sqr = seq(1, 100, by=2)
sqr.squared = NULL
for (n in 1:50)
{
sqr.squared[n] = sqr[n]^2
}
I came accross the loop above, for a beginner this was simple enough. To further understand r what was the precise purpose of the second line? For my research I gather it has something to do with resetting the vector. If someone could elaborate it'd be much appreciated.
sqr.squared <- NULL
is one of many ways initialize the empty vector sqr.squared prior to running it through a loop. In general, when the length of the resulting vector is known, it is much better practice to allocate the vector's length. So here,
sqr.squared <- vector("integer", 50)
would be much better practice. And faster too. This way you are not building the new vector in the loop. But since ^ is vectorized, you could also simply do
sqr[1:50] ^ 2
and ditch the loop all together.
Another way to think about it is to remember that everything in r is a function call, and functions need input (usually).
say you calculated y and want to store that value somewhere. You can do x <- y without initializing an x object (r does this for you unlike in other languages, c for example), but say you want to store it in a specific place in x.
So note that <- (or = in your example) is a function
y <- 1
x[2] <- y
# Error in x[2] <- y : object 'x' not found
This is a different function than <-. Since you want to put y at x[2], you need the function [<-
`[<-`(x, 2, y)
# Error: object 'x' not found
But this still doesn't work because we need the object x to use this function, so initialize x to something.
(x <- numeric(5))
# [1] 0 0 0 0 0
# and now use the function
`[<-`(x, 2, y)
# [1] 0 1 0 0 0
This prefix notation is easier for computers to parse (eg, + 1 1) but harder for humans (me at least), so we prefer infix notation (eg, 1 + 1). R makes such functions easier to use x[2] <- y rather than how I did above.
The first answer is correct, when you assign a NULL value to a variable, the purpose is to initialize a vector. In many cases, when you are working checking numbers or with different types of variables, you will need to set NULL this arrays, matrix, etc.
For example, in you want to create a some type of element, in some cases you will need to put something inside them. This is the purpose of to use NULL. In addition, sometimes you will require NA instead of NULL.
Related
This question is somewhat two-fold. First, I'm trying to figure out the best way to achieve a simple implementation of a function over a vector. Second, I'm trying to understand the meaning of FUN.VALUE in vapply.
I've defined a simple halfwave rectification function:
Fun = function(x) x[x<0]=0
I then define a vector:
A = -10:10
I wish to apply this function to the vector. I've tried a number of approaches, but can't figure out how to do it easily.
I could always do
A[A<0]=0
but that defeats the purpose of using a function.
I've tried defining the function so that it applies to each element of the vector, rather than the vector as a whole:
Fun = function(x) {if (x<0) {x=0} }
I can't use apply, since apply only seems to work when the object that you're applying the function to has dimensions, and a vector doesn't have dimensions in R for some reason.
If I use lapply, I get NULL, or 0 for each element of the output object, depending on which of the two above functions I use.
For vapply, I can't make any sense out of the documentation for FUN.VALUE
Here's what it says:
FUN.VALUE
a (generalized) vector; a template for the return value from FUN. See ‘Details’.
and in Details:
vapply returns a vector or array of type matching the FUN.VALUE. If length(FUN.VALUE) == 1 a vector of the same length as X is returned, otherwise an array. If FUN.VALUE is not an array, the result is a matrix with length(FUN.VALUE) rows and length(X) columns, otherwise an array a with dim(a) == c(dim(FUN.VALUE), length(X)).
This makes no sense to me. In my case, A is a vector with 20 elements, and clearly I want the output to be a 20 element vector. So how do I specify a generalized 20 element vector template? And wouldn't the length of such a template be 20, and not 1?
I'd really appreciate some help here. I'm finding the documentation to be incredibly frustrating, and it's hard to tell whether it's because I'm missing something obvious, or whether the documentation is vague.
Your function has a bug in it. Instead of
Fun = function(x) x[x<0]=0
You actually want:
correct_fun = function(x){
x[x<0]=0
return(x)
}
As is, your function is not returning anything at the moment. It modifies x but doesn't return it so it just dies with the function. Executing correct_fun on A I get:
[1] 0 0 0 0 0 0 0 0 0 0 0 1 2 3 4 5 6 7 8 9 10
I just finished reading about scoping in the R intro, and am very curious about the <<- assignment.
The manual showed one (very interesting) example for <<-, which I feel I understood. What I am still missing is the context of when this can be useful.
So what I would love to read from you are examples (or links to examples) on when the use of <<- can be interesting/useful. What might be the dangers of using it (it looks easy to loose track of), and any tips you might feel like sharing.
<<- is most useful in conjunction with closures to maintain state. Here's a section from a recent paper of mine:
A closure is a function written by another function. Closures are
so-called because they enclose the environment of the parent
function, and can access all variables and parameters in that
function. This is useful because it allows us to have two levels of
parameters. One level of parameters (the parent) controls how the
function works. The other level (the child) does the work. The
following example shows how can use this idea to generate a family of
power functions. The parent function (power) creates child functions
(square and cube) that actually do the hard work.
power <- function(exponent) {
function(x) x ^ exponent
}
square <- power(2)
square(2) # -> [1] 4
square(4) # -> [1] 16
cube <- power(3)
cube(2) # -> [1] 8
cube(4) # -> [1] 64
The ability to manage variables at two levels also makes it possible to maintain the state across function invocations by allowing a function to modify variables in the environment of its parent. The key to managing variables at different levels is the double arrow assignment operator <<-. Unlike the usual single arrow assignment (<-) that always works on the current level, the double arrow operator can modify variables in parent levels.
This makes it possible to maintain a counter that records how many times a function has been called, as the following example shows. Each time new_counter is run, it creates an environment, initialises the counter i in this environment, and then creates a new function.
new_counter <- function() {
i <- 0
function() {
# do something useful, then ...
i <<- i + 1
i
}
}
The new function is a closure, and its environment is the enclosing environment. When the closures counter_one and counter_two are run, each one modifies the counter in its enclosing environment and then returns the current count.
counter_one <- new_counter()
counter_two <- new_counter()
counter_one() # -> [1] 1
counter_one() # -> [1] 2
counter_two() # -> [1] 1
It helps to think of <<- as equivalent to assign (if you set the inherits parameter in that function to TRUE). The benefit of assign is that it allows you to specify more parameters (e.g. the environment), so I prefer to use assign over <<- in most cases.
Using <<- and assign(x, value, inherits=TRUE) means that "enclosing environments of the supplied environment are searched until the variable 'x' is encountered." In other words, it will keep going through the environments in order until it finds a variable with that name, and it will assign it to that. This can be within the scope of a function, or in the global environment.
In order to understand what these functions do, you need to also understand R environments (e.g. using search).
I regularly use these functions when I'm running a large simulation and I want to save intermediate results. This allows you to create the object outside the scope of the given function or apply loop. That's very helpful, especially if you have any concern about a large loop ending unexpectedly (e.g. a database disconnection), in which case you could lose everything in the process. This would be equivalent to writing your results out to a database or file during a long running process, except that it's storing the results within the R environment instead.
My primary warning with this: be careful because you're now working with global variables, especially when using <<-. That means that you can end up with situations where a function is using an object value from the environment, when you expected it to be using one that was supplied as a parameter. This is one of the main things that functional programming tries to avoid (see side effects). I avoid this problem by assigning my values to a unique variable names (using paste with a set or unique parameters) that are never used within the function, but just used for caching and in case I need to recover later on (or do some meta-analysis on the intermediate results).
One place where I used <<- was in simple GUIs using tcl/tk. Some of the initial examples have it -- as you need to make a distinction between local and global variables for statefullness. See for example
library(tcltk)
demo(tkdensity)
which uses <<-. Otherwise I concur with Marek :) -- a Google search can help.
On this subject I'd like to point out that the <<- operator will behave strangely when applied (incorrectly) within a for loop (there may be other cases too). Given the following code:
fortest <- function() {
mySum <- 0
for (i in c(1, 2, 3)) {
mySum <<- mySum + i
}
mySum
}
you might expect that the function would return the expected sum, 6, but instead it returns 0, with a global variable mySum being created and assigned the value 3. I can't fully explain what is going on here but certainly the body of a for loop is not a new scope 'level'. Instead, it seems that R looks outside of the fortest function, can't find a mySum variable to assign to, so creates one and assigns the value 1, the first time through the loop. On subsequent iterations, the RHS in the assignment must be referring to the (unchanged) inner mySum variable whereas the LHS refers to the global variable. Therefore each iteration overwrites the value of the global variable to that iteration's value of i, hence it has the value 3 on exit from the function.
