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-boun...@r-project.org> On Behalf Of Marna Wagley
Sent: Wednesday, April 18, 2018 12:56 PM
To: r-help mailing list <r-help@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,
[[alternative HTML version deleted]]
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