Related
function(q,b,Data1,Data2){
x<-sum(
ifelse(Data1[13+q,b]/Data1[12+q,b]>Data2[13+q,1]/Data2[12+q,1],1,0)+
ifelse(Data1[13+q,b]/Data1[11+q,b]>Data2[13+q,1]/Data2[11+q,1],1,0)+
ifelse(Data1[13+q,b]/Data1[10+q,b]>Data2[13+q,1]/Data2[10+q,1],1,0)+
ifelse(Data1[13+q,b]/Data1[9+q,b]>Data2[13+q,1]/Data2[9+q,1],1,0)+
ifelse(Data1[13+q,b]/Data1[8+q,b]>Data2[13+q,1]/Data2[8+q,1],1,0)+
ifelse(Data1[13+q,b]/Data1[7+q,b]>Data2[13+q,1]/Data2[7+q,1],1,0)+
ifelse(Data1[13+q,b]/Data1[6+q,b]>Data2[13+q,1]/Data2[6+q,1],1,0)+
ifelse(Data1[13+q,b]/Data1[5+q,b]>Data2[13+q,1]/Data2[5+q,1],1,0)+
ifelse(Data1[13+q,b]/Data1[4+q,b]>Data2[13+q,1]/Data2[4+q,1],1,0)+
ifelse(Data1[13+q,b]/Data1[3+q,b]>Data2[13+q,1]/Data2[3+q,1],1,0)+
ifelse(Data1[13+q,b]/Data1[2+q,b]>Data2[13+q,1]/Data2[2+q,1],1,0)+
ifelse(Data1[13+q,b]/Data1[1+q,b]>Data2[13+q,1]/Data2[1+q,1],1,0)
)/12
}
Is there a way to simplify this? (no characters, only numbers in the data sets)
Thank you
Two pieces of knowledge you can combine to improve your code:
Firstly, you can divide a single number by a vector and R will return a vector with elementwise divisions. For example:
5 / c(1,2,3,4,5,6)
# [1] 5.0000000 2.5000000 1.6666667 1.2500000 1.0000000 0.8333333
The numerator on both sides of the inequality are the same all the time, you can use the above. So instead of explicitly calling it for every inequality, you can just call it once.
Secondly, an expression with TRUE or FALSE will be coerced to 1 and 0 when you try to perform arithmetic operations (in your case division, or calculating a mean). Inequalities return TRUE or FALSE values. Explicitly telling R to convert them to 0 and 1 is wasted energy, because R will automatically do it in your last step.
Putting this together in a simplified function:
function(q, b, Data1, Data2){
qseq <- (1:12) + q # Replaces all "q+1", "q+2", ... , "q+12"
dat1 <- Data1[qseq, b] # Replaces all "Data1[q+1, b]", ... "Data1[q+12, b]"
dat2 <- Data2[qseq, 1] # Replaces all "Data2[q+1, 1]", ... "Data2[q+12, 1]"
mean( Data1[13+q, b]/dat1 > Data2[13+q, 1]/dat2 )
this simplify a bit:
function(q,b,Data1,Data2){
data1_num <- Data1[13+q,b]
data2_num <- Data2[13+q,1]
x <- 0
for (i in 1:12) {
x <- x + ((data1_num/Data1[i+q,b]) > (data2_num /Data2[i+q,1]))
}
x <- x /12
#return(x)
}
But If you provide data example, and the output your expecting, i'm sure there is way to simplify it better
I have a list of lists, with each sub-list containing 3 values. My goal is to cycle through every value of this nested list in a systematic way (i.e. start with list 1, go through all 3 values, go to list 2, and so on), applying a function to each. But my function hits missing values and breaks and I've traced the problem to the indexing itself, which doesn't behave in the way I am expecting. The lists are constructed as:
pop <- 1:100
treat.temp <- NULL
treat <- NULL
## Generate 5 samples of pop
for (i in 1:5){
treat.temp <- sample(pop, 3)
treat[[i]] <- treat.temp
}
## Create a list with which to index mapply
iterations <- (1:5)
Illustrative function and results.
test.function <- function(j, k){
for (n in 1:3){
print(k[[n]][j])
}
}
results <- mapply(test.function, iterations, treat)
[1] 61
[1] 63
[1] 73
[1] NA
[1] NA
[1] NA
[1] NA
[1] NA
<snipped>
For the first cycle through 'j', this works. But after that it throws NAs. But if I do it manually, it returns the values I would expect.
