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

[David L Carlson]

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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