subset
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
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