[R] How to replace numeric value in the column contains Text (Factor)?

[Marna Wagley]

Hi David and Jeff,
Both ways worked for my table. it helped me a lot (I was really struggling
on it to change the values and thinking to do manually). You are great.
Thank you

On Wed, Apr 18, 2018 at 11:13 AM, David L Carlson <dcarlson using tamu.edu> wrote:

> The simplest would be to convert precip to character and then back to a
> factor if you really want it to be a factor. This will also remove the
> levels that no longer exist.
>
> str(dat)
> # 'data.frame':   5 obs. of  3 variables:
> #  $ Sites : Factor w/ 5 levels "Site1","Site2",..: 1 2 3 4 5
> #  $ temp  : num  14 15 12 12.5 17
> #  $ precip: Factor w/ 5 levels "15","34","high",..: 3 4 5 2 1
>
> dat$precip <- as.character(dat$precip)
> dat[4:5, 3] <-"20"
> dat$precip <- factor(dat$precip)
>
> str(dat)
> # 'data.frame':   5 obs. of  3 variables:
> #  $ Sites : Factor w/ 5 levels "Site1","Site2",..: 1 2 3 4 5
> #  $ temp  : num  14 15 12 12.5 17
> #  $ precip: Factor w/ 4 levels "20","high","low",..: 2 3 4 1 1
>
> ----------------------------------------
> David L Carlson
> Department of Anthropology
> Texas A&M University
> College Station, TX 77843-4352
>
> -----Original Message-----
> From: R-help <r-help-bounces using r-project.org> On Behalf Of Marna Wagley
> Sent: Wednesday, April 18, 2018 12:56 PM
> To: r-help mailing list <r-help using r-project.org>
> Subject: [R] How to replace numeric value in the column contains Text
> (Factor)?
>
> Hi R user,
> Would you mind to help me on how I can change a value in a specific column
> and row in a big table? but the column of the table is a factor (not
> numeric).
> Here is an example. I want to change dat[4:5,3]<-"20" but it generated NA>
> do you have any suggestions for me?
>
> dat<-structure(list(Sites = structure(1:5, .Label = c("Site1", "Site2",
> "Site3", "Site4", "Site5"), class = "factor"), temp = c(14, 15, 12, 12.5,
> 17), precip = structure(c(3L, 4L, 5L, 2L, 1L), .Label = c("15", "34",
> "high", "low", "medium"), class = "factor")), .Names = c("Sites", "temp",
> "precip"), class = "data.frame", row.names = c(NA, -5L
> ))
> > dat[4:5, 3] <-"20"
> Warning message:
> In `[<-.factor`(`*tmp*`, iseq, value = c("20", "20")) :
>   invalid factor level, NA generated
> Thanks,
>
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>
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