> print(treat[[1]][1])
[1] 61
> print(treat[[1]][2])
[1] 63
> print(treat[[1]][3])
[1] 73
> print(treat[[2]][1])
[1] 59
> print(treat[[2]][2])
[1] 6
> print(treat[[2]][3])
[1] 75
<snipped>
I'm sure this is a basic question, but I can't seem to find the right search terms to find an answer here or on Google. Thanks in advance!
Edited to Add: MrFlick's answer works well for my problem. I have multiple list inputs (hence mapply) in my actual use. A more detailed example, with a few notes.
pop <- 1:100
years <- seq.int(2000, 2014, 1)
treat.temp <- NULL
treat <- NULL
year.temp <- NULL
year <- NULL
## Generate 5 samples of treated states, control states and treatment years
for (i in 1:5){
treat.temp <- sample(pop, 20)
treat[[i]] <- treat.temp
year.temp <- sample(years, 1)
year[[i]] <- year.temp
}
## Create a list with which to index mapply
iterations <- (1:5)
## Define function
test.function <- function(j, k, l){
for (n in 1:3){
## Cycles treat through each value of jXn
print(k[n])
## Holds treat (k) fixed for each 3 cycle set of n (using first value in each treat sub-list); cycles through sub-lists as j changes
print(k[1])
## Same as above, but with 2nd value in each sub-list of treat
print(k[2])
## Holds year (l) fixed for each 3 cycle set of n, cycling through values of year each time j changes
print(l[1])
## Functionally equivalent to
print(l)
}
}
results <- mapply(test.function, iterations, treat, year)
Well, you might be misunderstanding how mapply works. The function will loop through both of the iterations you pass as parameters, which means treat will also be subset each iteration. Essentially, the functions being called are
test.function(iterations[1], treat[[1]])
test.function(iterations[2], treat[[2]])
test.function(iterations[3], treat[[3]])
...
and you seem to treat the k variable as if it were the entire list. Also, you have your indexes backwards as well. But just to get your test working, you can do
test.function <- function(j, k){
for (n in 1:3) print(k[n])
}
results <- mapply(test.function, iterations, treat)
but this isn't really a super awesome way to iterate a list. What exactly are you trying to accomplish?
I have a function with two arguments. The first argument takes vector, and the second argument takes a scalar. I want to apply this function to each row of a matrix, but this function takes different second argument every time. I tried the following, it didn't work. I expected to calculate the p.value for each row and then divide the p.value by the row number. I expected the result to be a vector, but I got a matrix instead. This is a pseudo example, but it illustrates my purpose.
> foo = matrix(rnorm(100),ncol=20)
> f = function (x,y) t.test(x[1:10],x[11:20])$p.value/y
> goo = 1:5
> apply(foo,1,f,y=goo)
[,1] [,2] [,3] [,4] [,5]
[1,] 0.9406881 0.6134117 0.5484542 0.11299535 0.20420786
[2,] 0.4703440 0.3067059 0.2742271 0.05649767 0.10210393
[3,] 0.3135627 0.2044706 0.1828181 0.03766512 0.06806929
[4,] 0.2351720 0.1533529 0.1371135 0.02824884 0.05105196
[5,] 0.1881376 0.1226823 0.1096908 0.02259907 0.04084157
The following for loop strategy produces the expected result, expect would be very slow for the real data.
> res = numeric(5)
> for (i in 1:5){
res[i]=f(foo[i,],i)
}
> res
[1] 0.94068810 0.30670585 0.18281807 0.02824884 0.04084157
Any suggestions would be appreciated!
If your real purpose is like your example, you can vectorize the division:
f <- function(x) t.test(x[1:10], x[11:20])$p.value
apply(foo, 1, f) / goo
Based on the comment, the above is not appropriate.
