims){
x[k]<-rep(NA,M[k])
X[k]<-rep(NA,M[k])
for(i in 1:M[k]){x[k][i]<-runif(1,min=0,max=1)
if(x[k][i]>=0 & x[i]<=0.1056379){
X[k][i]<-rlnorm(1, 6.228244, 0.3565041)}
else{
X[k][i]<-rlnorm(1, 8.910837, 1.1890874)
}
}
}
The error appears to be that x[k] is not a valid name for a variable. Any way to make this work?
Thanks a lot :)
I've edited your R script slightly to get it working and make it reproducible. To do this I had to assume that eks_2016_kasko was an integer value of 10.
require(MASS)
sims<-10
# Because you R is not zero indexed add one
M<-rnegbin(sims, 10*exp(-2.17173), 840.1746) + 1
# Create a list
x <- list()
X <- list()
for(k in 1:sims){
x[[k]]<-rep(NA,M[k])
X[[k]]<-rep(NA,M[k])
for(i in 1:M[k]){
x[[k]][i]<-runif(1,min=0,max=1)
if(x[[k]][i]>=0 & x[[k]][i]<=0.1056379){
X[[k]][i]<-rlnorm(1, 6.228244, 0.3565041)}
else{
X[[k]][i]<-rlnorm(1, 8.910837, 1.1890874)
}
}
This will work and I think is what you were trying to do, BUT is not great R code. I strongly recommend using the lapply family instead of for loops, learning to use data.table and parallelisation if you need to get things to scale. Additionally if you want to read more about indexing in R and subsetting Hadley Wickham has a comprehensive break down here.
Hope this helps!
Let me start with a few remarks and then show you, how your problem can be solved using R.
In R, there is most of the time no need to use a for loop in order to assign several values to a vector. So, for example, to fill a vector of length 100 with uniformly distributed random variables, you do something like:
set.seed(1234)
x1 <- rep(NA, 100)
for (i in 1:100) {
x1[i] <- runif(1, 0, 1)
}
(set.seed() is used to set the random seed, such that you get the same result each time.) It is much simpler (and also much faster) to do this instead:
x2 <- runif(100, 0, 1)
identical(x1, x2)
## [1] TRUE
As you see, results are identical.
The reason that x[k]<-rep(NA,M[k]) does not work is that indeed x[k] is not a valid variable name in R. [ is used for indexing, so x[k] extracts the element k from a vector x. Since you try to assign a vector of length larger than 1 to a single element, you get an error. What you probably want to use is a list, as you will see in the example below.
So here comes the code that I would use instead of what you proposed in your post. Note that I am not sure that I correctly understood what you intend to do, so I will also describe below what the code does. Let me know if this fits your intentions.
# define M
library(MASS)
eks_2016_kasko <- 486689.1
sims<-10
M<-rnegbin(sims, eks_2016_kasko*exp(-2.17173), 840.1746)
# define the function that calculates X for a single value from M
calculate_X <- function(m) {
x <- runif(m, min=0,max=1)
X <- ifelse(x > 0.1056379, rlnorm(m, 6.228244, 0.3565041),
rlnorm(m, 8.910837, 1.1890874))
}
# apply that function to each element of M
X <- lapply(M, calculate_X)
As you can see, there are no loops in that solution. I'll start to explain at the end:
lapply is used to apply a function (calculate_X) to each element of a list or vector (here it is the vector M). It returns a list. So, you can get, e.g. the third of the vectors with X[[3]] (note that [[ is used to extract elements from a list). And the contents of X[[3]] will be the result of calculate_X(M[3]).
The function calculate_X() does the following: It creates a vector of m uniformly distributed random values (remember that m runs over the elements of M) and stores that in x. Then it creates a vector X that contains log normally distributed random variables. The parameters of the distribution depend on the value x.
Related
I am trying to generate the term frequency matrix of a document and subsequently look up the frequency of a certain word in a given query in that matrix. In the end I want to sum the frequencies found of the words in the query.
However, I am coping with the error message: Error in feature[i] <- x : replacement has length zero
I do not have a lot of coding experience in general, and this is my first time working with R, thus I am having difficulties solving this error. I presume it has something to do with a null-value. I already tried to avoid the nested for-loop with an apply function because I thought that might help (not sure though), but I could not quite get the hang of how to convert the for-loop into an apply function.
termfreqname <- function(queries,docs){
n <- length(queries)
feature <- vector(length=n)
for(i in 1:n){
query <- queries[i]
documentcorpus <- c(docs[i])
tdm <- TermDocumentMatrix(tm_corpus) #creates the term frequency matrix per document
m <- sapply(strsplit(query, " "), length) #length of the query in words
totalfreq <- list(0) #initialize list
freq_counter <- rowSums(as.matrix(tdm)) #counts the occurrence of a given word in the tdm matrix
for(j in 1:m){
freq <- freq_counter[word(query,j)] #finds frequency of each word in the given query, in the term frequency matrix
totalfreq[[j]] <- freq #adds this frequency to position j in the list
}
x <- reduce(totalfreq,'+') #sums all the numbers in the list
feature[i] <- x #adds this number to feature list
feature
}
}
It depends on your needs, but bottom line you need to add some if statement. How you use it depends on whether you want the default value of the vector to persist. In your code, while feature starts as a logical vector, it is likely coerced to integer or numeric once you overwrite its first value with a number. In that case, the default value in all positions of the vector will be 0 (or 0L, if integer). That's going to influence your decision on how to use the if statement.
if (length(x)) feature[i] <- x
This will only attempt to overwrite the ith value of feature if the x objects has length (that's equivalent to if (length(x) > 0)). In this case, since the default value in the vector will be zero, this means when you are done that you will not be able to distinguish between an element known to be 0 and an element that failed to find anything.
The alternative (and my preference/recommendation):
feature[i] <- if (length(x)) x else NA
In this case, when you are done, you can clearly distinguish between known-zero (0) and uncertain/unknown values (NA). When doing math operations on that vector, you might want/need na.rm=TRUE ... but it all depends on your use.
BTW, as MartinGal noted, your use of reduce(totalfreq, '+') is a little flawed: 'x' may not be (is not?) recognized as a known function. The first fix to this is to use backticks around the function, so
totalfreq <- 5:7
reduce(totalfreq, '+')
# NULL
reduce(totalfreq, `+`)
# [1] 18
sum(totalfreq)
# [1] 18
There the last is the much-more-preferred method. Why? With a vector of length 4, for instance, it takes the first two and adds them, then takes that result and adds it to the third, then takes that result and adds to the fourth. Three operations. When you have 100 elements, it will make 99 individual additions. sum does it once, and this does have an effect on performance (asymptotically).
However, if totalfreq is instead a list, then this changes slightly:
totalfreq <- as.list(5:7)
reduce(totalfreq, `+`)
# [1] 18
sum(totalfreq)
# Error in sum(totalfreq) : invalid 'type' (list) of argument
# x
sum(unlist(totalfreq))
# [1] 18
The reduce code still works, and the sum by itself fails, but we can unlist the list first, effectively creating a vector, and then call sum on that. Much much faster asymptotically. And perhaps clearer, more declarative.
(I'm assuming purrr::reduce, btw ...)
I am using the example of calculating the length of the arc around a circle and the area under the arc around a circle based on the radius of the circle (r) and the angle of the the arc(theta). The area and the length are both based on r and theta, and you can calculate them simultaneously in python.
In python, I can assign two values at the same time by doing this.
from math import pi
def circle_set(r, theta):
return theta * r, .5*theta*r*r
arc_len, arc_area = circle_set(1, .5*pi)
Implementing the same structure in R gives me this.
circle_set <- function(r, theta){
return(theta * r, .5 * theta * r *r)
}
arc_len, arc_area <- circle_set(1, .5*3.14)
But returns this error.
arc_len, arc_area <- circle_set(1, .5*3.14)
Error: unexpected ',' in "arc_len,"
Is there a way to use the same structure in R?
No, you can't do that in R (at least, not in base or any packages I'm aware of).
The closest you could come would be to assign objects to different elements of a list. If you really wanted, you could then use list2env to put the list elements in an environment (e.g., the global environment), or use attach to make the list elements accessible, but I don't think you gain much from these approaches.
If you want a function to return more than one value, just put them in a list. See also r - Function returning more than one value.
You can assign multiple variables the same value as below. Even here, I think the code is unusual and less clear, I think this outweighs any benefits of brevity. (Though I suppose it makes it crystal clear that all of the variables are the same value... perhaps in the right context it makes sense.)
x <- y <- z <- 1
# the above is equivalent to
x <- 1
y <- 1
z <- 1
As Gregor said, there's no way to do it exactly as you said and his method is a good one, but you could also have a vector represent your two values like so:
# Function that adds one value and returns a vector of all the arguments.
plusOne <- function(vec) {
vec <- vec + 1
return(vec)
}
# Creating variables and applying the function.
x <- 1
y <- 2
z <- 3
vec <- c(x, y, z)
vec <- plusOne(vec)
So essentially you could make a vector and have your function return vectors, which is essentially filling 3 values at once. Again, not what you want exactly, just a suggestion.
I'm trying to understand the answer to this question using R and I'm struggling a lot.
The dataset for the R code can be found with this code
library(devtools)
install_github("genomicsclass/GSE5859Subset")
library(GSE5859Subset)
data(GSE5859Subset) ##this loads the three tables you need
Here is the question
Write a function that takes a vector of values e and a binary vector group coding two groups, and returns the p-value from a t-test: t.test( e[group==1], e[group==0])$p.value.
