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.
Related
I am trying to analyse a dataframe using hierarchical clustering hclust function in R.
I would like to pass in a vector of p values I'll write beforehand (maybe something like c(5/4, 3/2, 7/4, 9/4)) and be able to have these specified as the different p value options with Minkowski distance when I use expand.grid. Ideally, when hyperparams is viewed, it would also be clear which value of p has been used for each minkowski, i.e. they should be labelled. So for example, where (if you run my code for hyperparams) there would currently just be one minkowski under Dists, for each of the methods in Meths, there would be, if I supplied the p vector as c(5/4, 3/2, 7/4, 9/4), now instead 4 rows for Minkowski distance: minkowski, p=5/4, minkowski, p=3/2, minkowski, p=7/4, minkowski, p=9/4 (or looking something like that, making the p values clear). Any ideas?
(Note: no packages please, only base R!)
Edit: I worded it poorly before, now rewritten. Let's take the following example instead:
acc <- function(x){
first = sum(x)
second = sum(x^2)
return(list(First=first,Second=second))
}
iris0 <- iris
iris1 <- cbind(log(iris[,1:4]),iris[5])
iris2 <- cbind(sqrt(iris[,1:4]),iris[5])
Now the important bit:
tests <- expand.grid(Dists=c("euclidean","maximum","manhattan","canberra","binary"),
DS=c("iris0","iris1","iris2"))
Table <- Map(function(x, ds){acc(table(ds$Species, cutree(hclust(dist(get(ds)[,1:4], method=x)),3)))},tests[[1]], tests[[2]])
This will work. But now if I want to include a term like "minkowski",p=3 in expand.grid, how would I do it?
tests <- expand.grid(Dists=c("euclidean","maximum","manhattan","canberra","binary","minkowski,p=3"),
DS=c("iris0","iris1","iris2"))
Table <- Map(function(x, ds){acc(table(ds$Species, cutree(hclust(dist(get(ds)[,1:4], method=x)),3)))},tests[[1]], tests[[2]])
This gives an error.
In reality there should be no p argument unless the method="minkowski". I have tried to use strsplit to get the first part of the expression into ds, and a switch with strsplit to get the second part and then use parse (it would return NULL if the length of the strsplit was not 2 -- this should pass no argument, I think). The issue seems to be that strsplit is not strsplit(x,",") fails to evaluate the vectorized x but rather tries to evaluate the character x which is not a string. Can anyone suggest any workaround/fix or other method for including the minkowski,p=1.6 terms and the like?
We can create a 'p' value column
tests <- expand.grid(Dists=c("euclidean","maximum","manhattan","canberra","binary",
"minkowski3", "minkowski4", "minkowski5"),
DS=c("iris0","iris1","iris2"))
Suppose, we have another column of 'p' values in 'tests', the above solution can be changed to
tests$p <- as.list(args(dist))$p # default value
i1 <- grepl("minkowski", tests$Dists)
tests$Dists <- sub("[0-9.]+$", "", tests$Dists)
tests$p[i1] <- rep(3:5, length.out = sum(i1))
Map(function(x, ds, p){
dist1 <- dist(get(ds)[, 1:4], method = x, p = p)
ct <- cutree(hclust(dist1), 3)
acc(table(get(ds)$Species, ct))},
as.character(tests[[1]]), as.character(tests[[2]]), tests$p )
I am trying to replace all values of r for which r<=10 with the value of the 1st observation in x (which is 1). This is just a very simplified example of what I am trying to do, so please do not question why I'm trying to do this in a complicated way because the full code is more complicated. The only thing I need help with is figuring out how to use the vector I created (p1) to replace r[p1] or equivalently r[c(1,2,3,4)] with x[ 1 ] (which is equal to 1). I can not write p1 explicitly because it will be generated in a loop (not shown in code).
x=c(1,2,3)
r=c(1,3,7,10,15)
assign(paste0("p", x[1]), which(r<=10))
p1
r[paste0("p", x[1])]=x[1]
In the code above, I tried using r[paste0("p", x[1])]=x[1] but this is the output I end up with
When instead I would like to see this output
Basically, I need to figure out a way to call p1 in this code r[??]=x[1] without explicitly typing p1.
I have included the full code I am attempting below in case context is needed.
