I think the
merge()
function should be adequate for this task. Here is an example.
A <- data.frame(day=1:5, x=runif(5))
B <- data.frame(day=3:7, x=runif(5))
A
day x
1 1 0.9764534
2 2 0.9693998
3 3 0.1324933
4 4 0.8311153
5 5 0.3264465
B <- data.frame(day=3:8, x=run
Sorry, the last one should be:
ix <- match(B$DayOfYear, A$DayOfYear)
A[ix, "x"] <- A[ix, "x"] - B$x
Again we are assuming B's days are a subset of A's.
On Sat, Jul 26, 2008 at 6:08 PM, Gabor Grothendieck
<[EMAIL PROTECTED]> wrote:
> Here is a third solution.
>
> A[B$DayOfYear, "x"] <- A[B$DayOfY
Here is a third solution.
A[B$DayOfYear, "x"] <- A[B$DayOfYear, "x"] - B$x
Its assumes B's days are a subset of A's but if that's not the case then
you would need to intersect them first: ?intersect
On Sat, Jul 26, 2008 at 5:26 PM, <[EMAIL PROTECTED]> wrote:
> I have two vectos (list) that repr
Here is a second solution. This one uses sqldf instead of zoo:
library(zoo)
sqldf("select A.x - ifnull(B.x, 0) from A left join B using(DayOfYear)")
See
http://sqldf.googlecode.com
On Sat, Jul 26, 2008 at 5:26 PM, <[EMAIL PROTECTED]> wrote:
> I have two vectos (list) that represent a years of
For the last statement we may prefer this
so it stays a zoo object:
> m <- merge(Az, Bz, fill = 0)
> m[,1] - m[,2]
12345
14290 30490 2219
On Sat, Jul 26, 2008 at 5:53 PM, Gabor Grothendieck
<[EMAIL PROTECTED]> wrote:
> Look at merge.zoo
>
>> library(zoo)
>> dput(A)
> st
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
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,]
Day
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