CODEWITHSUNDEEP
r
user3627010
user3627010

Reputation: 31

Understanding Indexing in R

I am just learning R, and trying to play around with indexing to help me understand. But, I try this code, and don't get what I expect:

> x 
[1] 3  6  1 NA  2
> x[!is.na(x[x>2])]
[1]  3  6 NA  2

Shouldn't the results of the second command be those elements of x that are not NA and are GT 2? In other words, it should return 3 6. What am I not understanding about this?

Upvotes: 0

Views: 119

Answers (2)

James King
James King

Reputation: 6365

This should do what you want:

x[!is.na(x) & x > 2]

Your command

x[!is.na(x[x>2])]

first returns all the x[x>2], which is

3  6 NA
!is.na(c(3,  6, NA)) 
TRUE  TRUE FALSE

So x[!is.na(x[x>2])] has the effect of dropping the third element of x.

Upvotes: 3

useR
useR

Reputation: 3082

As

> !is.na(x[x>2])
[1]  TRUE  TRUE FALSE

While the length of your vector is 5 Therefore it will give 

[1] TRUE TRUE FALSE TRUE TRUE

Thus, x[c(1,2,4,5)] are returns

My approach will be

>x <- x[!is.na(s)]
>x[x>2]
[1] 3 6

I am still thinking whether one line code is able to do the task.

Okay @6pool have provide a better answer 

x[!is.na(x) & x > 2]

Upvotes: 1

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