Silvano -
If you always have exactly one med1, one med2, and one
med3
for each combination of Grupos, Dias, and Rato, you can
use
aggregate(Esp.Inter.Trac~Grupos+Dias+ Rato,mean,data=x)
Grupos Dias Rato Esp.Inter.Trac
1 C 3 1 100.55667
2 GFC 3 1 87.62333
3 C 3 2 80.37000
4 GFC 3 2 101.33000
5 C 3 3 66.97333
6 GFC 3 3 101.52667
7 C 3 4 84.84000
8 GFC 3 4 103.97333
9 C 3 5 76.99333
10 GFC 3 5 121.67000
11 C 3 6 70.78333
12 GFC 3 6 91.41000
If there are multiples of any of med1, med2, or med3
within
any of the combinations, it would be a little trickier:
one =
aggregate(Esp.Inter.Trac~Grupos+Dias+Rato+blocos,mean,data=x)
two =
aggregate(Esp.Inter.Trac~Grupos+Dias+Rato,mean,data=one)
That would still assume there was at least one value for
each of
med1, med2, and med3 for each combination.
- Phil Spector
Statistical Computing Facility
Department of Statistics
UC Berkeley
spector at stat.berkeley.edu
On Thu, 16 Dec 2010, Silvano wrote:
Hi,
I have a file, like below, and I want create a new
data.frame with variables:
Grupos Dias Rato Esp.Inter.Trac, but Esp.Inter.Trac
will be the mean of
med1 med2 and med3 for each Rato.
How can I do this?
Grupos Dias Rato blocos Esp.Inter.Trac
GFC 3 1 med1 85.99
GFC 3 2 med1 112.78
GFC 3 3 med1 105.43
GFC 3 4 med1 86.18
GFC 3 5 med1 135.66
GFC 3 6 med1 76.25
GFC 3 1 med2 91.08
GFC 3 2 med2 100.57
GFC 3 3 med2 131.79
GFC 3 4 med2 138.46
GFC 3 5 med2 129.78
GFC 3 6 med2 107.92
GFC 3 1 med3 85.80
GFC 3 2 med3 90.64
GFC 3 3 med3 67.36
GFC 3 4 med3 87.28
GFC 3 5 med3 99.57
GFC 3 6 med3 90.06
C 3 1 med1 81.19
C 3 2 med1 94.74
C 3 3 med1 49.18
C 3 4 med1 105.76
C 3 5 med1 82.71
C 3 6 med1 80.10
C 3 1 med2 121.30
C 3 2 med2 82.77
C 3 3 med2 99.57
C 3 4 med2 73.66
C 3 5 med2 72.89
C 3 6 med2 76.47
C 3 1 med3 99.18
C 3 2 med3 63.60
C 3 3 med3 52.17
C 3 4 med3 75.10
C 3 5 med3 75.38
C 3 6 med3 55.78
--------------------------------------
Silvano Cesar da Costa
Departamento de Estat?stica
Universidade Estadual de Londrina
Fone: 3371-4346