R help on Vector access - r

I am new to R and trying to understand the effect of the following code.
> x <- c(1, 2)
> x[0]
numeric(0)
> x[FALSE]
numeric(0
> x[c(FALSE, TRUE)]
[1] 2
Specifically, having extensive background in C and C++, I am interesting in knowing what R does internally when accessing an element at index 0. I know that R has 1 based array indexing. But in this specific case, does it access the vector and then remove the result (numeric(0)) or does it remove 0 from the vector and show the results?
So, I want to know what is the definitive way to know about this? What should I type in R as part of '?' or 'help' command?

Based on comments from Roland and G. Grothendieck, I did a quick readup of the R language definition. The answer is right there in $3.4.1
A special case is the zero index, which has null effects: x[0] is an
empty vector and otherwise including zeros among positive or negative
indices has the same effect as if they were omitted.

Related

Why does 'out of bounds' indexing differ between a matrix and a data.frame?

I'm sure this is kind of basic, but I'd just like to really understand the logic of R data structures here.
If I subset a matrix by index out of bounds, I get exactly that error:
m <- matrix(data = c("foo", "bar"), nrow = 1)
m[2,]
# Error in m[2, ] : subscript out of bounds
If I do the same do a data frame, however, I get all NA rows:
df <- data.frame(foo = "foo", bar = "bar")
df[2,]
# foo bar
# NA <NA> <NA>
If I subset into a non-existent data frame column I get the familiar
df[, 3]
# Error in `[.data.frame`(df, , 3) : undefined columns selected
I know (roughly) that data frame rows are weird and to be treated carefully, but I don't quite see the connection to the above behavior.
Can someone explain why R behaves in this way for non-existent df rows?
Update
To be sure, giving NA on out-of-bounds subsets, is normal R behavior for 1D vectors:
vec <- c("foo", "bar")
vec[3]
# [1] NA
So in a way, the weird one out here is matrix subsetting, not dataframe subsetting, depending from where you're starting out.
Still the different 2D subsetting behavior (m[2, ] vs df[2, ]) might strike a dense user (as I am right now) as inconsistent.
Can someone explain why R behaves in this way[?]
Short answer: No, probably not.
Longer answer:
Once upon a time I was thinking about something similar and read this thread on R-devel: Definition of [[. Basically it boils down to:
The semantics of [ and [[ don't seem to be fully specified in the Reference manual. [...] I assume that these are features, not bugs, but I can't find documentation for them
Duncan Murdoch, a former member of the R core team gives a very nice reply:
There is more documentation in the man page for Extract, but I think it is incomplete. The most complete documentation is of course the source code*, but it may not answer the question of what's intentional and what's accidental
As mentioned in the R-devel thread, the only description in the manual is 3.4.1 Indexing by vectors:
If i is positive and exceeds length(x) then the corresponding selection is NA
But, this applies to "indexing of simple vectors". Similar out of bounds indexing for "non-simple" vectors does not seem to be described. Duncan Murdoch again:
So what is a simple vector? That is not explicitly defined, and it probably should be.
Thus, it may seem like no one knows the answer to your why question.
See also "8.2.13 nonexistent value in subscript" in the excellent R Inferno by Patrick Burns, and the section "Missing/out of bounds indices" in Hadley's book.
*Source code for the [ subset operator. A search for R_MSG_subs_o_b (which corresponds to error message "subscript out of bounds") provides no obvious clue why OOB [ indexing of matrices and when using [[ give an error, whereas OOB [ indexing of "simple vectors" results in NA.

