Here is a particular way to solve the problem:> test3 <- seq(from=as.Date("1/1/2000","%m/%d/%Y"),to=as.Date("1/3/2000",+ "%m/%d/%Y"),length=3)> test3[1] "2000-01-01" "2000-01-02" "2000-01-03"> zoo(table(dog.df$V1,dog.df$V2)[,1],order=test3)2000-01-01 2000-01-02 2000-01-03 1 1 0> cat.zoo <- zoo(table(dog.df$V1,dog.df$V2)[,1],order=test3) > dog.zoo <- zoo(table(dog.df$V1,dog.df$V2)[,1],order=test3) > dog.zoo <- zoo(table(dog.df$V1,dog.df$V2)[,2],order=test3) > tree.zoo <- zoo(table(dog.df$V1,dog.df$V2)[,3],order=test3) > dog.zoo2000-01-01 2000-01-02 2000-01-03 1 0 1> tree.zoo2000-01-01 2000-01-02 2000-01-03 thanks though! sincerely, Erin -- Erin Hodgess Associate Professor Department of Computer and Mathematical Sciences University of Houston - Downtown mailto: erinm.hodgess at gmail.com
Am 08.06.2010 17:04, schrieb Erin Hodgess:> Here is a particular way to solve the problem: >If you solve your own problem then please reply to your own message otherwise things get confused. How should one know what your problem was without knowing your first e-mail - if you reply your own e-mail it gets sorted as such otherwise not. However> >> cat.zoo <- zoo(table(dog.df$V1,dog.df$V2)[,1],order=test3) >> dog.zoo <- zoo(table(dog.df$V1,dog.df$V2)[,1],order=test3) >>that must be an error, a line to much bur you "heal" it:>> dog.zoo <- zoo(table(dog.df$V1,dog.df$V2)[,2],order=test3) >> tree.zoo <- zoo(table(dog.df$V1,dog.df$V2)[,3],order=test3) >This is way to complicating, you should put the things into one object, then you can e.g. plot them or analyse them. You can do so after this more complicating code with: all<-merge(cat.zoo,dog.zoo,tree.zoo)