search for: myday

...nse_variable =  (c/r)*log(1+exp(r*(Day-tt))), where "Day" is time-variable my response variable rl <- c(2,1.5,1.8,2,2,2.5,2.6,1.5,2.4,1.7,2.3,2.4,2.2,2.6,                  2.8,2,2.5,1.8,2.4,2.4,2.3,2.6,3,2,2.6,1.8,2.5,2.5,                  2.3,2.7,3,2.2,2.6,1.8,2.5,2.5,2.3,2.7,3,2.2) myday <- rep(c(3,5,7,9,10), each = 8) # creating my predictor time-variable mydata <- data.frame(rl,myday) # data object # fitting model equation in "nls" function, when i assigned initial value for tt = 0.6, CASE-I: > mytest <- nls(rl ~ (c/r)*log(1+exp(r*(myday-tt))), data = my...

...ariable =  (c/r)*log(1+exp(r*(Day-tt))), where "Day" is time-variable   ## my response variable   rl <- c(2,1.5,1.8,2,2,2.5,2.6,1.5,2.4,1.7,2.3,2.4,2.2,2.6,                  2.8,2,2.5,1.8,2.4,2.4,2.3,2.6,3,2,2.6,1.8,2.5,2.5,                  2.3,2.7,3,2.2,2.6,1.8,2.5,2.5,2.3,2.7,3,2.2) myday <- rep(c(3,5,7,9,10), each = 8) # creating my predictor time-variable mydata <- data.frame(rl,myday) # data object ### fitting model equation in "nls" function ### when i assigned initial value for tt = 0.6,   CASE-I:   > mytest <- nls(rl ~ (c/r)*log(1+exp(r*(myday-tt))), da...

Dear all, can anyone please tell me how to generate a sequence of days continuously, however without considering weekends i.e. Saturday and Sunday? I am aware of following code: > seq(as.Date("2010-01-01"), as.Date("2010-02-01"), by="1 day")  [1] "2010-01-01" "2010-01-02" "2010-01-03" "2010-01-04" "2010-01-05"

Hi Can someone help me with this? How can I apply a function to a list of variables. something like this listvar=list("Monday","Tuesday","Wednesday") func=function(x){x[which(x<=10)]=NA} lapply(listvar, func) were Monday=[213,56,345,33,34,678,444] Tuesday=[213,56,345,33,34,678,444] ... in my case I have a neverending list of vectors. Thanks!