R help - Jun 2023 - OK, next Q - a sort of factorial on a vector

? Wed, 21 Jun 2023 03:13:52 +1000
Philip Rhoades via R-help <r-help at r-project.org> ?????:
> This:
> 
>    !(1,2,3,4,5)
> 
> would give this:
> 
>    (2,3,4,5, 6,8,10, 12,15, 20)
Do you mean taking a product of every element of the vector with all
following vector elements? A relatively straightforward way would be
(given your vector stored in `x`):

unlist(lapply(seq_along(x), function(i) x[i] * x[-(1:i)]))

(I'm sure it could be golfed further.)
> and this:
> 
>    !(1,2,NA,4,5)
> 
> would give this:
> 
>    (2,4,5, 8,10, 20)
The previous solution seems to give your vector interspersed a bunch of
NAs, so one way to continue would be to filter it using v[!is.na(v)].

-- 
Best regards,
Ivan

Ivan,


On 2023-06-21 03:32, Ivan Krylov wrote:
> ? Wed, 21 Jun 2023 03:13:52 +1000
> Philip Rhoades via R-help <r-help at r-project.org> ?????:
> 
>> This:
>> 
>>    !(1,2,3,4,5)
>> 
>> would give this:
>> 
>>    (2,3,4,5, 6,8,10, 12,15, 20)
> 
> Do you mean taking a product of every element of the vector with all
> following vector elements? A relatively straightforward way would be
> (given your vector stored in `x`):
> 
> unlist(lapply(seq_along(x), function(i) x[i] * x[-(1:i)]))

Perfect!

> (I'm sure it could be golfed further.)

I will look at Sarah's suggestion too.

>> and this:
>> 
>>    !(1,2,NA,4,5)
>> 
>> would give this:
>> 
>>    (2,4,5, 8,10, 20)
> 
> The previous solution seems to give your vector interspersed a bunch of
> NAs, so one way to continue would be to filter it using v[!is.na(v)].

Exactly!

Thanks people - it would have taken forever to work that out myself (it 
has been decades since I looked at R).

Phil.
-- 
Philip Rhoades

PO Box 896
Cowra  NSW  2794
Australia
E-mail:  phil at pricom.com.au