Or write `a[(1:3)[1:2]] = 1` instead.
I'm also very unclear on what's going on in R since R does not return a
view object. Moreover, we have this:
> a <- c(0,0,0,0,0)
> a[1:3][1:2] <- 1
> a
[1] 1 1 0 0 0
Yet we also have this:
> a <- c(0,0,0,0,0)
> b <- a[1:3]
> b[1:2] <- 1
> a
[1] 0 0 0 0 0
Huh? So apparently the `a[1:3]` part of `a[1:3][1:2]` cannot be considered
a proper subexpression. In other words, `a[1:3]` here does not do what
`a[1:3]` by itself does – it is inextricably connected to the following
`[1:2]` in some fashion. This interpretation seems to be supported by the
fact that attempting to parenthesize this part is an error:
> a <- c(0,0,0,0,0)
> (a[1:3])[1:2] <- 1
Error in (a[1:3])[1:2] <- 1 : could not find function "(<-"
Can any R expert explain this behavior?
On Sun, Aug 21, 2016 at 4:43 AM, 'Greg Plowman' via julia-users <
julia-users@googlegroups.com> wrote:
> But also note that a[1:3] = 1.0 will modify a (rather than a copy)
>
> On Sunday, August 21, 2016 at 6:34:28 PM UTC+10, Kristoffer Carlsson wrote:
>>
>> Range indexing produces copies yes. Either just write the loops or use
>> "view" or "sub" to refer to the original memory.
>
>
