R rep function returns the wrong number of elements

Question

I ran the following code in R version 3.1.2:

> 7 - (.05+.1)/.05
[1] 4
> rep(NA, 4)
[1] NA NA NA NA
> rep(NA, 7 - (.05+.1)/.05)
[1] NA NA NA

Note that the last rep call returned 3 NAs instead of 4 even though 7 - (.05+.1)/.05 = 4. Can others replicate this same error? Is there a simple explanation for it?

`(7 - (.05+.1)/.05) ==4#[1] FALSE` and `rep(NA,3.9)#[1] NA NA NA` — akrun, Feb 24 '15 at 17:18
This is R FAQ 7.31: http://stackoverflow.com/search?q=[r]+faq+7.31 — Ben Bolker, Feb 24 '15 at 17:20
Try `sprintf("%.30f", 7 - (.05+.1)/.05)`. As you can see it's not 4 but 3.9999999... so when `rep` internally calls `as.integer(3.99999)` it gets 3... — digEmAll, Feb 24 '15 at 17:21

score 2 · Accepted Answer · answered Feb 24 '15 at 17:28

7.31 Why doesn’t R think these numbers are equal?

The only numbers that can be represented exactly in R’s numeric type are integers and fractions whose denominator is a power of 2. Other numbers have to be rounded to (typically) 53 binary digits accuracy. As a result, two floating point numbers will not reliably be equal unless they have been computed by the same algorithm, and not always even then. For example
R> a <- sqrt(2)
R> a * a == 2
[1] FALSE
R> a * a - 2
[1] 4.440892e-16
The function all.equal() compares two objects using a numeric tolerance of .Machine$double.eps ^ 0.5. If you want much greater accuracy than this you will need to consider error propagation carefully.

If you're expecting a result to be an integer, you could force it to be a perfect one by using the round function:

7L - (.05+.1)/.05 == 4
[1] FALSE

round(7L - (.05+.1)/.05) == 4
[1] TRUE

R rep function returns the wrong number of elements

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