Here's a super newbie question. I'm working through The Art of R Programming and came across this simple code:
z <- NULL
for (i in 1:10) if (i %% 2 == 0) z <- c(z,i)
z
the output is c(2,4,6,8,10).
What I don't understand is why you need c(z,i) to populate the vector z. Why wouldn't c(i) work. When I do that, it skips straight to the last iteration and fills vector z with the last associated value. Is there some form of rule that governs populating vectors? Try as I might, I didn't find a clear answer anywhere, just a bunch of samples that, while they made sense, didn't help me solve the above.
Related
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.
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've tried a couple ways of doing this problem but am having trouble with how to write it. I think I did the first three steps correctly, but now I have to fill the vector z with numbers from y that are divisible by four, not divisible by three, and have an odd number of digits. I know that I'm using the print function in the wrong way, I'm just at a loss on what else to use ...
This is different from that other question because I'm not using a while loop.
#Step 1: Generate 1,000,000 random, uniformly distributed numbers between 0
#and 1,000,000,000, and name as a vector x. With a seed of 1.
set.seed(1)
x=runif(1000000, min=0, max=1000000000)
#Step 2: Generate a rounded version of x with the name y
y=round(x,digits=0)
#Step 3: Empty vector named z
z=vector("numeric",length=0)
#Step 4: Create for loop that populates z vector with the numbers from y that are divisible by
#4, not divisible by 3, with an odd number of digits.
for(i in y) {
if(i%%4==0 && i%%3!=0 && nchar(i,type="chars",allowNA=FALSE,keepNA=NA)%%2!=0){
print(z,i)
}
}
NOTE: As per #BenBolker's comment, a loop is an inefficient way to solve your problem here. Generally, in R, try to avoid loops where possible to maximise the efficiency of your code. #SymbolixAU has provided an example of doing so here in the comments. Having said that, in aid of helping you learn the ins-and-outs of loops and vectors, here's a solution which only requires a change to one line of your code:
You've got the vector created before the loop, that's a good start. Now, inside your loop, you need to populate that vector. To do so, you've currently got print(z,i), which won't really do too much. What you need to to change the vector itself:
z <- c( z, i )
Should work for you (just replace that print line in your loop).
What's happening here is that we're taking the existing z vector, binding i to the end of it, and making that new vector z again. So every time a value is added, the vector gets a little longer, such that you'll end up with a complete vector.
where you have print put this instead:
z <- append(z, i)
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.
How do I grab elements from a table in R?
My data looks like this:
V1 V2
1 12.448 13.919
2 22.242 4.606
3 24.509 0.176
etc...
I basically just want to grab elements individually. I'm getting confused with all the R terminology, like vectors, and I just want to be able to get at the individual elements.
Is there a function where I can just do like data[v1][1] and get the element in row 1 column 1?
Try
data[1, "V1"] # Row first, quoted column name second, and case does matter
Further note: Terminology in discussing R can be crucial and sometimes tricky. Using the term "table" to refer to that structure leaves open the possibility that it was either a 'table'-classed, or a 'matrix'-classed, or a 'data.frame'-classed object. The answer above would succeed with any of them, while #BenBolker's suggestion below would only succeed with a 'data.frame'-classed object.
There is a ton of free introductory material for beginners in R: CRAN: Contributed Documentation
?"[" pretty much covers the various ways of accessing elements of things.
Under usage it lists these:
x[i]
x[i, j, ... , drop = TRUE]
x[[i, exact = TRUE]]
x[[i, j, ..., exact = TRUE]]
x$name
getElement(object, name)
x[i] <- value
x[i, j, ...] <- value
x[[i]] <- value
x$i <- value
The second item is sufficient for your purpose
Under Arguments it points out that with [ the arguments i and j can be numeric, character or logical
So these work:
data[1,1]
data[1,"V1"]
As does this:
data$V1[1]
and keeping in mind a data frame is a list of vectors:
data[[1]][1]
data[["V1"]][1]
will also both work.
So that's a few things to be going on with. I suggest you type in the examples at the bottom of the help page one line at a time (yes, actually type the whole thing in one line at a time and see what they all do, you'll pick up stuff very quickly and the typing rather than copypasting is an important part of helping to commit it to memory.)
Maybe not so perfect as above ones, but I guess this is what you were looking for.
data[1:1,3:3] #works with positive integers
data[1:1, -3:-3] #does not work, gives the entire 1st row without the 3rd element
data[i:i,j:j] #given that i and j are positive integers
Here indexing will work from 1, i.e,
data[1:1,1:1] #means the top-leftmost element