Thank you. Sarah Goslee. I am rather new in learning R. So people like you
are great support. Really appreciate you, taking the time to correct my
mistakes. Thanks
On Thu 9 Apr, 2015 6:54 pm Sarah Goslee <sarah.gos...@gmail.com> wrote:
> Hi,
>
> Please don't put quotes around your code. It makes it hard to copy and
> paste. Alternatively, don't post in HTML, because it screws up your
> code.
>
> On Wed, Apr 8, 2015 at 8:57 PM, samarvir singh <samarvir1...@gmail.com>
> wrote:
> > So I have a list that contains certain characters as shown below
> >
> > `list <- c("MY","GM+" ,"TY","RS","LG")`
>
> That's a character vector, not a list. A list is a specific type of object
> in R.
>
> > And I have a variable named "CODE" in the data frame as follows
> >
> > `code <- c("MY GM+", ,"LGTY", "RS","TY")`
>
> That doesn't work, and I have no idea what you expect to have there,
> so I'm deleting the extra comma. Also, your vector is named code, not
> CODE.
>
> code <- c("MY GM+", "LGTY", "RS","TY")
> x <- c(1:4)
>
> > 'x <- c(1:5)
> > `df <- data.frame(x,code)`
>
> You problably actually want
> mydf <- data.frame(x, code, stringsAsFactors=FALSE)
>
> Note I changed the name, because df() is a base R function.
>
>
> > Now I want to create 5 new variables named "MY","GM+","TY","RS","LG"
> >
> > Which takes binary value, 1 if there's a match case in the CODE variable
> >
> > df
> > x code MY GM+ TY RS LG
> > 1 MY GM+ 1 1 0 0 0
> > 2 0 0 0 0 0
> > 3 LGTY 0 0 1 0 1
> > 4 RS 0 0 0 1 0
> > 5 TY 0 0 1 0 0
>
> grepl() will give you a logical match
>
> data.frame(mydf, sapply(code, function(x)grepl(x, mydf$code)),
> stringsAsFactors=FALSE, check.names=FALSE)
>
> Sarah
>
>
> --
> Sarah Goslee
> http://www.functionaldiversity.org
>
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