For the last statement we may prefer this
so it stays a zoo object:
> m <- merge(Az, Bz, fill = 0)
> m[,1] - m[,2]
1 2 3 4 5
1429 0 3049 0 2219
On Sat, Jul 26, 2008 at 5:53 PM, Gabor Grothendieck
<[EMAIL PROTECTED]> wrote:
> Look at merge.zoo
>
>> library(zoo)
>> dput(A)
> structure(list(DayOfYear = 1:5, x = c(1429L, 3952L, 3049L, 2844L,
> 2219L)), .Names = c("DayOfYear", "x"), class = "data.frame", row.names =
> c("1",
> "2", "3", "4", "5"))
>> B <- A[c(2,4),]
>> Az <- zoo(A$x, A$DayOfYear)
>> Bz <- zoo(B$x, B$DayOfYear)
>> merge(Az, Bz, fill = 0)
> Az Bz
> 1 1429 0
> 2 3952 3952
> 3 3049 0
> 4 2844 2844
> 5 2219 0
>> merge(Az, Bz, fill = 0) %*% c(1, -1)
> [,1]
> [1,] 1429
> [2,] 0
> [3,] 3049
> [4,] 0
> [5,] 2219
>
>
> On Sat, Jul 26, 2008 at 5:26 PM, <[EMAIL PROTECTED]> wrote:
>> I have two vectos (list) that represent a years of data. Each "row" is
>> represented by the day of year and the quantity that was sold for that day.
>> I would like to form a new vector that is the difference between the two
>> years of data. A sample of A (and similarly B) looks like:
>>
>>> A[1:5,]
>> DayOfYear x
>> 1 1 1429
>> 2 2 3952
>> 3 3 3049
>> 4 4 2844
>> 5 5 2219
>>>
>>
>> D <- A - B
>>
>> This works just fine if A and B are both the same length. How is the best
>> way to handle the situation where A and B are of different lengths? If the
>> day of year exists in both vectors (lists) then I just want the coorespondng
>> "row" in D to be the difference btween A and B values. If the "row" doesn't
>> exist in either A or B then the difference should be treated as if the
>> missing "row" was zero. Is this feasible?
>>
>> Thank you.
>>
>> Kevin
>>
>> ______________________________________________
>> [email protected] mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
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