In the case of the example, you might observe that the diagonal of the returned matrix is the desired result:
f = function (x,y) t.test(x[1:10],x[11:20])$p.value/y
goo = 1:5
diag(apply(foo,1,f,y=goo))
Besides being inefficient in time or space, this suffers from another problem. It is a result of the operation on y being vectorized that this is correct for the example. And in that case, the former solution is better. So I suspect that in your actual problem, your operation is not vectorized.
Sometimes a for loop really is the best answer. The apply family of functions are not magical; they are still loops.
Here is an sapply solution. It won't beat for for time (probably won't lose either) but it doesn't have a high space overhead. The idea is to apply the row index and use that to extract the row of foo and the element of goo to pass to f
sapply(seq(nrow(foo)), function(i) f(foo[i,], goo[i]))
f <- function (x,y) t.test(x[1:10],x[11:20])$p.value/y
f2 <- function(a, b){
tt <- t.test(x = a[1:10], y = a[11:20])$p.value
tt/b
}
f3 <- function() {
res <- numeric(5)
for (i in 1:5){
res[i] <- f(foo[i,],i)
}
res
}
f4 <- function(x) t.test(x[1:10], x[11:20])$p.value
set.seed(101)
foo <- matrix(rnorm(100),ncol=20)
goo <- 1:5
library(rbenchmark)
benchmark(
apply(foo, 1, f4) / goo,
mapply(f,split(foo,row(foo)),goo),
f2(foo,goo),
f3(),replications=1000,
sapply(seq(nrow(foo)), function(i) f(foo[i,], goo[i])),
columns=c("test","replications","elapsed","relative"))
## test replications elapsed relative
## 1 apply(foo, 1, f4)/goo 1000 1.581 5.528
## 3 f2(foo, goo) 1000 0.286 1.000
## 4 f3() 1000 1.458 5.098
## 2 mapply(...) 1000 1.599 5.591
## 5 sapply(...) 1000 1.486 5.196
The direct division is best (but not actually applicable); for this example there's not much difference between the other solutions, but for loop is better than sapply which is better than mapply. You should try this on a more realistic example to see how it's going to scale for your problem.
I have two lists looking like this:
mylist <- list(a=c(1:5),
b = c(5:12),
c = c(2:8))
list.id <- list(a=2, b=8, c=5)
I want to count the number of elements in mylist that are higher than the corresponding element in list.id and divide the result for the length of element in mylist. I have written this function.
perm.fun <- perm.fun2 = function(x,y){length(which(x[[i]] < y[[i]]))/length(x[[i]])}
However, when I do: lapply(mylist, perm.fun, list.id) I do not obtain the expected result.
Thanks
Using lapply, you would need to loop on the indices (1, 2, 3) so they can be used to extract the elements from both mylist and list.id:
perm.fun <- function(i, x, y) mean(x[[i]] > y[[i]])
lapply(seq_along(mylist), perm.fun, mylist, list.id)
But mapply is a much better tool for that task. From the doc:
mapply applies FUN to the first elements of each ... argument, the second elements, the third elements, and so on.
So your code can just be:
mapply(function(x, y) mean(x > y), mylist, list.id)
# a b c
# 0.6000000 0.5000000 0.4285714
Still trying to get into the R logic... what is the "best" way to unpack (on LHS) the results from a function returning multiple values?
I can't do this apparently:
R> functionReturningTwoValues <- function() { return(c(1, 2)) }
R> functionReturningTwoValues()
[1] 1 2
R> a, b <- functionReturningTwoValues()
Error: unexpected ',' in "a,"
R> c(a, b) <- functionReturningTwoValues()
Error in c(a, b) <- functionReturningTwoValues() : object 'a' not found
must I really do the following?
R> r <- functionReturningTwoValues()
R> a <- r[1]; b <- r[2]
or would the R programmer write something more like this:
R> functionReturningTwoValues <- function() {return(list(first=1, second=2))}
R> r <- functionReturningTwoValues()
R> r$first
[1] 1
R> r$second
[1] 2
--- edited to answer Shane's questions ---
I don't really need giving names to the result value parts. I am applying one aggregate function to the first component and an other to the second component (min and max. if it was the same function for both components I would not need splitting them).
(1) list[...]<- I had posted this over a decade ago on r-help. Since then it has been added to the gsubfn package. It does not require a special operator but does require that the left hand side be written using list[...] like this:
library(gsubfn) # need 0.7-0 or later
list[a, b] <- functionReturningTwoValues()
If you only need the first or second component these all work too:
list[a] <- functionReturningTwoValues()
list[a, ] <- functionReturningTwoValues()
list[, b] <- functionReturningTwoValues()
(Of course, if you only needed one value then functionReturningTwoValues()[[1]] or functionReturningTwoValues()[[2]] would be sufficient.)
See the cited r-help thread for more examples.
(2) with If the intent is merely to combine the multiple values subsequently and the return values are named then a simple alternative is to use with :
myfun <- function() list(a = 1, b = 2)
list[a, b] <- myfun()
a + b
# same
with(myfun(), a + b)
(3) attach Another alternative is attach:
attach(myfun())
a + b
ADDED: with and attach
I somehow stumbled on this clever hack on the internet ... I'm not sure if it's nasty or beautiful, but it lets you create a "magical" operator that allows you to unpack multiple return values into their own variable. The := function is defined here, and included below for posterity:
':=' <- function(lhs, rhs) {
frame <- parent.frame()
lhs <- as.list(substitute(lhs))
if (length(lhs) > 1)
lhs <- lhs[-1]
if (length(lhs) == 1) {
do.call(`=`, list(lhs[[1]], rhs), envir=frame)
return(invisible(NULL))
}
if (is.function(rhs) || is(rhs, 'formula'))
rhs <- list(rhs)
if (length(lhs) > length(rhs))
rhs <- c(rhs, rep(list(NULL), length(lhs) - length(rhs)))
for (i in 1:length(lhs))
do.call(`=`, list(lhs[[i]], rhs[[i]]), envir=frame)
return(invisible(NULL))
}
With that in hand, you can do what you're after:
functionReturningTwoValues <- function() {
return(list(1, matrix(0, 2, 2)))
}
c(a, b) := functionReturningTwoValues()
a
#[1] 1
b
# [,1] [,2]
# [1,] 0 0
# [2,] 0 0
I don't know how I feel about that. Perhaps you might find it helpful in your interactive workspace. Using it to build (re-)usable libraries (for mass consumption) might not be the best idea, but I guess that's up to you.
... you know what they say about responsibility and power ...
Usually I wrap the output into a list, which is very flexible (you can have any combination of numbers, strings, vectors, matrices, arrays, lists, objects int he output)
so like:
func2<-function(input) {
a<-input+1
b<-input+2
output<-list(a,b)
return(output)
}
output<-func2(5)
for (i in output) {
print(i)
}
[1] 6
[1] 7
I put together an R package zeallot to tackle this problem. zeallot includes a multiple assignment or unpacking assignment operator, %<-%. The LHS of the operator is any number of variables to assign, built using calls to c(). The RHS of the operator is a vector, list, data frame, date object, or any custom object with an implemented destructure method (see ?zeallot::destructure).
Here are a handful of examples based on the original post,
library(zeallot)
functionReturningTwoValues <- function() {
return(c(1, 2))
}
c(a, b) %<-% functionReturningTwoValues()
a # 1
b # 2
functionReturningListOfValues <- function() {
return(list(1, 2, 3))
}
c(d, e, f) %<-% functionReturningListOfValues()
d # 1
e # 2
f # 3
functionReturningNestedList <- function() {
return(list(1, list(2, 3)))
}
c(f, c(g, h)) %<-% functionReturningNestedList()
f # 1
g # 2
h # 3
functionReturningTooManyValues <- function() {
return(as.list(1:20))
}
c(i, j, ...rest) %<-% functionReturningTooManyValues()
i # 1
j # 2
rest # list(3, 4, 5, ..)
Check out the package vignette for more information and examples.
functionReturningTwoValues <- function() {
results <- list()
results$first <- 1
results$second <-2
return(results)
}
a <- functionReturningTwoValues()
I think this works.
