[R] R help!

[peter dalgaard]

On May 3, 2012, at 14:05 , Nutter, Benjamin wrote:

> So long as x is a character vector, I tend to use the following for this problem.
> 
>> x <- c("12/31/11 23:45", "1/1/12 2:15")
>> 
>> x.split <- strsplit(x, " ")
>> 
>> x.date <- sapply(x.split, function(y) return(y[1]))
>> x.time <- sapply(x.split, function(y) if (length(y) > 1) return(y[2]) else NA)
>> 
>> x.date
> [1] "12/31/11" "1/1/12"  
>> x.time
> [1] "23:45" "2:15" 
>> 
> 
>   Benjamin Nutter |  Biostatistician     |  Quantitative Health Sciences
>   Cleveland Clinic    |  9500 Euclid Ave.  |  Cleveland, OH 44195  | (216) 445-1365
> 


I think this can be simplified to

> sapply(x.split,`[`,1)
[1] "12/31/11" "1/1/12"  
> sapply(x.split,`[`,2)
[1] "23:45" "2:15" 

It's a bit inefficient, though. Other ideas:

> sub(" .*$", "", x)
[1] "12/31/11" "1/1/12"  
> sub("^.* ", "", x)
[1] "23:45" "2:15" 
> read.table(text=x)
        V1    V2
1 12/31/11 23:45
2   1/1/12  2:15


-- 
Peter Dalgaard, Professor
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Email: pd.mes at cbs.dk  Priv: PDalgd at gmail.com