Another approach... not exactly more direct, but perhaps more robust and more
general:
tmp <- aggregate(x,list(lvl=cumsum(abs(diff(c(FALSE,x>0))))), FUN=sum)
ans <- tmp[0==tmp$lvl,"x"]
abs(diff()) finds transitions, FALSE forces level zero to represent negative
numbers
cumsum marks levels (groups of positive and not-positive numbers)
aggregate does the summation
ans may be empty if x started with positive numbers.
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Sent from my phone. Please excuse my brevity.
"R. Michael Weylandt" <michael.weyla...@gmail.com> wrote:
>I'd do something like this:
>
>x <- sample(seq(-10, 10))
>
>sum(x[seq_len(which.max(x > 0)])
>
>Though others might have more direct solutions.
>
>which.max() gets you the index of the first time x > 0 -- seq_len
>gives you numbers 1 to that index -- then just subset and sum like
>normal.
>
>Best,
>Michael
>
>On Tue, Aug 7, 2012 at 7:37 AM, number10 <havana_dr...@ymail.com>
>wrote:
>> Hi, Is it possible to avoid using do and while loops to calculate the
>sum of
>> the elements of a vector until the appearance of the first positive
>element.
>>
>>
>>
>> --
>> View this message in context:
>http://r.789695.n4.nabble.com/Sum-of-vector-elements-tp4639395.html
>> Sent from the R help mailing list archive at Nabble.com.
>>
>> ______________________________________________
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