There's no right answer to this question. I really depends on what you're doing with the data. In the simple example above, I would strongly suggest:
Keep things as simple as possible.
Wherever possible, it's a best practice to keep your functions vectorized. That provides the greatest amount of flexibility and speed in the long run.
Is it important that the values 1 and 2 above have names? In other words, why is it important in this example that 1 and 2 be named a and b, rather than just r[1] and r[2]? One important thing to understand in this context is that a and b are also both vectors of length 1. So you're not really changing anything in the process of making that assignment, other than having 2 new vectors that don't need subscripts to be referenced:
> r <- c(1,2)
> a <- r[1]
> b <- r[2]
> class(r)
[1] "numeric"
> class(a)
[1] "numeric"
> a
[1] 1
> a[1]
[1] 1
You can also assign the names to the original vector if you would rather reference the letter than the index:
> names(r) <- c("a","b")
> names(r)
[1] "a" "b"
> r["a"]
a
1
[Edit] Given that you will be applying min and max to each vector separately, I would suggest either using a matrix (if a and b will be the same length and the same data type) or data frame (if a and b will be the same length but can be different data types) or else use a list like in your last example (if they can be of differing lengths and data types).
> r <- data.frame(a=1:4, b=5:8)
> r
a b
1 1 5
2 2 6
3 3 7
4 4 8
> min(r$a)
[1] 1
> max(r$b)
[1] 8
If you want to return the output of your function to the Global Environment, you can use list2env, like in this example:
myfun <- function(x) { a <- 1:x
b <- 5:x
df <- data.frame(a=a, b=b)
newList <- list("my_obj1" = a, "my_obj2" = b, "myDF"=df)
list2env(newList ,.GlobalEnv)
}
myfun(3)
This function will create three objects in your Global Environment:
> my_obj1
[1] 1 2 3
> my_obj2
[1] 5 4 3
> myDF
a b
1 1 5
2 2 4
3 3 3
Lists seem perfect for this purpose. For example within the function you would have
x = desired_return_value_1 # (vector, matrix, etc)
y = desired_return_value_2 # (vector, matrix, etc)
returnlist = list(x,y...)
} # end of function
main program
x = returnlist[[1]]
y = returnlist[[2]]
Yes to your second and third questions -- that's what you need to do as you cannot have multiple 'lvalues' on the left of an assignment.
How about using assign?
functionReturningTwoValues <- function(a, b) {
assign(a, 1, pos=1)
assign(b, 2, pos=1)
}
You can pass the names of the variable you want to be passed by reference.
> functionReturningTwoValues('a', 'b')
> a
[1] 1
> b
[1] 2
If you need to access the existing values, the converse of assign is get.
[A]
If each of foo and bar is a single number, then there's nothing wrong with c(foo,bar); and you can also name the components: c(Foo=foo,Bar=bar). So you could access the components of the result 'res' as res[1], res[2]; or, in the named case, as res["Foo"], res["BAR"].
[B]
If foo and bar are vectors of the same type and length, then again there's nothing wrong with returning cbind(foo,bar) or rbind(foo,bar); likewise nameable. In the 'cbind' case, you would access foo and bar as res[,1], res[,2] or as res[,"Foo"], res[,"Bar"]. You might also prefer to return a dataframe rather than a matrix:
data.frame(Foo=foo,Bar=bar)
and access them as res$Foo, res$Bar. This would also work well if foo and bar were of the same length but not of the same type (e.g. foo is a vector of numbers, bar a vector of character strings).
[C]
If foo and bar are sufficiently different not to combine conveniently as above, then you shuld definitely return a list.
For example, your function might fit a linear model and
also calculate predicted values, so you could have
LM<-lm(....) ; foo<-summary(LM); bar<-LM$fit
and then you would return list(Foo=foo,Bar=bar) and then access the summary as res$Foo, the predicted values as res$Bar
source: http://r.789695.n4.nabble.com/How-to-return-multiple-values-in-a-function-td858528.html
Year 2021 and this is something I frequently use.
tidyverse package has a function called lst that assigns name to the list elements when creating the list.
