Consider removing those elements from a vector that match a certain set if criteria. The expected behaviour is to remove those that match, and, in particular, if none match then remove none:
> d = 1:20
> d
[1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
> d[-which(d > 10)]
[1] 1 2 3 4 5 6 7 8 9 10
> d[-which(d > 100)]
integer(0)
We see here that the final statement has both done something very unexpected and silently hidden the error without even a warning.
I initially thought that this was an undesirable (but consistent) consequence of the choice that an empty index selects all elements of a vector
http://stat.ethz.ch/R-manual/R-devel/library/base/html/Extract.html
as is commonly used to e.g. select the first column of a matrix, m, by writing
m[ , 1]
However the behaviour observed here is consistent with interpreting an empty vector as "no elements", not "all elements":
> a = integer(0)
selecting "no elements" works exactly as expected:
> v[a]
numeric(0)
however removing "no elements" does not:
> v[-a]
numeric(0)
For an empty vector to both select no elements and remove all elements requires inconsistency.
Obviously it is possible to work around this issue, either by checking that the which() returns non-zero length or using a logical expression as covered here In R, why does deleting rows or cols by empty index results in empty data ? Or, what's the 'right' way to delete?
but my two questions are:
Why is the behaviour inconsistent?
Why does it silently do the wrong thing without an error or warning?
This doesn't work because which(d > 100) and -which(d > 100) are the same object: there is no difference between an empty vector and the negative of that empty vector.
For example, imagine you did:
d = 1:10
indexer = which(d > 100)
negative_indexer = -indexer
The two variables would be the same (which is the only consistent behavior- turning all the elements of an empty vector negative leaves it the same since it has no elements).
indexer
#> integer(0)
negative_indexer
#> integer(0)
identical(indexer, negative_indexer)
#> [1] TRUE
At that point, you couldn't expect d[indexer] and d[negative_indexer] to give different results. There is also no place to provide an error or warning: it doesn't know when passed an empty vector that you "meant" the negative version of that empty vector.
The solution is that for subsetting there's no reason you need which() at all: you could use d[d > 10] instead of your original example. You could therefore use !(d > 100) or d <= 100 for your negative indexing. This behaves as you'd expect because d > 10 or !(d > 100) are logical vectors rather than vectors of indices.
Related
I want to test if an object is a vector in R. I'm confused as to why
is.vector(c(0.1))
returns TRUE and so does
is.vector(0.1)
I would like it to return false when it is just a number and true when it is a vector. Can anyone offer any help on this please?
Many thanks in advance.
in R there doesn't exist a single number or string alone. They are vectors of length 1. Or embedded in some more complex structures.
is.vector(c(0.1)) and is.vector(0.1) are in R absolutely identical.
That is also the reason, why length("this is a string/character") returns 1 - because length() in this case measures the number of elements in the vector.
And you see it if you type "this is a string/character" into R console:
It returns [1] "this is a string/character" - the [1] indicates: vector of length 1.
So you have to do nchar("this is a string/character") to get the length of the first element - the charater string - returning 26.
nchar(c("this is a string/character", "and this another string"))
## [1] 26 23
## nchar is vectorized as you see ...
This is an important difference to Python, where strings and numbers can stand alone.
So len("this") returns 4 in Python. len(["this"]) however 1 (1 element in list, thus length of list is 1).
As already mentioned by #RHertel, R considers c(0.1) a vector of length 1. You may want to test for length as well. E.g.
> x <- 1
> y <- 1:2
> is.vector(x) & length(x) > 1
[1] FALSE
> is.vector(y) & length(y) > 1
[1] TRUE
Suppose I have a vector x = 1:10, and it is constructed by concatenating two other vectors a = integer(0) and b = 1:10 together (this is an edge case). I want to split up the combined vector again into a and b later on. I would have thought I could safely separate them with:
i = seq_along(a)
x[i]
x[-i]
But I discovered that when I use x[-integer(0)] I get integer(0) returned, instead of x itself as I naively thought. What is the best way to do this sort of thing?
If you want to use negative indexing and the index may degenerate to integer(0) (for example, the index is computed from which), pad a large "out-of-bound" value to the index. Removing an "out-of-bound" value has no side effect.
x <- 1:10
i <- integer(0)
x[-c(i, 11)] ## position 11 is "out-of-bound"
# [1] 1 2 3 4 5 6 7 8 9 10
If you bother setting this "out-of-bound" value, here is a canonical choice: 2 ^ 31, because this value has exceeded representation range of 32-bit signed integer, yet it is not Inf.
An alternative way is to do an if test on length(i). For example:
if (length(i)) x[-i] else x
Caution: don't use function ifelse for this purpose.
This question already has answers here:
Is there an R function for finding the index of an element in a vector?
(4 answers)
Closed 4 years ago.
