ep("Q1_", names(scs.c2))) {
scs.c2[, iCol] <- factor(scs.c2[, iCol], levels=c(...))
}
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> -----Original Message-
> From: Lopez, Dan [mailto:lopez...@llnl.gov]
> Sent: Thursday, February 21, 2013 2:51 PM
&
2L, 2L, 2L, 2L, 2L), .Label = c("", "yes"
), class = "factor"), Q16 = structure(c(4L, 4L, 4L, 3L, 3L,
3L, 4L, 4L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 3L, 4L, 3L,
3L, 3L, 4L, 3L, 4L), .Label = c("", "did not meet expectations",
,"Bill",..: 1 2 3
- attr(*, "Date")= POSIXlt, format: "2013-02-21"
> str(f3(x)) # mangles column names, drops unused levels, drops Date attribute
'data.frame': 3 obs. of 3 variables:
$ No.Yes: Factor w/ 1 level "Yes": 1 1 1
$ Size
r(*, "Date")= POSIXlt, format: "2013-02-21"
> str(f3(x)) # mangles column names, drops unused levels, drops Date attribute
'data.frame': 3 obs. of 3 variables:
$ No.Yes: Factor w/ 1 level "Yes": 1 1 1
$ Size : Ord.factor
roject.org)
Subject: Re: [R] Having trouble converting a dataframe of character vectors to
factors
Pleaser re-read ?sapply and pay particular attention to the "simplify" argument.
The following should help explain the issues:
> z <- data.frame(a=letters[1:3],b=letters[4:6],stringsAsFacto
How about this?
scs2<-data.frame(lapply(scs2, factor))
From: "Lopez, Dan"
To: "R help (r-help@r-project.org)"
Sent: Wednesday, February 20, 2013 7:09 PM
Subject: [R] Having trouble converting a dataframe of character vectors to
fact
Pleaser re-read ?sapply and pay particular attention to the "simplify" argument.
The following should help explain the issues:
> z <- data.frame(a=letters[1:3],b=letters[4:6],stringsAsFactors=FALSE)
> sapply(z,class)
a b
"character" "character"
> z1 <- sapply(z,as.factor)
> sa
R Experts,
I have a dataframe made up of character vectors--these are results from survey
questions. I need to convert them to factors.
I tried the following which did not work:
scs2<-sapply(scs2,as.factor)
also this didn't work:
scs2<-sapply(scs2,function(x) as.factor(x))
After doing either of
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