On Aug 27, 2012, at 12:18 PM, Bert Gunter wrote:
Well ...See below.
-- Cheers, Bert
On Mon, Aug 27, 2012 at 9:19 AM, David Winsemius <dwinsem...@comcast.net
> wrote:
On Aug 27, 2012, at 3:09 AM, Fridolin wrote:
What is a smart way to change an entry inside a column of a
dataframe or
matrix which is of type "factor"?
Here is my script incl. input data:
#set working directory:
setwd("K:/R")
#read in data:
input<-read.table("Exampleinput.txt", sep="\t", header=TRUE)
#check data:
input
Ind M1 M2 M3
1 1 96/98 120/120 0/0
2 2 102/108 120/124 305/305
3 3 96/108 120/120 0/0
4 4 0/0 116/120 300/305
5 5 96/108 120/130 300/305
6 6 98/98 116/120 300/305
7 7 98/108 120/120 305/305
8 8 98/108 120/120 305/305
9 9 98/102 120/124 300/300
10 10 108/108 120/120 305/305
str(input)
'data.frame': 10 obs. of 4 variables:
$ Ind: int 1 2 3 4 5 6 7 8 9 10
$ M1 : Factor w/ 8 levels "0/0","102/108",..: 5 2 4 1 4 8 7 7 6 3
$ M2 : Factor w/ 4 levels "116/120","120/120",..: 2 3 2 1 4 1 2 2
3 2
$ M3 : Factor w/ 4 levels "0/0","300/300",..: 1 4 1 3 3 3 4 4 2 4
#replace 0/0 by 999/999:
for (r in 1:10)
+ for (c in 2:4)
+ if (input[r,c]=="0/0") input[r,c]<-"999/999"
Warnmeldungen:
1: In `[<-.factor`(`*tmp*`, iseq, value = "999/999") :
invalid factor level, NAs generated
2: In `[<-.factor`(`*tmp*`, iseq, value = "999/999") :
invalid factor level, NAs generated
3: In `[<-.factor`(`*tmp*`, iseq, value = "999/999") :
invalid factor level, NAs generated
input
Ind M1 M2 M3
1 1 96/98 120/120 <NA>
2 2 102/108 120/124 305/305
3 3 96/108 120/120 <NA>
4 4 <NA> 116/120 300/305
5 5 96/108 120/130 300/305
6 6 98/98 116/120 300/305
7 7 98/108 120/120 305/305
8 8 98/108 120/120 305/305
9 9 98/102 120/124 300/300
10 10 108/108 120/120 305/305
I want to replace all "0/0" by "999/999". My code should work for
columns
of
type "character" and "integer". But to make it work for a "factor"-
column
I
would need to add the new level of "999/999" at first, I guess.
How do I
add
a new level?
?levels
levels(input$M1) <- c(levels(input$M1), "999/999")
This adds an additional level; then you have to replace the "0/0"
level with this one; then you have to call levels again to remove the
"0/0" level.
Then do it this way (different from what I thought was originally
desired):
> x <- factor(letters[1:3])
> levels(x) <- c("d", levels(x)[2:3])
> x
[1] d b c
Levels: d b c
I think the following slight tweak may be preferred, as illustrated
with a little example (opinions?):
x <- factor(letters[1:3])
x
[1] a b c
Levels: a b c
## create a new levels vector
newlvl <- levels(x)
newlvl[newlvl == "a"] <- "d"
## Create the new factor and replace the old with it
x <- factor(newlvl[x])
x
[1] d b c
Levels: b c d
Note, however, as Bill D. said, in either case your level ordering --
which will be used, e.g. in printing and displaying -- will be weird.
So the above method might be what you expect. Several options are now
available to the questioner.
--
David.
--
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
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--
Bert Gunter
Genentech Nonclinical Biostatistics
Internal Contact Info:
Phone: 467-7374
Website:
http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm
David Winsemius, MD
Alameda, CA, USA
______________________________________________
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