Hope this helps someone - this stumped me for a couple of hours today! (BTW, just replace <<- with <- and the function works as expected).
f <- function(n, x0) {x <- x0; replicate(n, (function(){x <<- x+rnorm(1)})())}
plot(f(1000,0),typ="l")
The <<- operator can also be useful for Reference Classes when writing Reference Methods. For example:
myRFclass <- setRefClass(Class = "RF",
fields = list(A = "numeric",
B = "numeric",
C = function() A + B))
myRFclass$methods(show = function() cat("A =", A, "B =", B, "C =",C))
myRFclass$methods(changeA = function() A <<- A*B) # note the <<-
obj1 <- myRFclass(A = 2, B = 3)
obj1
# A = 2 B = 3 C = 5
obj1$changeA()
obj1
# A = 6 B = 3 C = 9
I use it in order to change inside map() an object in the global environment.
a = c(1,0,0,1,0,0,0,0)
Say I want to obtain a vector which is c(1,2,3,1,2,3,4,5), that is if there is a 1, let it 1, otherwise add 1 until the next 1.
map(
.x = seq(1,(length(a))),
.f = function(x) {
a[x] <<- ifelse(a[x]==1, a[x], a[x-1]+1)
})
a
[1] 1 2 3 1 2 3 4 5
The following function is used to multiply a sequence 1:x by y
f1<-function(x,y){return (lapply(1:x, function(a,b) b*a, b=y))}
Looks like a is used to represent the element in the sequence 1:x, but I do not know how to understand this parameter passing mechanism. In other OO languages, like Java or C++, there have call by reference or call by value.
Short answer: R is call by value. Long answer: it can do both.
Call By Value, Lazy Evaluation, and Scoping
You'll want to read through: the R language definition for more details.
R mostly uses call by value but this is complicated by its lazy evaluation:
So you can have a function:
f <- function(x, y) {
x * 3
}
If you pass in two big matrixes to x and y, only x will be copied into the callee environment of f, because y is never used.
But you can also access variables in parent environments of f:
y <- 5
f <- function(x) {
x * y
}
f(3) # 15
Or even:
y <- 5
f <- function() {
x <- 3
g <- function() {
x * y
}
}
f() # returns function g()
f()() # returns 15
Call By Reference
There are two ways for doing call by reference in R that I know of.
One is by using Reference Classes, one of the three object oriented paradigms of R (see also: Advanced R programming: Object Oriented Field Guide)
The other is to use the bigmemory and bigmatrix packages (see The bigmemory project). This allows you to create matrices in memory (underlying data is stored in C), returning a pointer to the R session. This allows you to do fun things like accessing the same matrix from multiple R sessions.
To multiply a vector x by a constant y just do
x * y
The (some prefix)apply functions works very similar to each other, you want to map a function to every element of your vector, list, matrix and so on:
x = 1:10
x.squared = sapply(x, function(elem)elem * elem)
print(x.squared)
[1] 1 4 9 16 25 36 49 64 81 100
It gets better with matrices and data frames because you can now apply a function over all rows or columns, and collect the output. Like this:
m = matrix(1:9, ncol = 3)
# The 1 below means apply over rows, 2 would mean apply over cols
row.sums = apply(m, 1, function(some.row) sum(some.row))
print(row.sums)
[1] 12 15 18
If you're looking for a simple way to multiply a sequence by a constant, definitely use #Fernando's answer or something similar. I'm assuming you're just trying to determine how parameters are being passed in this code.
lapply calls its second argument (in your case function(a, b) b*a) with each of the values of its first argument 1, 2, ..., x. Those values will be passed as the first parameter to the second argument (so, in your case, they will be argument a).
Any additional parameters to lapply after the first two, in your case b=y, are passed to the function by name. So if you called your inner function fxn, then your invocation of lapply is making calls like fxn(1, b=4), fxn(2, b=4), .... The parameters are passed by value.
You should read the help of lapply to understand how it works. Read this excellent answer to get and a good explanation of different xxpply family functions.
From the help of laapply:
lapply(X, FUN, ...)
Here FUN is applied to each elementof X and ... refer to:
... optional arguments to FUN.
Since FUN has an optional argument b, We replace the ... by , b=y.
You can see it as a syntax sugar and to emphasize the fact that argument b is optional comparing to argument a. If the 2 arguments are symmetric maybe it is better to use mapply.
