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function with multiple outputs
This seems like an easy question, but I can't figure it out and I haven't had luck in the R manuals I've looked at. I want to find dim(x), but I want to assign dim(x)[1] to a and dim(x)[2] to b in a single line.
I've tried [a b] <- dim(x) and c(a, b) <- dim(x), but neither has worked. Is there a one-line way to do this? It seems like a very basic thing that should be easy to handle.
This may not be as simple of a solution as you had wanted, but this gets the job done. It's also a very handy tool in the future, should you need to assign multiple variables at once (and you don't know how many values you have).
Output <- SomeFunction(x)
VariablesList <- letters[1:length(Output)]
for (i in seq(1, length(Output), by = 1)) {
assign(VariablesList[i], Output[i])
}
Loops aren't the most efficient things in R, but I've used this multiple times. I personally find it especially useful when gathering information from a folder with an unknown number of entries.
EDIT: And in this case, Output could be any length (as long as VariablesList is longer).
EDIT #2: Changed up the VariablesList vector to allow for more values, as Liz suggested.
You can also write your own function that will always make a global a and b. But this isn't advisable:
mydim <- function(x) {
out <- dim(x)
a <<- out[1]
b <<- out[2]
}
The "R" way to do this is to output the results as a list or vector just like the built in function does and access them as needed:
out <- dim(x)
out[1]
out[2]
R has excellent list and vector comprehension that many other languages lack and thus doesn't have this multiple assignment feature. Instead it has a rich set of functions to reach into complex data structures without looping constructs.
Doesn't look like there is a way to do this. Really the only way to deal with it is to add a couple of extra lines:
temp <- dim(x)
a <- temp[1]
b <- temp[2]
It depends what is in a and b. If they are just numbers try to return a vector like this:
dim <- function(x,y)
return(c(x,y))
dim(1,2)[1]
# [1] 1
dim(1,2)[2]
# [1] 2
If a and b are something else, you might want to return a list
dim <- function(x,y)
return(list(item1=x:y,item2=(2*x):(2*y)))
dim(1,2)[[1]]
[1] 1 2
dim(1,2)[[2]]
[1] 2 3 4
EDIT:
try this: x <- c(1,2); names(x) <- c("a","b")
Related
(1,2,2,3,3,3,4,4,4,4,...,n,...,n)
I want to make the above vector by for loop, but not using rep function or the other functions. This may not be good question to ask in stackoverflow, but since I am a newbie for R, I dare ask here to be helped.
(You can suppose the length of the vector is 10)
With a for loop, it can be done with
n <- 10
out <- c()
for(i in seq_len(n)){
for(j in seq_len(i)) {
out <- c(out, i)
}
}
In R, otherwise, this can be done as
rep(seq_len(n), seq_len(n))
I have been beaten by #akrun by seconds, even so I'd like to give you a few hints if using rep would have been possible which may help you with R in general. (Without rep usage, just look at #akrun)
Short answer using rep
rep(1:n, 1:n)
Long Answer using rep
Before posting a question you should try to develop your own solutions and share them.
Trying googling a bit and sharing what you already found is usually good as well. Please, have a look at "help/how-to-ask"
Let's try to do it together.
First of all, we should try to have a look at official sources:
R-project "getting help", here you can see the standard way to get a function's documentation is just typing ?func_name in your R console
R-project "official manuals" offer a good introduction to R. Try looking at the first topic, "An Introduction to R"
From the previous two (and other sources as well) you will find two interesting functions:
: operator: it can be used to generate a sequence of integers from a to b like a:b. Typing 1:3, for instance, gives you the 1, 2, 3 vector
rep(x, t) is a function which can be used to replicate the item(s) x t times.
You also need to know R is "vector-oriented", that is it applies functions over vectors without you typing explicits loops.
For instance, if you call repl(1:3, 2), it's (almost) equivalent to running:
for(i in 1:3)
rep(i, 2)
By combining the previous two functions and the notion R is "vector-oriented", you get the rep(1:n, 1:n) solution.
I am not sure why you don't want to use rep, but here is a method of not using it or any functions similar to rep within the loop.
`for (i in 1:10){
a<-NA
a[1:i] <- i
if (i==1){b<-a}
else if (i >1){b <- c(b,a)}
assign("OutputVector",b,envir = .GlobalEnv)
}`
`OutputVector`
Going for an n of ten seemed subjective so I just did the loop for numbers 1 through 10 and you can take the first 10 numbers in the vector if you want. OutputVector[1:10]
You can do this with a single loop, though it's a while rather than a for
n <- 10
x <- 1;
i <- 2;
while(i <= n)
{
x <- c(x, 1/i);
if(sum(x) %% 1 == 0) i = i + 1;
}
1/x
How do I modify an argument being passed to a function in R? In C++ this would be pass by reference.
g=4
abc <- function(x) {x<-5}
abc(g)
I would like g to be set to 5.
There are ways as #Dason showed, but really - you shouldn't!
The whole paradigm of R is to "pass by value". #Rory just posted the normal way to handle it - just return the modified value...
Environments are typically the only objects that can be passed by reference in R.
But lately new objects called reference classes have been added to R (they use environments). They can modify their values (but in a controlled way). You might want to look into using them if you really feel the need...
There has got to be a better way to do this but...
abc <- function(x){eval(parse(text = paste(substitute(x), "<<- 5")))}
g <- 4
abc(g)
g
gives the output
[1] 5
I have a solution similar to #Dason's, and I am curious if there are good reasons not to use this or if there are important pitfalls I should be aware of:
changeMe = function(x){
assign(deparse(substitute(x)), "changed", env=.GlobalEnv)
}
I think that #Dason's method is the only way to do it theoretically, but practically I think R's way already does it.
