Hello,
Try the following.
agg <- aggregate(buddleiat ~ samplet + datet, data = traffic, FUN = mean)
mrg <- merge(encounters, agg,
by.x = c("samplec", "datec"),
by.y = c("samplet", "datet"))
mrg$Div <- with(mrg, Bladen/buddleiat)
Hope this helps,
Rui Barradas
Em 18-09-2012 12:17, Marta Miguel escreveu:
Dear all,
I have two different data frames, that have two common variables: date and
sample. Here is a small extract of both of them
head(traffic)
datet sessiont samplet buddleiat
1 07-08-2012 1 1 1
2 07-08-2012 1 1 1
3 07-08-2012 1 1 1
4 07-08-2012 1 2 3
5 07-08-2012 1 2 1
6 07-08-2012 1 2 2
7 07-08-2012 2 3 4
8 07-08-2012 2 3 5
9 07-08-2012 2 3 5
10 07-08-2012 2 4 8
11 07-08-2012 2 4 4
12 07-08-2012 2 4 6
13 08-08-2012 1 1 9
14 08-08-2012 1 1 12
15 08-08-2012 1 1 7
(...)
head(encounters)
datec samplec Bladen
1 07-08-2012 1 9
2 07-08-2012 1 6
3 07-08-2012 1 8
4 07-08-2012 1 8
5 07-08-2012 1 5
6 07-08-2012 1 4
7 07-08-2012 1 7
8 07-08-2012 1 6
9 07-08-2012 1 4
10 07-08-2012 1 2
11 07-08-2012 1 7
12 07-08-2012 1 8
(...)
They don't have the same length.
Using the function tapply, I managed to calculate the mean of the column
buddleiat (from traffic) by sample and date.
tapply(buddleiat, list(samplet,datet), mean)
Now I want to divide each different value of Bladen by the value that has
the same number of sample and date of the buddleiat mean calculated. Being
that, different values from Bladen can be divided by the same value of
buddleiat mean.
I tried using this code:
budt=(Bladen/tapply(buddleiat, list(samplet,datet), mean))
Error: dims [product 68] do not match the length of object [1360]
But they don't have the same length, and I am not sure that it will do what
I am asking.
Do you have any suggestions? Is there a function I could use or will it be
easier to do it "manually" by creating a third data frame with both
variables and repeated values of the buddleiat means?
I hope I made myself understand.
Thanks in advance,
Marta Miguel
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