[R] I think a simple question
Jeffrey Robert Spies
jspies at nd.edu
Mon Nov 13 00:04:30 CET 2006
Note, the wrapped sort is a really tight solution but won't
necessarily always give you an "intertwined" array. For example, if
x=31 (or basically, tempin[2] or greater), then you would get 1, 31,
32, etc. instead of 1, 32, 31.... Depends on what you were looking for.
Jeff.
On Nov 12, 2006, at 5:45 PM, Gavin Simpson wrote:
> On Sun, 2006-11-12 at 17:20 -0500, Leeds, Mark (IED) wrote:
>> I have index ( of a vector ) values of say
>>
>> tempin<-c(1 31 61 91 121 all the way upto 1411)
>>
>> What I want is a function that takes in a number say, x = 5, and
>> gives
>> me an new vector
>> of
>>
>> tempout<-1 6 31 36 91 96 121 126 .......... 1411 1416
>>
>> This can't be so hard but I can't get it and I've honestly tried.
>> Obviously, tempin + 5 gives me the missing values but I don't know
>> how
>> to interwine them in the order above. Thanks for any help
>> you can provide.
>>
>> mark
>
> Hi Mark,
>
> ?sort, as in, use sort on the concatenated vector of data and (data +
> 5):
>
> ## dummy data
>> dat <- floor(seq(1, 1411, length = 20))
>> dat
> [1] 1 75 149 223 297 372 446 520 594 668 743
> [12] 817 891 965 1039 1114 1188 1262 1336 1411
> ## simply sort dat and dat + 5, concatenated together
>> sort(c(dat, dat + 5))
> [1] 1 6 75 80 149 154 223 228 297 302 372
> [12] 377 446 451 520 525 594 599 668 673 743 748
> [23] 817 822 891 896 965 970 1039 1044 1114 1119 1188
> [34] 1193 1262 1267 1336 1341 1411 1416
>
> which if you want a function, a suitable wrapper would be:
>
> foo <- function(x, inc) {
> sort(c(x, x + inc))
> }
>
>> foo(dat, 5)
> [1] 1 6 75 80 149 154 223 228 297 302 372
> [12] 377 446 451 520 525 594 599 668 673 743 748
> [23] 817 822 891 896 965 970 1039 1044 1114 1119 1188
> [34] 1193 1262 1267 1336 1341 1411 1416
>
> Of course, checking for suitability of input data for foo is left
> up to
> you.
>
> HTH
>
> G
>
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