Also, BTW, dat.num() is matrix, but if you use lapply(), it is still a
dataframe. Anyway, it depends on what the OP really wants as output.
dat.num <- apply(dat, 2, FUN = function(x) as.numeric(as.character(x)))
dat[] <- lapply(dat,function(x) as.numeric(as.character(x)))
str(dat)
'data.fram
Firstly, please make sure to reply-all so the r-help list also receives
these emails.
Second, I have just run this sequence as it provides an exact copy with
each as numeric. Use the apply function, it iterates over each column and
converts each to numeric.
dat <- read.table(text="a
I'm not honestly sure why data.matrix didn't work off hand. Perhaps
another user can shed some light on this. An alternative is the following:
apply(dat, 2, FUN = function(x) as.numeric(as.character(x)))
On Thu, Oct 10, 2013 at 8:26 AM, arun wrote:
> Did you mean to apply it like this or is
data.matrix() should do the job for you
Charles
On Thu, Oct 10, 2013 at 8:02 AM, arun wrote:
> Hi,
> It is not clear whether all the variables are factor or only a few are..
>
> dat<- read.table(text="acoef
> coef.l coef.h
> 1 1 0.005657825001254 0.00300612956
Hi,
It is not clear whether all the variables are factor or only a few are..
dat<- read.table(text="a coef coef.l
coef.h
1 1 0.005657825001254 0.00300612956318132 0.00830952043932667
2 2 0.00634505314577229 0.00334102345418614 0.00934908283735844
foo <- as.numeric(as.character(your_factors) ).
It's a common mistake to forget the first conversion, in which case you end
up with an integer sequence rather than the desired values.
--
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