Any logical test to distinguish between make-up of numerical objects

I was wondering if there a way for R to detect the existence or absence of the sign * as used in the following objects?
In other words, can R understand that a has a * sign but b doesn't?
a = 3*4
b = 12
If you keep the expressions unevaluated, R can understand their internal complexity. Under normal circumstances, though, R evaluates expressions immediately, so there is no way to tell the difference between a <- 3*4 and b <- 12 once the assignments have been made. That means that the answer to your specific question is No.
Dealing with unevaluated expressions can get a bit complex, but quote() is one simple way to keep e.g. 3*4 from being evaluated:
> length(quote(3*4))
[1] 3
> length(quote(12))
[1] 1
If you're working inside a function, you can use substitute to retrieve the unevaluated form of the function arguments:
> f <- function(a) {
+ length(substitute(a))
+ }
> f(12)
[1] 1
> f(3*4)
[1] 3
In case you're pursuing this farther, you should be aware that counting complexity might not be as easy as you think:
> f(sqrt(2*3+(7*19)^2))
[1] 2
What's going on is that R stores expressions as a tree; the top level here is made up of sqrt and <the rest of the expression>, which has length 2. If you want to measure complexity you'll need to do some kind of collapsing or counting down the branches of the tree ...
Furthermore, if you first assign a <- 3*4 and then call f(a) you get 1, not 3, because substitute() gives you back just the symbol a, which has length 1 ... the information about the difference between "12" and "3*4" gets lost as soon as the expression is evaluated, which happens when the value is assigned to the symbol a. The bottom line is that you have to be very careful in controlling when expressions get evaluated, and it's not easy.
Hadley Wickham's chapter on expressions might be a good place to read more.

Does a dot after numbers have any particular meaning within arithmetic statements?

I'm maintaining some code where a previous author has used statements like:
x <- 1. * a + b
or:
if (y < 1.e-3)
or:
z[z < 0.] <- 0
or:
f <- aa + bb / 2.
The dots in these statements aren't appearing within parameter or function names, and they're not appearing within formulas, so I'm having trouble figuring out whether they have any significance. As far as I can tell, similar statements evaluated with constants substituted for the variables don't evaluate any differently.
I thought that perhaps the periods were inserted to coerce the result to a float, but the statements don't seem to be ambiguous in this regard, and I wasn't under the impression that R needed any help in this regard with different numeric types. he only other explanation I can come up with is that the values originally were floats and the previous author got lazy removing the decimal points when they were changed to integers.
Is there any other possible use for the dot that could be relevant, or am I safe cleaning up these statements?

vectorize a bidimensional function in R

I have a some true and predicted labels
truth <- factor(c("+","+","-","+","+","-","-","-","-","-"))
pred <- factor(c("+","+","-","-","+","+","-","-","+","-"))
and I would like to build the confusion matrix.
I have a function that works on unary elements
f <- function(x,y){ sum(y==pred[truth == x])}
however, when I apply it to the outer product, to build the matrix, R seems unhappy.
outer(levels(truth), levels(truth), f)
Error in outer(levels(x), levels(x), f) :
dims [product 4] do not match the length of object [1]
What is the recommended strategy for this in R ?
I can always go through higher order stuff, but that seems clumsy.
I sometimes fail to understand where outer goes wrong, too. For this task I would have used the table function:
> table(truth,pred) # arguably a lot less clumsy than your effort.
pred
truth - +
- 4 2
+ 1 3
In this case, you are test whether a multivalued vector is "==" to a scalar.
outer assumes that the function passed to FUN can take vector arguments and work properly with them. If m and n are the lengths of the two vectors passed to outer, it will first create two vectors of length m*n such that every combination of inputs occurs, and pass these as the two new vectors to FUN. To this, outer expects, that FUN will return another vector of length m*n
The function described in your example doesn't really do this. In fact, it doesn't handle vectors correctly at all.
One way is to define another function that can handle vector inputs properly, or alternatively, if your program actually requires a simple matching, you could use table() as in #DWin 's answer
If you're redefining your function, outer is expecting a function that will be run for inputs:
f(c("+","+","-","-"), c("+","-","+","-"))
and per your example, ought to return,
c(3,1,2,4)
There is also the small matter of decoding the actual meaning of the error:
Again, if m and n are the lengths of the two vectors passed to outer, it will first create a vector of length m*n, and then reshapes it using (basically)
dim(output) = c(m,n)
This is the line that gives an error, because outer is trying to shape the output into a 2x2 matrix (total 2*2 = 4 items) while the function f, assuming no vectorization, has given only 1 output. Hence,
Error in outer(levels(x), levels(x), f) :
dims [product 4] do not match the length of object [1]

How to correctly use lists?

Brief background: Many (most?) contemporary programming languages in widespread use have at least a handful of ADTs [abstract data types] in common, in particular,
string (a sequence comprised of characters)
list (an ordered collection of values), and
map-based type (an unordered array that maps keys to values)
In the R programming language, the first two are implemented as character and vector, respectively.
When I began learning R, two things were obvious almost from the start: list is the most important data type in R (because it is the parent class for the R data.frame), and second, I just couldn't understand how they worked, at least not well enough to use them correctly in my code.
For one thing, it seemed to me that R's list data type was a straightforward implementation of the map ADT (dictionary in Python, NSMutableDictionary in Objective C, hash in Perl and Ruby, object literal in Javascript, and so forth).
For instance, you create them just like you would a Python dictionary, by passing key-value pairs to a constructor (which in Python is dict not list):
x = list("ev1"=10, "ev2"=15, "rv"="Group 1")
And you access the items of an R List just like you would those of a Python dictionary, e.g., x['ev1']. Likewise, you can retrieve just the 'keys' or just the 'values' by:
names(x) # fetch just the 'keys' of an R list
# [1] "ev1" "ev2" "rv"
unlist(x) # fetch just the 'values' of an R list
# ev1 ev2 rv
# "10" "15" "Group 1"
x = list("a"=6, "b"=9, "c"=3)
sum(unlist(x))
# [1] 18
but R lists are also unlike other map-type ADTs (from among the languages I've learned anyway). My guess is that this is a consequence of the initial spec for S, i.e., an intention to design a data/statistics DSL [domain-specific language] from the ground-up.
three significant differences between R lists and mapping types in other languages in widespread use (e.g,. Python, Perl, JavaScript):
first, lists in R are an ordered collection, just like vectors, even though the values are keyed (ie, the keys can be any hashable value not just sequential integers). Nearly always, the mapping data type in other languages is unordered.
second, lists can be returned from functions even though you never passed in a list when you called the function, and even though the function that returned the list doesn't contain an (explicit) list constructor (Of course, you can deal with this in practice by wrapping the returned result in a call to unlist):
x = strsplit(LETTERS[1:10], "") # passing in an object of type 'character'
class(x) # returns 'list', not a vector of length 2
# [1] list
A third peculiar feature of R's lists: it doesn't seem that they can be members of another ADT, and if you try to do that then the primary container is coerced to a list. E.g.,
x = c(0.5, 0.8, 0.23, list(0.5, 0.2, 0.9), recursive=TRUE)
class(x)
# [1] list
my intention here is not to criticize the language or how it is documented; likewise, I'm not suggesting there is anything wrong with the list data structure or how it behaves. All I'm after is to correct is my understanding of how they work so I can correctly use them in my code.
Here are the sorts of things I'd like to better understand:
What are the rules which determine when a function call will return a list (e.g., strsplit expression recited above)?
If I don't explicitly assign names to a list (e.g., list(10,20,30,40)) are the default names just sequential integers beginning with 1? (I assume, but I am far from certain that the answer is yes, otherwise we wouldn't be able to coerce this type of list to a vector w/ a call to unlist.)
Why do these two different operators, [], and [[]], return the same result?
x = list(1, 2, 3, 4)
both expressions return "1":
x[1]
x[[1]]
why do these two expressions not return the same result?
x = list(1, 2, 3, 4)
x2 = list(1:4)
Please don't point me to the R Documentation (?list, R-intro)--I have read it carefully and it does not help me answer the type of questions I recited just above.
(lastly, I recently learned of and began using an R Package (available on CRAN) called hash which implements conventional map-type behavior via an S4 class; I can certainly recommend this Package.)
Just to address the last part of your question, since that really points out the difference between a list and vector in R:
Why do these two expressions not return the same result?
x = list(1, 2, 3, 4); x2 = list(1:4)
A list can contain any other class as each element. So you can have a list where the first element is a character vector, the second is a data frame, etc. In this case, you have created two different lists. x has four vectors, each of length 1. x2 has 1 vector of length 4:
> length(x[[1]])
[1] 1
> length(x2[[1]])
[1] 4
So these are completely different lists.
R lists are very much like a hash map data structure in that each index value can be associated with any object. Here's a simple example of a list that contains 3 different classes (including a function):
> complicated.list <- list("a"=1:4, "b"=1:3, "c"=matrix(1:4, nrow=2), "d"=search)
> lapply(complicated.list, class)
$a
[1] "integer"
$b
[1] "integer"
$c
[1] "matrix"
$d
[1] "function"
Given that the last element is the search function, I can call it like so:
> complicated.list[["d"]]()
[1] ".GlobalEnv" ...
As a final comment on this: it should be noted that a data.frame is really a list (from the data.frame documentation):
A data frame is a list of variables of the same number of rows with unique row names, given class ‘"data.frame"’
That's why columns in a data.frame can have different data types, while columns in a matrix cannot. As an example, here I try to create a matrix with numbers and characters:
> a <- 1:4
> class(a)
[1] "integer"
> b <- c("a","b","c","d")
> d <- cbind(a, b)
> d
a b
[1,] "1" "a"
[2,] "2" "b"
[3,] "3" "c"
[4,] "4" "d"
> class(d[,1])
[1] "character"
Note how I cannot change the data type in the first column to numeric because the second column has characters:
> d[,1] <- as.numeric(d[,1])
> class(d[,1])
[1] "character"
Regarding your questions, let me address them in order and give some examples:
1) A list is returned if and when the return statement adds one. Consider
R> retList <- function() return(list(1,2,3,4)); class(retList())
[1] "list"
R> notList <- function() return(c(1,2,3,4)); class(notList())
[1] "numeric"
R>
2) Names are simply not set:
R> retList <- function() return(list(1,2,3,4)); names(retList())
NULL
R>
3) They do not return the same thing. Your example gives
R> x <- list(1,2,3,4)
R> x[1]
[[1]]
[1] 1
R> x[[1]]
[1] 1
where x[1] returns the first element of x -- which is the same as x. Every scalar is a vector of length one. On the other hand x[[1]] returns the first element of the list.
4) Lastly, the two are different between they create, respectively, a list containing four scalars and a list with a single element (that happens to be a vector of four elements).
Just to take a subset of your questions:
This article on indexing addresses the question of the difference between [] and [[]].
In short [[]] selects a single item from a list and [] returns a list of the selected items. In your example, x = list(1, 2, 3, 4)' item 1 is a single integer but x[[1]] returns a single 1 and x[1] returns a list with only one value.
> x = list(1, 2, 3, 4)
> x[1]
[[1]]
[1] 1
> x[[1]]
[1] 1
One reason lists work as they do (ordered) is to address the need for an ordered container that can contain any type at any node, which vectors do not do. Lists are re-used for a variety of purposes in R, including forming the base of a data.frame, which is a list of vectors of arbitrary type (but the same length).
Why do these two expressions not return the same result?
x = list(1, 2, 3, 4); x2 = list(1:4)
To add to #Shane's answer, if you wanted to get the same result, try:
x3 = as.list(1:4)
Which coerces the vector 1:4 into a list.
Just to add one more point to this:
R does have a data structure equivalent to the Python dict in the hash package. You can read about it in this blog post from the Open Data Group. Here's a simple example:
> library(hash)
> h <- hash( keys=c('foo','bar','baz'), values=1:3 )
> h[c('foo','bar')]
<hash> containing 2 key-value pairs.
bar : 2
foo : 1
In terms of usability, the hash class is very similar to a list. But the performance is better for large datasets.
You say:
For another, lists can be returned
from functions even though you never
passed in a List when you called the
function, and even though the function
doesn't contain a List constructor,
e.g.,
x = strsplit(LETTERS[1:10], "") # passing in an object of type 'character'
class(x)
# => 'list'
And I guess you suggest that this is a problem(?). I'm here to tell you why it's not a problem :-). Your example is a bit simple, in that when you do the string-split, you have a list with elements that are 1 element long, so you know that x[[1]] is the same as unlist(x)[1]. But what if the result of strsplit returned results of different length in each bin. Simply returning a vector (vs. a list) won't do at all.
For instance:
stuff <- c("You, me, and dupree", "You me, and dupree",
"He ran away, but not very far, and not very fast")
x <- strsplit(stuff, ",")
xx <- unlist(strsplit(stuff, ","))
In the first case (x : which returns a list), you can tell what the 2nd "part" of the 3rd string was, eg: x[[3]][2]. How could you do the same using xx now that the results have been "unraveled" (unlist-ed)?
This is a very old question, but I think that a new answer might add some value since, in my opinion, no one directly addressed some of the concerns in the OP.
Despite what the accepted answer suggests, list objects in R are not hash maps. If you want to make a parallel with python, list are more like, you guess, python lists (or tuples actually).
It's better to describe how most R objects are stored internally (the C type of an R object is SEXP). They are made basically of three parts:
an header, which declares the R type of the object, the length and some other meta data;
the data part, which is a standard C heap-allocated array (contiguous block of memory);
the attributes, which are a named linked list of pointers to other R objects (or NULL if the object doesn't have attributes).
From an internal point of view, there is little difference between a list and a numeric vector for instance. The values they store are just different. Let's break two objects into the paradigm we described before:
x <- runif(10)
y <- list(runif(10), runif(3))
For x:
The header will say that the type is numeric (REALSXP in the C-side), the length is 10 and other stuff.
The data part will be an array containing 10 double values.
The attributes are NULL, since the object doesn't have any.
For y:
The header will say that the type is list (VECSXP in the C-side), the length is 2 and other stuff.
The data part will be an array containing 2 pointers to two SEXP types, pointing to the value obtained by runif(10) and runif(3) respectively.
The attributes are NULL, as for x.
So the only difference between a numeric vector and a list is that the numeric data part is made of double values, while for the list the data part is an array of pointers to other R objects.
What happens with names? Well, names are just some of the attributes you can assign to an object. Let's see the object below:
z <- list(a=1:3, b=LETTERS)
The header will say that the type is list (VECSXP in the C-side), the length is 2 and other stuff.
The data part will be an array containing 2 pointers to two SEXP types, pointing to the value obtained by 1:3 and LETTERS respectively.
The attributes are now present and are a names component which is a character R object with value c("a","b").
From the R level, you can retrieve the attributes of an object with the attributes function.
The key-value typical of an hash map in R is just an illusion. When you say:
z[["a"]]
this is what happens:
the [[ subset function is called;
the argument of the function ("a") is of type character, so the method is instructed to search such value from the names attribute (if present) of the object z;
if the names attribute isn't there, NULL is returned;
if present, the "a" value is searched in it. If "a" is not a name of the object, NULL is returned;
if present, the position of the first occurence is determined (1 in the example). So the first element of the list is returned, i.e. the equivalent of z[[1]].
The key-value search is rather indirect and is always positional. Also, useful to keep in mind:
in hash maps the only limit a key must have is that it must be hashable. names in R must be strings (character vectors);
in hash maps you cannot have two identical keys. In R, you can assign names to an object with repeated values. For instance:
names(y) <- c("same", "same")
is perfectly valid in R. When you try y[["same"]] the first value is retrieved. You should know why at this point.
In conclusion, the ability to give arbitrary attributes to an object gives you the appearance of something different from an external point of view. But R lists are not hash maps in any way.
x = list(1, 2, 3, 4)
x2 = list(1:4)
all.equal(x,x2)
is not the same because 1:4 is the same as c(1,2,3,4).
If you want them to be the same then:
x = list(c(1,2,3,4))
x2 = list(1:4)
all.equal(x,x2)
Although this is a pretty old question I must say it is touching exactly the knowledge I was missing during my first steps in R - i.e. how to express data in my hand as an object in R or how to select from existing objects. It is not easy for an R novice to think "in an R box" from the very beginning.
So I myself started to use crutches below which helped me a lot to find out what object to use for what data, and basically to imagine real-world usage.
Though I not giving exact answers to the question the short text below might help the reader who just started with R and is asking similar questions.
Atomic vector ... I called that "sequence" for myself, no direction, just sequence of same types. [ subsets.
Vector ... the sequence with one direction from 2D, [ subsets.
Matrix ... bunch of vectors with the same length forming rows or columns, [ subsets by rows and columns, or by sequence.
Arrays ... layered matrices forming 3D
Dataframe ... a 2D table like in excel, where I can sort, add or remove rows or columns or make arit. operations with them, only after some time I truly recognized that data frame is a clever implementation of list where I can subset using [ by rows and columns, but even using [[.
List ... to help myself I thought about the list as of tree structure where [i] selects and returns whole branches and [[i]] returns item from the branch. And because it is tree like structure, you can even use an index sequence to address every single leaf on a very complex list using its [[index_vector]]. Lists can be simple or very complex and can mix together various types of objects into one.
So for lists you can end up with more ways how to select a leaf depending on situation like in the following example.
l <- list("aaa",5,list(1:3),LETTERS[1:4],matrix(1:9,3,3))
l[[c(5,4)]] # selects 4 from matrix using [[index_vector]] in list
l[[5]][4] # selects 4 from matrix using sequential index in matrix
l[[5]][1,2] # selects 4 from matrix using row and column in matrix
This way of thinking helped me a lot.
Regarding vectors and the hash/array concept from other languages:
Vectors are the atoms of R. Eg, rpois(1e4,5) (5 random numbers), numeric(55) (length-55 zero vector over doubles), and character(12) (12 empty strings), are all "basic".
Either lists or vectors can have names.
> n = numeric(10)
> n
[1] 0 0 0 0 0 0 0 0 0 0
> names(n)
NULL
> names(n) = LETTERS[1:10]
> n
A B C D E F G H I J
0 0 0 0 0 0 0 0 0 0
Vectors require everything to be the same data type. Watch this:
> i = integer(5)
> v = c(n,i)
> v
A B C D E F G H I J
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
> class(v)
[1] "numeric"
> i = complex(5)
> v = c(n,i)
> class(v)
[1] "complex"
> v
A B C D E F G H I J
0+0i 0+0i 0+0i 0+0i 0+0i 0+0i 0+0i 0+0i 0+0i 0+0i 0+0i 0+0i 0+0i 0+0i 0+0i
Lists can contain varying data types, as seen in other answers and the OP's question itself.
I've seen languages (ruby, javascript) in which "arrays" may contain variable datatypes, but for example in C++ "arrays" must be all the same datatype. I believe this is a speed/efficiency thing: if you have a numeric(1e6) you know its size and the location of every element a priori; if the thing might contain "Flying Purple People Eaters" in some unknown slice, then you have to actually parse stuff to know basic facts about it.
Certain standard R operations also make more sense when the type is guaranteed. For example cumsum(1:9) makes sense whereas cumsum(list(1,2,3,4,5,'a',6,7,8,9)) does not, without the type being guaranteed to be double.
As to your second question:
Lists can be returned from functions even though you never passed in a List when you called the function
Functions return different data types than they're input all the time. plot returns a plot even though it doesn't take a plot as an input. Arg returns a numeric even though it accepted a complex. Etc.
(And as for strsplit: the source code is here.)
If it helps, I tend to conceive "lists" in R as "records" in other pre-OO languages:
they do not make any assumptions about an overarching type (or rather the type of all possible records of any arity and field names is available).
their fields can be anonymous (then you access them by strict definition order).
The name "record" would clash with the standard meaning of "records" (aka rows) in database parlance, and may be this is why their name suggested itself: as lists (of fields).
why do these two different operators, [ ], and [[ ]], return the same result?
x = list(1, 2, 3, 4)
[ ] provides sub setting operation. In general sub set of any object
will have the same type as the original object. Therefore, x[1]
provides a list. Similarly x[1:2] is a subset of original list,
therefore it is a list. Ex.
x[1:2]
[[1]] [1] 1
[[2]] [1] 2
[[ ]] is for extracting an element from the list. x[[1]] is valid
and extract the first element from the list. x[[1:2]] is not valid as [[ ]]
does not provide sub setting like [ ].
x[[2]] [1] 2
> x[[2:3]] Error in x[[2:3]] : subscript out of bounds
you can try something like,
set.seed(123)
l <- replicate(20, runif(sample(1:10,1)), simplify = FALSE)
out <- vector("list", length(l))
for (i in seq_along(l)) {
out[[i]] <- length(unique(l[[i]])) #length(l[[i]])
}
unlist(out)
unlist(lapply(l,length))
unlist(lapply(l, class))
unlist(lapply(l, mean))
unlist(lapply(l, max))

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