Displaying 1 result from an estimated 1 matches for "elsplit6".
2013 Feb 19
1
data format
...brary(plyr)
elsplit3<-lapply(elsplit2,function(x) join(datetrial,x,by="date1",type="full"))
?elsplit4<-lapply(elsplit3,function(x) {x[,2][is.na(x[,2])]<- "-9999.000";x})
elsplit5<-lapply(elsplit4,function(x) {x[,1]<-format(x[,1],"%Y.%m.%d");x})
elsplit6<-lapply(elsplit5,function(x){substr(x[,1],6,6)<-ifelse(substr(x[,1],6,6)==0," ",substr(x[,1],6,6));substr(x[,1],9,9)<- ifelse(substr(x[,1],9,9)==0," ",substr(x[,1],9,9));x})
?elsplit6[[1]][1:4,]
#?????? date1 discharge
#1 1930. 1. 1 -9999.000
#2 1930. 1. 2 -9999.000
#3...