I'm a little unclear, but maybe ?is.na .
As in:
> x <- c(1:3,8888)
> x
[1] 1 2 3 8888
> is.na(x) <- x==8888 ## 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."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Thu, Oct 17, 2019 at 6:38 PM ani jaya <gaaa...@gmail.com> wrote:
> Dear R-Help,
>
> I have a list of data frame that I import from excel file using read.xlsx
> command.
>
> sheets <- openxlsx::getSheetNames("rainfall.xlsx")
> test <- lapply(sheets,function(i) read.xlsx("rainfall.xlsx", sheet=i,
> startRow=8, cols=1:2))
> names(test) <- sprintf("%i", 1986:2015)
>
> And I got a data.frame with 365 rows and two columns contains date and
> rainfall data.
> There is a value in rainfall data that i want to change as NA (8888)
>
> test[1]$`1986`
> Date RR
> 1 01-01-1986 0
> 2 02-01-1986 7
> 3 03-01-1986 72
> 4 04-01-1986 4
> 5 05-01-1986 19
> 6 06-01-1986 4
> 7 07-01-1986 16
> 8 08-01-1986 21
> 9 09-01-1986 34
> 10 10-01-1986 72
> 11 11-01-1986 93
> 12 12-01-1986 178
> 13 13-01-1986 86
> 14 14-01-1986 11
> 15 15-01-1986 0
> 16 16-01-1986 31
> 17 17-01-1986 22
> 18 18-01-1986 18
> 19 19-01-1986 3
> 20 20-01-1986 0
> 21 21-01-1986 31
> 22 22-01-1986 46
> 23 23-01-1986 4
> 24 24-01-1986 40
> 25 25-01-1986 63
> 26 26-01-1986 125
> 27 27-01-1986 33
> 28 28-01-1986 44
> 29 29-01-1986 6
> 30 30-01-1986 0
> 31 31-01-1986 0
> 32 01-02-1986 2
> 33 02-02-1986 71
> 34 03-02-1986 8888
> 35 04-02-1986 0
> 36 05-02-1986 0
> 37 06-02-1986 56
> 38 07-02-1986 19
> 39 08-02-1986 3
> 40 09-02-1986 7
> 41 10-02-1986 24
> 42 11-02-1986 55
> 43 12-02-1986 0
> 44 13-02-1986 0
>
> [[alternative HTML version deleted]]
>
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>
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