Now define g to code cases (1) and controls (0) like this g <- factor(sampleInfo$group)
Next use the function apply to run a t-test for each row of geneExpression and obtain the p-value. What is smallest p-value among all these t-tests?
The answer provided is
myttest <- function(e,group){
x <- e[group==1]
y <- e[group==0]
return( t.test(x,y)$p.value )
}
g <- factor(sampleInfo$group)
pvals <- apply(geneExpression,1,myttest, group=g)
min( pvals )
Which gives you the answer of 1.406803e-21.
What exactly is the input of the "e" argument of the myttest function when you run this? Is it possible to write this function as a formula like
t.test(DV ~ sampleInfo$group)
The t test is comparing the gene expression values of the 24 people (the values of which I believe are in the "geneExpression" matrix) by what group they were
in which you can find in sampleInfo's "group" column. I've run t tests so many times in R, but for some reason I can't wrap my mind around what's going on in this code.
You question seems to be about understanding the function apply().
For the technical description, see ?apply.
My quick explanation: the apply() line of code in your question applies the following function to each of the rows of geneExpression
myttest(e=x, group=g)
where x is a placeholder for each row.
To help make sense of it, a for loop version of that apply() line would look something like:
N <- nrows(geneExpression) #so we don't have to type this twice
pvals <- numeric(N) #empty vector to store results
# what 'apply' does (but it does it very quickly and with less typing from us)
for(i in 1:N) {
pvals[i] <- myttest(geneExpression[i,], group=g[i])
}
Say I have a function func that takes two scalar numeric inputs and delivers a scalar numeric result, and I have the following code to calculate a result vector u, based on input numeric vector v and initial value u0 for the result vector:
u<-rep(u0,1+length(v))
for (k in 2:length(u)){
u[k]<-func(u[k-1],v[k-1])
}
Note how a component of the result vector depends not only on the corresponding element of the input vector but also on the immediately prior element of the result vector. I can see no obvious way to vectorise this.
It is common to do this sort of thing in financial simulations, for instance when projecting forward company accounts, rolling them up with interest or inflation and adding in operational cash flows each year.
For some specific instances, it is possible to find a case-specific, non-iterative coding, but I would like to know if there's a general solution.
The problem can also be coded by recursion, as follows:
calc.u<-function(v,u0){
if (length(v)<2){
func(u0,v[1]) }
else {
u.prior<-func(u0,v[-length(v),drop=FALSE])
c(u.prior,func(u.prior[length(u.prior)],v[length(v)]) )
}
u<-calc.u(v,u0)
Is there an R tactic for doing this without using either iteration or recursion, ie for vectorising it?
Answered: Thank you #MrFlick for introducing me to the Reduce function, which does exactly what I was wanting. I see that
Reduce('+',v,0,accumulate=T)[-1]
gives me
cumsum(v)
and
Reduce('*',v,0,accumulate=T)[-1]
gives me
cumprod(v)
as expected, where the [-1] is to discard the initial value.
Very nice indeed! Thanks again.
If you have this example
u0 <- 5
v <- (1:5)*2
func <- function(u,v) {u/2+v}
u <- rep(u0,1+length(v))
for (k in 2:length(u)){
u[k]<-func(u[k-1],v[k-1])
}
this is equivalent to
w <- Reduce(func, v, u0, accumulate=TRUE)
And we can check that
all(u==w)
# [1] TRUE
I have this parameter:
L_inf <- seq(17,20,by=0.1)
and this function:
fun <- function(x){
L_inf*(1-exp(-B*(x-0)))}
I would to apply this function for a range of value of L_inf.
I tried with loop for, like this:
A <- matrix() # maybe 10 col and 31 row or vice versa
for (i in L_inf){
A[i] <- fun(1:10)
}
Bur R respond: longer object length is not a multiple of shorter object length.
My expected output is a matrix (or data frame, or list maybe) with 10 result (fun(1:10)) for each value of the vector L_inf (lenght=31).
How can to do it?
You are trying to put a vector of 10 elements into one of the matrix cell. You want to assign it to the matrix row instead (you can access the ith row with A[i,]).
But using a for loop in this case is inefficient and it is quite straightforward to use one of the "apply" function. Apply functions typically return a list (which is the most versatile container since there is basically no constraint).
Here sapply is an apply function which tries to Simplify its result to a convenient data structure. In this case, since all results have the same length (10), sapply will simplify the result to a matrix.
Note that I modified your function to make it explicitly depend on L_inf. Otherwise it will not do what you think it should do (see keyword "closures" if you want more info).
L_inf_range <- seq(17,20,by=0.1)
B <- 1
fun <- function(x, L_inf) {
L_inf*(1-exp(-B*(x-0)))
}
sapply(L_inf_range, function(L) fun(1:10, L_inf=L))