R help - Feb 2009 - factors to integers preserving value in a dataframe

I want to produce a dataframe with integer columns for elements of  
string pairs:

pairs <- c("10 21","23 45")
pairs.split <- lapply(pairs,function(x)strsplit(x," "))
pdf <- as.data.frame(pairs.split)
names(pdf) <- c("p","q")

-- at this point things look good, except the columns are factors, as  
I didn't change the default stringsAsFactors parameter to the  
as.data.frame.

Now if I want to convert columns to integers, I get

 > typeof(pdf$p)
[1] "integer"
 > pdf$p
[1] 10 21
Levels: 10 21
 > as.integer(pdf$p)
[1] 1 2

-- being factor levels instead of the original values.  I could have  
used stringsAsFactors=F and then convert the strings to integers with  
as.integer all the same; what other ways are there -- e.g., is there a  
way to convert integer-looking factors to integers directly, without  
substituting levels?

Cheers,
Alexy

R-FAQ 7.10
:
http://cran.r-project.org/doc/FAQ/R-FAQ.html#How-do-I-convert-factors-to-numeric_003f

--  
David Winsemius
On Feb 27, 2009, at 2:56 PM, Alexy Khrabrov wrote:
> I want to produce a dataframe with integer columns for elements of  
> string pairs:
>
> pairs <- c("10 21","23 45")
> pairs.split <- lapply(pairs,function(x)strsplit(x," "))
> pdf <- as.data.frame(pairs.split)
> names(pdf) <- c("p","q")
>
> -- at this point things look good, except the columns are factors,  
> as I didn't change the default stringsAsFactors parameter to the  
> as.data.frame.
>
> Now if I want to convert columns to integers, I get
>
> > typeof(pdf$p)
> [1] "integer"
> > pdf$p
> [1] 10 21
> Levels: 10 21
> > as.integer(pdf$p)
> [1] 1 2
>
> -- being factor levels instead of the original values.  I could have  
> used stringsAsFactors=F and then convert the strings to integers  
> with as.integer all the same; what other ways are there -- e.g., is  
> there a way to convert integer-looking factors to integers directly,  
> without substituting levels?
>
> Cheers,
> Alexy
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.

have a look at R FAQ 7.10, e.g., try

as.numeric(levels(pdf$p))[as.integer(pdf$p)]


Best,
Dimitris


Alexy Khrabrov wrote:
> I want to produce a dataframe with integer columns for elements of 
> string pairs:
> 
> pairs <- c("10 21","23 45")
> pairs.split <- lapply(pairs,function(x)strsplit(x," "))
> pdf <- as.data.frame(pairs.split)
> names(pdf) <- c("p","q")
> 
> -- at this point things look good, except the columns are factors, as I 
> didn't change the default stringsAsFactors parameter to the
as.data.frame.
> 
> Now if I want to convert columns to integers, I get
> 
>  > typeof(pdf$p)
> [1] "integer"
>  > pdf$p
> [1] 10 21
> Levels: 10 21
>  > as.integer(pdf$p)
> [1] 1 2
> 
> -- being factor levels instead of the original values.  I could have 
> used stringsAsFactors=F and then convert the strings to integers with 
> as.integer all the same; what other ways are there -- e.g., is there a 
> way to convert integer-looking factors to integers directly, without 
> substituting levels?
> 
> Cheers,
> Alexy
> 
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide 
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
> 
-- 
Dimitris Rizopoulos
Assistant Professor
Department of Biostatistics
Erasmus University Medical Center

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