Hai Rui,
It seems doesnt work for me, the "" still there.
So I used this one (Bert suggestion),
test<-lapply(test,function(x){x$RR[x$RR==] <- NA; x})
Best,
Ani
On Sat, Oct 19, 2019 at 6:55 PM Rui Barradas wrote:
> Hello,
>
> Why not use read.xlsx argument 'na.strings', an argumen
Hello,
Why not use read.xlsx argument 'na.strings', an argument that exists in
many file reading functions? (read.table, and derivatives.)
test <- lapply(sheets,function(i) {
read.xlsx("rainfall.xlsx", sheet = i,
startRow = 8, cols = 1:2,
na.strings = "")
})
Ho
Hi ani,
Sorry, a typo in the function - should be:
makeNA(x)<-function(x,varname,value) {
x[,varname][x[,varname]==value]<-NA
return(x)
}
Jim
On Fri, Oct 18, 2019 at 2:01 PM Jim Lemon wrote:
>
> Hi ani,
> You say you want to replace with NA, so:
>
> # it will be easier if you don't use n
Hi ani,
You say you want to replace with NA, so:
# it will be easier if you don't use numbers for the names of the data frames
names(test) <- paste0("Y",1986:2015)
makeNA(x)<-function(x,varname,value) {
x[,varname][x[,varname]<-value]<-NA
return(x)
}
lapply(test,makeNA,list("RR",))
War
Thank you Mr. Bert, but my data frame is in the list,
here 'test' list of data frame have 30 data frames (elements), names '1986'
~ '2015', and each data frame contain two variables, date and R.
>a2<-rbind(test$`1987`)
>is.na(a2$RR)<- a2$RR==
Above is good enough but only for '1987'. Is it po
I'm a little unclear, but maybe ?is.na .
As in:
> x <- c(1:3,)
> x
[1]123
> is.na(x) <- x== ## rhs is an "index vector" of logicals
> x
[1] 1 2 3 NA
Bert Gunter
"The trouble with having an open mind is that people keep coming along and
sticking things into it."
-- O
6 matches
Mail list logo
