how to Subset based on partial matching of columns?
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.goslee at 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 <samarvir1996 at 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