So I have a list that contains certain characters as shown below
`list <- c("MY","GM+" ,"TY","RS","LG")`
And I have a variable named "CODE" in the data frame as follows
`code <- c("MY GM+", ,"LGTY", "RS","TY")`
'x <- c(1:5)
`df <- data.frame(x,code)`
df
x code
1 MY GM+
2
3 LGTY
4 RS
5 TY
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
Really appreciate your help. Thank you.
how to Subset based on partial matching of columns?
4 messages · samarvir singh, Sarah Goslee, David L Carlson
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$code)), stringsAsFactors=FALSE, check.names=FALSE) Sarah
Sarah Goslee http://www.functionaldiversity.org
Thank you. Sarah Goslee. I am rather new in learning R. So people like you are great support. Really appreciate you, taking the time to correct my mistakes. Thanks
On Thu 9 Apr, 2015 6:54 pm Sarah Goslee <sarah.goslee at gmail.com> wrote:
Hi, Please don't put quotes around your code. It makes it hard to copy and paste. Alternatively, don't post in HTML, because it screws up your code. On Wed, Apr 8, 2015 at 8:57 PM, samarvir singh <samarvir1996 at gmail.com> wrote:
So I have a list that contains certain characters as shown below
`list <- c("MY","GM+" ,"TY","RS","LG")`
That's a character vector, not a list. A list is a specific type of object in R.
And I have a variable named "CODE" in the data frame as follows
`code <- c("MY GM+", ,"LGTY", "RS","TY")`
That doesn't work, and I have no idea what you expect to have there,
so I'm deleting the extra comma. Also, your vector is named code, not
CODE.
code <- c("MY GM+", "LGTY", "RS","TY")
x <- c(1:4)
'x <- c(1:5) `df <- data.frame(x,code)`
You problably actually want mydf <- data.frame(x, code, stringsAsFactors=FALSE) Note I changed the name, because df() is a base R function.
Now I want to create 5 new variables named "MY","GM+","TY","RS","LG"
Which takes binary value, 1 if there's a match case in the CODE variable
df
x code MY GM+ TY RS LG
1 MY GM+ 1 1 0 0 0
2 0 0 0 0 0
3 LGTY 0 0 1 0 1
4 RS 0 0 0 1 0
5 TY 0 0 1 0 0
grepl() will give you a logical match data.frame(mydf, sapply(code, function(x)grepl(x, mydf$code)), stringsAsFactors=FALSE, check.names=FALSE) Sarah -- Sarah Goslee http://www.functionaldiversity.org
From Sarah's data frame you can get what you want directly with the table() function which will create a table object, mydf.tbl. If you want a data frame you need to convert the table using as.data.frame.matrix() to make mydf.df. Finally combine the two data frames if your x column consists of unique values in ascending order to make mydf.all.
mydf.tbl <- table(mydf$x, mydf$code) mydf.tbl
LGTY MY GM+ RS TY 1 0 1 0 0 2 1 0 0 0 3 0 0 1 0 4 0 0 0 1
mydf.df <- as.data.frame.matrix(mydf.tbl) mydf.df
LGTY MY GM+ RS TY 1 0 1 0 0 2 1 0 0 0 3 0 0 1 0 4 0 0 0 1
mydf.all <- data.frame(mydf, mydf.df) mydf.all
x code LGTY MY.GM. RS TY 1 1 MY GM+ 0 1 0 0 2 2 LGTY 1 0 0 0 3 3 RS 0 0 1 0 4 4 TY 0 0 0 1 -----Original Message----- From: R-help [mailto:r-help-bounces at r-project.org] On Behalf Of samarvir singh Sent: Thursday, April 9, 2015 8:50 AM To: Sarah Goslee Cc: r-help Subject: Re: [R] 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
______________________________________________ R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.