For example, when you do the following:
y <- c(1,2)
x <- y
x is really just a pointer to a the value c(1,2). Similarly, when you do
abc <- function(x) {x <- 5; x}
g <- abc(g)
It is not that you are spending time copying g to the function and then copying the result back into g. I think what R does with the code
g <- abc(g)
is:
The right side is looked at first. An environment for the function abc is set up.
A pointer is created in that environment called x.
x points to the same value that g points to.
Then x points to 5
The function returns the pointer x
g now points to the same value that x pointed to at the time of return.
Thus, it is not that there is a whole bunch of unnecessary copying of large options.
I hope that someone can confirm/correct this.
Am I missing something as to why you can't just do this?
g <- abc(g)
I'm pretty new in R, I only know its foundamental concepts.
I have the v vector:
v<-c(1,2,3,4)
and I would like to append to v four NA values, obtaining:
v(1,2,3,4,NA,NA,NA,NA)
T o solve this I can use a for loop:
for(i in 1:4){
v<-append(v, NA)
}
My question is: are there clever and/or faster R solutions I could use?
From the above comments we had found some useful answers where every new OP can view in aswer window rather than comment sections, thanks OP for your valuable answers
v <- c(v, rep(NA, 4)) # joel.wilson
length(v)<-length(v)+4 # Nicola
'length<-'(v, 8) # akrun
Please note:
in general the Joel.Wilson's option is the good one 'cause can be used to append several times a specific value (numeric, character, boolean, etc.), while other two solutions only NA values as they play on the length property.
I'm very new to R, and much more new to programming in R. I have the following question and its answer (which is not mine). I've trying to understand why some values, from where they are obtained, why they are used, etc.
Question: Make the vector 3 5 7 9 11 13 15 17 with a for loop. Start
with x=numeric() and fill this vector with the for loop
I know I have to create x=numeric() so I can fill it with the result obtained from the loop.
The answer from a classmate was:
> x <- numeric()
> for(i in 1:8){
if(i==1){ ## Why ==1 and not 0, or any other value
x[i] <- 3
}else{
x[i] <- x[i-1]+2 ### And why i-1
}
I'm having similar problems in questions like:
Make a for loop that adds the second element of a vector to the first,
subtracts the third element from the result, adds the fourth again and
so on for the entire length of the vector
So far, I created the vector and the empty vector
> y = c(5, 10, 15, 20, 25, 30)
> answer <- 0
And then, when I try to do the for loop, I get stuck here:
for(i in 1:length(y)){
if(i...){ ### ==1? ==0?
answer = y[i] ###and here I really don't know how to continue.
}else if()
}
Believe me when I tell you I've read several replies to questions here, like in How to make a vector using a for loop, plus pages and pages about for loop, but cannot really figure how to solve these (and other) problems.
I repeat, I'm very new, so I'm struggling trying to understand it. Any help would be much appreciated.
First, I will annotate the loop to answer what the loop is doing.
# Initialize the vector
x <- numeric()
for(i in 1:8){
# Initialize the first element of the vector, x[1]. Remember, R indexes start at 1, not 0.
if(i==1){
x[i] <- 3
} else {
# Define each additional element in terms of the previous one (x[i - 1]
# is the element of x before the current one.
x[i] <- x[i-1]+2 ### And why i-1
}
}
A better solution that uses a loop and grows it (like the instructions state) is something like this:
x <- numeric()
for(i in 1:8){
x[i] <- 2 * i + 1
}
This is still not a good way to do things because growing a vector inside a loop is very slow. To fix this, you can preallocate the vector by telling numeric the length of the vector you want:
x <- numeric(8)
The best way to solve this would be:
2 * 1:8 + 1
using vectorized operations.
To help you solve your other problem, I suggest writing out each step of the loop as a table. For example, for my solution, the table would be
i | x[i]
------------------
1 | 2 * 1 + 1 = 3
2 | 2 * 2 + 1 = 5
and so on. This will give you an idea of what the for loop is doing at each iteration.
This is intentionally not an answer because there are better ways to solve the alternating sign summation problem than a for-loop. I suppose there could be value in getting comfortable with for-loops but the vectorized approaches in R should be learned as well. R has "argument recycling" for many of its operations, including the "*" (multiplication) operation: Look at:
(1:10)*c(1,-1)
Then take an arbitrary vector, say vec and try:
sum( vec*c(1,-1) )
The more correct answer after looking at that result would be:
vvec[1] + sum( vec[-1]*c(1,-1) )
Which has the educational advantage of illustrating R's negative indexing. Look up "argument recycling" in your documentation. The shorter objects are automagically duplicatied/triplicated/however-many-needed-cated to exactly match the length of the longest vector in the mathematical or logical expression.
I am just beginning to learn R and am having an issue that is leaving me fairly confused. My goal is to create an empty vector and append elements to it. Seems easy enough, but solutions that I have seen on stackoverflow don't seem to be working.
To wit,
> a <- numeric()
> append(a,1)
[1] 1
> a
numeric(0)
I can't quite figure out what I'm doing wrong. Anyone want to help a newbie?
append does something that is somewhat different from what you are thinking. See ?append.
In particular, note that append does not modify its argument. It returns the result.
You want the function c:
> a <- numeric()
> a <- c(a, 1)
> a
[1] 1
Your a vector is not being passed by reference, so when it is modified you have to store it back into a. You cannot access a and expect it to be updated.
You just need to assign the return value to your vector, just as Matt did:
> a <- numeric()
> a <- append(a, 1)
> a
[1] 1
Matt is right that c() is preferable (fewer keystrokes and more versatile) though your use of append() is fine.