I am confused about the which function. Basically I thought that it checks at which position of an input object (e.g., a vector) a logical condition is true. As seen in the documentation:
which(LETTERS == "R")
[1] 18
In other words, it goes through all LETTERS values and checks if value == R. But this seems to be a misunderstanding. If I input
a <- c("test","test2","test3","test4")
b <- c("test","test3")
which(a==b)
[1] 1
it returns [1] 1 although test3 does also appear in both vectors. Also, if I input a shorter vector for a, it returns a warning:
a <- c("test","test2","test3")
b <- c("test","test3")
which(a==b)
[1] 1
Warning message:
In a == b : longer object length is not a multiple of shorter object length
My question here is twofold:
How can I return the positions of a character vector a that match a character vector b?
How does which() operate because I obviously misunderstand the function.
Thank you for your answers
Edit: thank you for your quick replies, you clarified my misunderstanding!
You need to give which an input that tells it what elements of a are in b:
which(a%in%b)
[1] 1 3
which essentially identifies which elements are TRUE in a logical vector.
== compares values 1 by 1 (a[1]==b[1]);(a[2]==b[2])..... and not as sets.
for set operations use %in%
use a[which(a %in% b)] to get [1] "test" "test3"
which() returns the index of TRUE expressions (!) not the value.
which(a %in% b) will return
[1] 1 3
the reason for the strange warning message is R's recycling
Warning message:
In a == b : longer object length is not a multiple of shorter object length
so when you compare a vector of length 4 with a vector of length 2, value by value (using == ), R 'recycles' the short vector. in 4 and 2 it works and you will get an answer for this question: (a1==b1,a2==b2,a3==b1,a4==b2). in case of length 4 and 3 - you get a warning message saying the short vector cannot be multiplied by an integer to get the long vector length.
If I have the following vector :
x
[1] 1 5 8 9 1 0 15 15
and I want to know how many elements are greater than 10, how can I proceed without using a loop ?
I would like to get :
2
as a result
Use length or sum:
> length(x[x > 10])
[1] 2
> sum(x > 10)
[1] 2
In the first approach, you would be creating a vector that subsets the values that matches your condition, and then retrieving the length of the vector.
In the second approach, you are simply creating a logical vector that states whether each value matches the condition (TRUE) or doesn't (FALSE). Since TRUE and FALSE equate to "1" and "0", you can simply use sum to get your answer.
Because the first approach requires indexing and subsetting before counting, I am almost certain that the second approach would be faster than the first.
Another way to do this:
> length(which(as.vector(x) > 10))
[1] 2
I have a script that has a bunch of quality control checksums and it got caught on a dataset that had no need to remove any samples (rows) due to quality control. However, this script gave me an unexpected result of a dataframe with zero rows. With example data, why does this work:
data(iris)
##get rid of those pesky factors
iris$Species <- NULL
med <- which(iris[, 1] < 4.9)
medtemp <- iris[-med, ]
dim(medtemp)
[1] 134 4
but this returns a dataframe of zero rows:
small <- which(iris[, 1] < 4.0)
smalltemp <- iris[-small, ]
dim(smalltemp)
[1] 0 4
As does this:
x <- 0
zerotemp <- iris[-x, ]
dim(zerotemp)
[1] 0 4
It seems that the smalltemp dataframe should be the same size as iris since there are no rows to remove at all. Why is this?
Copied verbatim from Patrick Burns's R Inferno p. 41 (I hope this constitutes "fair use" -- if someone objects I'll remove it)
negative nothing is something
> x2 <- 1:4
> x2[-which(x2 == 3)]
[1] 1 2 4
The command above returns all of the values in x2 not equal to 3.
> x2[-which(x2 == 5)]
numeric(0)
The hope is that the above command returns all of x2 since no elements are
equal to 5. Reality will dash that hope. Instead it returns a vector of length
zero.
There is a subtle difference between the two following statements:
x[]
x[numeric(0)]
Subtle difference in the input, but no subtlety in the difference in the output.
There are at least three possible solutions for the original problem.
out <- which(x2 == 5)
if(length(out)) x2[-out] else x2
Another solution is to use logical subscripts:
x2[!(x2 %in% 5)]
Or you can, in a sense, work backwards:
x2[ setdiff(seq along(x2), which(x2 == 5)) ]
Could it be that in your second example, small evaluates to 0?
Taking the zeroth element of a vector will always return the empty vector:
> foo <- 1:3
> foo
[1] 1 2 3
> foo[0]
integer(0)
>
Instead of using which to get your indices, I would use a boolean vector and negate it. That way you can do this:
small <- iris[, 1] < 4.0
smalltemp <- iris[!small, ]
dim(smalltemp)
[1] 150 4
EDIT: I don't think a negative index of 0 (as in your case) is allowed since there is no 0th index and thus R can't exclude that index from your selection. Negative indexing can be interpreted as: "give me back all rows except those with these indices".
It is because of the rules of what to do with an index that is zero. Only strictly positive or strictly negative indices are allowed. As [0] returns nothing, and
R> -0 == 0
[1] TRUE
Hence you get nothing where you expected it to drop nothing.
The identical(0) issue is treated as indexing by a NULL and this is documented to work as if indexing by 0 and hence the same behaviour.
This is discussed in the R Language Definition manual