##Creates a function to generate discrete random values from a specified pmf
##n is the number of random values you wish to generate
##x is a vector of discrete values (e.g. c(1,2,3))
##pmf is the associated pmf for the discrete values (e.g. c(.3,.2,.5))
r.dscrt.pmf=function(n,x,pmf){
set.seed(1)
##Generate n uniform random values from 0 to 1
r=runif(n)
high=0
low=0
for (i in 1:length(x)){
##High will establish the appropriate upper bound to consider
high=high+pmf[i]
if (i==1){
##Creates the variable p1 which contains the positions of all
##r less than or equal to the first value of pmf
assign(paste0("p", x[i]), which(r<=pmf[i]))
} else {
##Creates the variable p2,p3,p4,etc. which contains the positions of all
##r between the appropriate interval of high and low
assign(paste0("p", x[i]), which(r>low & r<=high))
}
##Low will establish the appropriate lower bound to consider
low=low+pmf[i]
}
for (i in 1:length(x)){
##Will loops to replace the values of r at the positions specified at
##p1,p2,p3,etc. with x[1],x[2],x[3],etc. respectively.
r[paste0("p", x[i])]=x[i]
}
##Returns the new r
r
}
##Example call of the function
r.dscrt.pmf(10,c(0,1,3),c(.3,.2,.5))
get is like assign, in that it lets you refer to variables by string instead of name.
r[get(paste0("p", x[1]))]=x[1]
But get is one of those "flags" of something that could be written in a much clearer and safer way.
Would this suit your needs?
ifelse(r<11, x[1], r)
[1] 1 1 1 1 15
The question is: Create a function that takes in a numeric vector. The output should be a vector with running mean values. The i-th element of the output vector should be the mean of the values in the input vector from 1 to i.
My main problem is in the for loop, which is as follows:
x1 <- c(2,4,6,8,10)
for (i in 2: length(x1)){
ma <- sum(x1[i-1] , x1[i]) / i
print(ma)
mresult <- rbind(ma)
}
View(ma)
I know there must be something wrong in it. But I am just not sure what it is.
As you have noticed there are more efficient ways using already available functions and packages to achieve what you are trying to do. But here is how you would go about fixing your loop
x1 <- c(2,4,6,8,10)
mresult = numeric(0) #Initiate mresult. Or maybe you'd want to initiate with 0
for (i in 2: length(x1)){
ma <- sum(x1[1:i])/i #You were originally dividing the sum of (i-1)th and ith value by i
print(ma) #This is optional
mresult <- c(mresult,ma) #Since you have only an array, there is no need to rbind
}
View(ma) #The last computed average
View(mresult) #All averages
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.
My first question here: how to apply an efficient routine that iterates values of two vectors (pairwise) of a given data frame?
To be more specific, consider the following example using the following data frame:
df0 <- data.frame(matrix(c(1,2,2,3,1,3,0.4,0.2,0.2,0.1,0.4,0.1),nrow=6,ncol=2))
colnames(df0) <- c("value","frequency")
The first column is a real value and the second column is a frequency (or weights). NOTICE: the weights have to be strictly positive, they might be repeated, they not necessarily add up to one (because of repetition).
I am performing the following LOOP to calculate my function P. This P is supposed to be a number between 0 and 1.
# Define two parameters
K = 1/2
alpha = 0
# LOOP
mattemp <- matrix(,nrow=length(df0$value), ncol=length(df0$value))
for(i in 1:length(df0$value)) {
for(j in 1:length(df0$value)) {
mattemp[i,j] <- df0$frequency[i]^(1+alpha) * df0$frequency[j] * abs(df0$value[i]-df0$value[j])
P <- K * sum(mattemp)
}
}
Basically, my function P is calculating:
P = K * (0.4^alpha * 0.2 * |1-2| + 0.4^alpha * 0.1 * |1-3| + ...
This code works perfectly well as long as the matrix is small.
However, I am trying to implement this routine for a big matrix (5400 x 5400) and this LOOP does not seem to find an end.
I already tried to loop it using a foreach command (using %dopar%), but it does not work as well.
Is there a smart and concise routine that R can handle??? It does not need to follow the above structure, as long as it is efficient.
Thank you very much
Try:
df$nval <- (df0$value - mean(df0$value)) / sd(df0$value)
ij <- combn(nrow(df0), 2)
foo <- sum(df0$frequency[ij[1, ]] ^ (1 + alpha) * df0$frequency[ij[2, ]] * abs(df0$nval[ij[1, ]] - df0$nval[ij[2, ]]))
P <- K*2*sum(foo)
Reasoning: Basically you are testing every possible permutation between frequencies and normalized values. We use combn to create half of those. We then just vectorize the whole thing. Since combn only gives unique combinations, we need to multiply by 2. [Keep in mind that we don't need the values on the diagonal, as abs(df0$value[i] - df0$value[i]) is equal to 0, and we are only missing cases where i=j and j=i, so that's why we multiply by 2.] We then multiply by K and get P.
It's not clear how you want to normalize, so I just substracted the mean and divided that by the standard deviation. If you meant something else, you yourself can change it accordingly.
Edit1: Big thanks to #alexis_laz for finding a mistake and suggesting improvements that almost double the speed!
Edit2: Adjusted script to fit changed requirements.