How to check if entire vector has no values other than NA (or NAN) in R ?

If I use is.na it returns a vector of TRUE / FALSE.

I need to check if there is single not NA element or not.

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# R – How to check if entire vector has no values other than NA (or NAN) in R

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How to check if entire vector has no values other than NA (or NAN) in R ?

If I use is.na it returns a vector of TRUE / FALSE.

I need to check if there is single not NA element or not.

R has many *apply functions which are ably described in the help files (e.g. `?apply`

). There are enough of them, though, that beginning useRs may have difficulty deciding which one is appropriate for their situation or even remembering them all. They may have a general sense that "I should be using an *apply function here", but it can be tough to keep them all straight at first.

Despite the fact (noted in other answers) that much of the functionality of the *apply family is covered by the extremely popular `plyr`

package, the base functions remain useful and worth knowing.

This answer is intended to act as a sort of **signpost** for new useRs to help direct them to the correct *apply function for their particular problem. Note, this is **not** intended to simply regurgitate or replace the R documentation! The hope is that this answer helps you to decide which *apply function suits your situation and then it is up to you to research it further. With one exception, performance differences will not be addressed.

**apply**-*When you want to apply a function to the rows or columns of a matrix (and higher-dimensional analogues); not generally advisable for data frames as it will coerce to a matrix first.*`# Two dimensional matrix M <- matrix(seq(1,16), 4, 4) # apply min to rows apply(M, 1, min) [1] 1 2 3 4 # apply max to columns apply(M, 2, max) [1] 4 8 12 16 # 3 dimensional array M <- array( seq(32), dim = c(4,4,2)) # Apply sum across each M[*, , ] - i.e Sum across 2nd and 3rd dimension apply(M, 1, sum) # Result is one-dimensional [1] 120 128 136 144 # Apply sum across each M[*, *, ] - i.e Sum across 3rd dimension apply(M, c(1,2), sum) # Result is two-dimensional [,1] [,2] [,3] [,4] [1,] 18 26 34 42 [2,] 20 28 36 44 [3,] 22 30 38 46 [4,] 24 32 40 48`

If you want row/column means or sums for a 2D matrix, be sure to investigate the highly optimized, lightning-quick

`colMeans`

,`rowMeans`

,`colSums`

,`rowSums`

.**lapply**-*When you want to apply a function to each element of a list in turn and get a list back.*This is the workhorse of many of the other *apply functions. Peel back their code and you will often find

`lapply`

underneath.`x <- list(a = 1, b = 1:3, c = 10:100) lapply(x, FUN = length) $a [1] 1 $b [1] 3 $c [1] 91 lapply(x, FUN = sum) $a [1] 1 $b [1] 6 $c [1] 5005`

**sapply**-*When you want to apply a function to each element of a list in turn, but you want a***vector**back, rather than a list.If you find yourself typing

`unlist(lapply(...))`

, stop and consider`sapply`

.`x <- list(a = 1, b = 1:3, c = 10:100) # Compare with above; a named vector, not a list sapply(x, FUN = length) a b c 1 3 91 sapply(x, FUN = sum) a b c 1 6 5005`

In more advanced uses of

`sapply`

it will attempt to coerce the result to a multi-dimensional array, if appropriate. For example, if our function returns vectors of the same length,`sapply`

will use them as columns of a matrix:`sapply(1:5,function(x) rnorm(3,x))`

If our function returns a 2 dimensional matrix,

`sapply`

will do essentially the same thing, treating each returned matrix as a single long vector:`sapply(1:5,function(x) matrix(x,2,2))`

Unless we specify

`simplify = "array"`

, in which case it will use the individual matrices to build a multi-dimensional array:`sapply(1:5,function(x) matrix(x,2,2), simplify = "array")`

Each of these behaviors is of course contingent on our function returning vectors or matrices of the same length or dimension.

**vapply**-*When you want to use*`sapply`

but perhaps need to squeeze some more speed out of your code or want more type safety.For

`vapply`

, you basically give R an example of what sort of thing your function will return, which can save some time coercing returned values to fit in a single atomic vector.`x <- list(a = 1, b = 1:3, c = 10:100) #Note that since the advantage here is mainly speed, this # example is only for illustration. We're telling R that # everything returned by length() should be an integer of # length 1. vapply(x, FUN = length, FUN.VALUE = 0L) a b c 1 3 91`

**mapply**-*For when you have several data structures (e.g. vectors, lists) and you want to apply a function to the 1st elements of each, and then the 2nd elements of each, etc., coercing the result to a vector/array as in*`sapply`

.This is multivariate in the sense that your function must accept multiple arguments.

`#Sums the 1st elements, the 2nd elements, etc. mapply(sum, 1:5, 1:5, 1:5) [1] 3 6 9 12 15 #To do rep(1,4), rep(2,3), etc. mapply(rep, 1:4, 4:1) [[1]] [1] 1 1 1 1 [[2]] [1] 2 2 2 [[3]] [1] 3 3 [[4]] [1] 4`

**Map**-*A wrapper to*`mapply`

with`SIMPLIFY = FALSE`

, so it is guaranteed to return a list.`Map(sum, 1:5, 1:5, 1:5) [[1]] [1] 3 [[2]] [1] 6 [[3]] [1] 9 [[4]] [1] 12 [[5]] [1] 15`

**rapply**-*For when you want to apply a function to each element of a***nested list**structure, recursively.To give you some idea of how uncommon

`rapply`

is, I forgot about it when first posting this answer! Obviously, I'm sure many people use it, but YMMV.`rapply`

is best illustrated with a user-defined function to apply:`# Append ! to string, otherwise increment myFun <- function(x){ if(is.character(x)){ return(paste(x,"!",sep="")) } else{ return(x + 1) } } #A nested list structure l <- list(a = list(a1 = "Boo", b1 = 2, c1 = "Eeek"), b = 3, c = "Yikes", d = list(a2 = 1, b2 = list(a3 = "Hey", b3 = 5))) # Result is named vector, coerced to character rapply(l, myFun) # Result is a nested list like l, with values altered rapply(l, myFun, how="replace")`

**tapply**-*For when you want to apply a function to***subsets**of a vector and the subsets are defined by some other vector, usually a factor.The black sheep of the *apply family, of sorts. The help file's use of the phrase "ragged array" can be a bit confusing, but it is actually quite simple.

A vector:

`x <- 1:20`

A factor (of the same length!) defining groups:

`y <- factor(rep(letters[1:5], each = 4))`

Add up the values in

`x`

within each subgroup defined by`y`

:`tapply(x, y, sum) a b c d e 10 26 42 58 74`

More complex examples can be handled where the subgroups are defined by the unique combinations of a list of several factors.

`tapply`

is similar in spirit to the split-apply-combine functions that are common in R (`aggregate`

,`by`

,`ave`

,`ddply`

, etc.) Hence its black sheep status.

The `%in%`

operator tells you which elements are among the numers to remove:

```
> a <- sample (1 : 10)
> remove <- c (2, 3, 5)
> a
[1] 10 5 2 7 1 6 3 4 8 9
> a %in% remove
[1] FALSE TRUE TRUE FALSE FALSE FALSE TRUE FALSE FALSE FALSE
> a [! a %in% remove]
[1] 10 7 1 6 4 8 9
```

Note that this will silently remove incomparables (stuff like `NA`

or `Inf)`

as well (while it will keep duplicate values in `a`

as long as they are not listed in `remove`

).

If

`a`

can contain incomparables, but`remove`

will not, we can use`match`

, telling it to return`0`

for non-matches and incomparables (`%in%`

is a conventient shortcut for`match`

):`> a <- c (a, NA, Inf) > a [1] 10 5 2 7 1 6 3 4 8 9 NA Inf > match (a, remove, nomatch = 0L, incomparables = 0L) [1] 0 3 1 0 0 0 2 0 0 0 0 0 > a [match (a, remove, nomatch = 0L, incomparables = 0L) == 0L] [1] 10 7 1 6 4 8 9 NA Inf`

`incomparables = 0`

is not*needed*as incomparables will anyways not match, but I'd include it for the sake of readability.

This is, btw., what`setdiff`

does internally (but without the`unique`

to throw away duplicates in`a`

which are not in`remove`

).If

`remove`

contains incomparables, you'll have to check for them individually, e.g.`if (any (is.na (remove))) a <- a [! is.na (a)]`

(This does not distinguish

`NA`

from`NaN`

but the R manual anyways warns that one should not rely on having a difference between them)For

`Inf`

/`-Inf`

you'll have to check both`sign`

and`is.finite`

## Best Solution

The function

`all()`

, when passed a Boolean vector, will tell you whether all of the values in it are`TRUE`

: