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
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--
Dimitris Rizopoulos
Assistant Professor
Department of Biostatistics
Erasmus University Medical Center
Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands
Tel: +31/(0)10/7043478
Fax: +31/(0)10/7043014
______________________________________________
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and provide commented, minimal, self-contained, reproducible code.