I have two integers a and b (with a less than b), as well as a function f(x). Is there a way of getting the vector
x<-(f(a), ..., f(b))
from R without having to explicitly having to write it out? as my a and b vary.
Thanks for your help.
You can try something like the following :
foo <- function(x) x+1
a <- 1
b <- 5
sapply(a:b, foo)
But note that if you need this kind of behavior, you should vectorize your function, ie make it accept a vector as argument instead of a single integer. In my previous example, the sapply is not needed at all : + is vectorized, so I can just do :
foo(a:b)
Consider the following simple function:
f <- function(x, value){print(x);print(substitute(value))}
Argument x will eventually be evaluated by print, but value never will. So we can get results like this:
> f(a, a)
Error in print(x) : object 'a' not found
> f(3, a)
[1] 3
a
> f(1+1, 1+1)
[1] 2
1 + 1
> f(1+1, 1+"one")
[1] 2
1 + "one"
Everything as expected.
Now consider the same function body in a replacement function:
'g<-' <- function(x, value){print(x);print(substitute(value))}
(the single quotes should be fancy quotes)
Let's try it:
> x <- 3
> g(x) <- 4
[1] 3
[1] 4
Nothing unusual so far...
> g(x) <- a
Error: object 'a' not found
This is unexpected. Name a should be printed as a language object.
> g(x) <- 1+1
[1] 4
1 + 1
This is ok, as x's former value is 4. Notice the expression passed unevaluated.
The final test:
> g(x) <- 1+"one"
Error in 1 + "one" : non-numeric argument to binary operator
Wait a minute... Why did it try to evaluate this expression?
Well the question is: bug or feature? What is going on here? I hope some guru users will shed some light about promises and lazy evaluation on R. Or we may just conclude it's a bug.
We can reduce the problem to a slightly simpler example:
g <- function(x, value)
'g<-' <- function(x, value) x
x <- 3
# Works
g(x, a)
`g<-`(x, a)
# Fails
g(x) <- a
This suggests that R is doing something special when evaluating a replacement function: I suspect it evaluates all arguments. I'm not sure why, but the comments in the C code (https://github.com/wch/r-source/blob/trunk/src/main/eval.c#L1656 and https://github.com/wch/r-source/blob/trunk/src/main/eval.c#L1181) suggest it may be to make sure other intermediate variables are not accidentally modified.
Luke Tierney has a long comment about the drawbacks of the current approach, and illustrates some of the more complicated ways replacement functions can be used:
There are two issues with the approach here:
A complex assignment within a complex assignment, like
f(x, y[] <- 1) <- 3, can cause the value temporary
variable for the outer assignment to be overwritten and
then removed by the inner one. This could be addressed by
using multiple temporaries or using a promise for this
variable as is done for the RHS. Printing of the
replacement function call in error messages might then need
to be adjusted.
With assignments of the form f(g(x, z), y) <- w the value
of z will be computed twice, once for a call to g(x, z)
and once for the call to the replacement function g<-. It
might be possible to address this by using promises.
Using more temporaries would not work as it would mess up
replacement functions that use substitute and/or
nonstandard evaluation (and there are packages that do
that -- igraph is one).
I think the key may be found in this comment beginning at line 1682 of "eval.c" (and immediately followed by the evaluation of the assignment operation's RHS):
/* It's important that the rhs get evaluated first because
assignment is right associative i.e. a <- b <- c is parsed as
a <- (b <- c). */
PROTECT(saverhs = rhs = eval(CADR(args), rho));
We expect that if we do g(x) <- a <- b <- 4 + 5, both a and b will be assigned the value 9; this is in fact what happens.
Apparently, the way that R ensures this consistent behavior is to always evaluate the RHS of an assignment first, before carrying out the rest of the assignment. If that evaluation fails (as when you try something like g(x) <- 1 + "a"), an error is thrown and no assignment takes place.
I'm going to go out on a limb here, so please, folks with more knowledge feel free to comment/edit.
Note that when you run
'g<-' <- function(x, value){print(x);print(substitute(value))}
x <- 1
g(x) <- 5
a side effect is that 5 is assigned to x. Hence, both must be evaluated. But if you then run
'g<-'(x,10)
both the values of x and 10 are printed, but the value of x remains the same.
Speculation:
So the parser is distinguishing between whether you call g<- in the course of making an actual assignment, and when you simply call g<- directly.