Post which I use list2env() to assign variable or use the list directly
library(tidyverse)
fun <- function(){
a<-1
b<-2
lst(a,b)
}
list2env(fun(), envir=.GlobalEnv)#unpacks list key-values to variable-values into the current environment
This is only for the sake of completeness and not because I personally prefer it. You can pipe %>% the result, evaluate it with curly braces {} and write variables to the parent environment using double-arrow <<-.
library(tidyverse)
functionReturningTwoValues() %>% {a <<- .[1]; b <<- .[2]}
UPDATE:
Your can also use the multiple assignment operator from the zeallot package:: %<-%
c(a, b) %<-% list(0, 1)
I will post a function that returns multiple objects by way of vectors:
Median <- function(X){
X_Sort <- sort(X)
if (length(X)%%2==0){
Median <- (X_Sort[(length(X)/2)]+X_Sort[(length(X)/2)+1])/2
} else{
Median <- X_Sort[(length(X)+1)/2]
}
return(Median)
}
That was a function I created to calculate the median. I know that there's an inbuilt function in R called median() but nonetheless I programmed it to build other function to calculate the quartiles of a numeric data-set by using the Median() function I just programmed. The Median() function works like this:
If a numeric vector X has an even number of elements (i.e., length(X)%%2==0), the median is calculated by averaging the elements sort(X)[length(X)/2] and sort(X)[(length(X)/2+1)].
If Xdoesn't have an even number of elements, the median is sort(X)[(length(X)+1)/2].
On to the QuartilesFunction():
QuartilesFunction <- function(X){
X_Sort <- sort(X) # Data is sorted in ascending order
if (length(X)%%2==0){
# Data number is even
HalfDN <- X_Sort[1:(length(X)/2)]
HalfUP <- X_Sort[((length(X)/2)+1):length(X)]
QL <- Median(HalfDN)
QU <- Median(HalfUP)
QL1 <- QL
QL2 <- QL
QU1 <- QU
QU2 <- QU
QL3 <- QL
QU3 <- QU
Quartiles <- c(QL1,QU1,QL2,QU2,QL3,QU3)
names(Quartiles) = c("QL (1)", "QU (1)", "QL (2)", "QU (2)","QL (3)", "QU (3)")
} else{ # Data number is odd
# Including the median
Half1DN <- X_Sort[1:((length(X)+1)/2)]
Half1UP <- X_Sort[(((length(X)+1)/2)):length(X)]
QL1 <- Median(Half1DN)
QU1 <- Median(Half1UP)
# Not including the median
Half2DN <- X_Sort[1:(((length(X)+1)/2)-1)]
Half2UP <- X_Sort[(((length(X)+1)/2)+1):length(X)]
QL2 <- Median(Half2DN)
QU2 <- Median(Half2UP)
# Methods (1) and (2) averaged
QL3 <- (QL1+QL2)/2
QU3 <- (QU1+QU2)/2
Quartiles <- c(QL1,QU1,QL2,QU2,QL3,QU3)
names(Quartiles) = c("QL (1)", "QU (1)", "QL (2)", "QU (2)","QL (3)", "QU (3)")
}
return(Quartiles)
}
This function returns the quartiles of a numeric vector by using three methods:
Discarding the median for the calculation of the quartiles when the number of elements of the numeric vector Xis odd.
Keeping the median for the calculation of the quartiles when the number of elements of the numeric vector Xis odd.
Averaging the results obtained by using methods 1 and 2.
When the number of elements in the numeric vector X is even, the three methods coincide.
The result of the QuartilesFunction() is a vector that depicts the first and third quartiles calculated by using the three methods outlined.
With R 3.6.1, I can do the following
fr2v <- function() { c(5,3) }
a_b <- fr2v()
(a_b[[1]]) # prints "5"
(a_b[[2]]) # prints "3"
To obtain multiple outputs from a function and keep them in the desired format you can save the outputs to your hard disk (in the working directory) from within the function and then load them from outside the function:
myfun <- function(x) {
df1 <- ...
df2 <- ...
save(df1, file = "myfile1")
save(df2, file = "myfile2")
}
load("myfile1")
load("myfile2")