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help with gsub and date pattern

3 messages · Tim Clark, Marc Schwartz, Gabor Grothendieck

Tim Clark
# 
Dear List,

I am having a problem using gsub to remove dates from a date/time string.

For example:

x<-c("5/31/2009 12:34:00","6/1/2009 1:14:00")

I would like to remove the date and have just the time.

I have tried:
gsub("[0-9+]/[0-9+]/[0-9+]","",x)

and various versions.  I think my problem is that the / is a special character and is telling it something that I don't mean.  I would appreciate any suggestions on how to proceed.

Thanks,

Tim



Tim Clark
Department of Zoology 
University of Hawaii
Marc Schwartz
#
On May 21, 2009, at 4:13 PM, Tim Clark wrote:

            

      
      
      
    

        
Switch the '+' to outside the brackets:

 > gsub("[0-9]+/[0-9]+/[0-9]+ ","",x)
[1] "12:34:00" "1:14:00"


A few other options:

# Use strsplit
 > sapply(strsplit(x, split = " "), "[", 2)
[1] "12:34:00" "1:14:00"


# Return the pattern contained within the parens
# See ?regex
 > gsub("^.* (.*)$", "\\1", x)
[1] "12:34:00" "1:14:00"


# Replace the characters up to the space with an empty vector
 > gsub("^.* ", "", x)
[1] "12:34:00" "1:14:00"


HTH,

Marc Schwartz

  
  


  


      

    

    

      
      

        


  

        

      Gabor Grothendieck
    

    

      
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Here are a few more to add to the list.

x <- c("5/31/2009 12:34:00","6/1/2009 1:14:00")

# 1
read.table(textConnection(x), as.is = TRUE)[, 2]

# 2
format(as.POSIXct(x, format = "%m/%d/%Y %H:%M:%S"), "%H:%M:%S")

# 3
library(gsubfn)
strapply(x, "[0-9]*:..:..", simplify = c)

Also see R News 4/1.

        
On Thu, May 21, 2009 at 5:34 PM, Marc Schwartz <marc_schwartz at me.com> wrote: