Hi, I have a table in which one column has the name of the objects as shown below. Name Budlamp-Woodcutter Complex - 15 to 60% slope (60/25/15) Budlamp-Woodcutter Complex - 15 to 60% slope (60/25/15) Terrarossa-Blacktail-Pyeatt Complex - 1 to 40% slope (40/35/15/10) Terrarossa-Blacktail-Pyeatt Complex - 1 to 40% slope (40/35/15/10) How can I split the single column into three columns like name (Budlamp-Woodcutter Complex), slope (15 to 60% slope) and percentage (60/25/15) thanks [[alternative HTML version deleted]]
Hi,
Try this:
dat1<-readLines(textConnection("Name
Budlamp-Woodcutter Complex - 15 to 60% slope (60/25/15)
Budlamp-Woodcutter Complex - 15 to 60% slope (60/25/15)
Terrarossa-Blacktail-Pyeatt Complex - 1 to 40% slope (40/35/15/10)
Terrarossa-Blacktail-Pyeatt Complex - 1 to 40% slope (40/35/15/10)
"))
dat3<-data.frame(dat2[-1,])
colnames(dat3)<-"Name"
col2<-gsub(".*-\\s(.*slope).*","\\1",dat3$Name)
?col3<-gsub(".*(\\(.*\\))","\\1",dat3$Name)
?col1<-gsub("(.*Complex).*","\\1",dat3$Name)
?dat3New<-data.frame(col1,col2,col3)
?dat3New
???????????????????????????????? col1??????????? col2????????? col3
1????????? Budlamp-Woodcutter Complex 15 to 60% slope??? (60/25/15)
2????????? Budlamp-Woodcutter Complex 15 to 60% slope??? (60/25/15)
3 Terrarossa-Blacktail-Pyeatt Complex? 1 to 40% slope (40/35/15/10)
4 Terrarossa-Blacktail-Pyeatt Complex? 1 to 40% slope (40/35/15/10)
A.K.
----- Original Message -----
From: Sapana Lohani <lohani.sapana at ymail.com>
To: "r-help at r-project.org" <r-help at r-project.org>
Cc:
Sent: Thursday, August 16, 2012 2:41 PM
Subject: [R] How to extract from a column in a table?
Hi,
I have a table in which one column has the name of the objects as shown below.
Name
Budlamp-Woodcutter Complex - 15 to 60% slope (60/25/15)
Budlamp-Woodcutter Complex - 15 to 60% slope (60/25/15)
Terrarossa-Blacktail-Pyeatt Complex - 1 to 40% slope (40/35/15/10)
Terrarossa-Blacktail-Pyeatt Complex - 1 to 40% slope (40/35/15/10)
How can I split the single column into three columns like name
(Budlamp-Woodcutter Complex), slope (15 to 60% slope) and percentage (60/25/15)
thanks
??? [[alternative HTML version deleted]]
______________________________________________
R-help at r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Hi,
Forgot the last part.
colnames(dat3New)<-c("name","slope","percentage")
?dat3New
???????????????????????????????? name?????????? slope??? percentage
1????????? Budlamp-Woodcutter Complex 15 to 60% slope??? (60/25/15)
2????????? Budlamp-Woodcutter Complex 15 to 60% slope??? (60/25/15)
3 Terrarossa-Blacktail-Pyeatt Complex? 1 to 40% slope (40/35/15/10)
4 Terrarossa-Blacktail-Pyeatt Complex? 1 to 40% slope (40/35/15/10)
A.K.
----- Original Message -----
From: Sapana Lohani <lohani.sapana at ymail.com>
To: "r-help at r-project.org" <r-help at r-project.org>
Cc:
Sent: Thursday, August 16, 2012 2:41 PM
Subject: [R] How to extract from a column in a table?
Hi,
I have a table in which one column has the name of the objects as shown below.
Name
Budlamp-Woodcutter Complex - 15 to 60% slope (60/25/15)
Budlamp-Woodcutter Complex - 15 to 60% slope (60/25/15)
Terrarossa-Blacktail-Pyeatt Complex - 1 to 40% slope (40/35/15/10)
Terrarossa-Blacktail-Pyeatt Complex - 1 to 40% slope (40/35/15/10)
How can I split the single column into three columns like name
(Budlamp-Woodcutter Complex), slope (15 to 60% slope) and percentage (60/25/15)
thanks
??? [[alternative HTML version deleted]]
______________________________________________
R-help at r-project.org mailing list
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.
Sapana Lohani <lohani.sapana@ymail.com> wrote on 08/16/2012 01:41:23 PM:> > Hi, > > I have a table in which one column has the name of the objects as shownbelow. You don't provide example data, so I'm not sure what you mean by a "table". I will assume that you mean a data frame.> Name > > Budlamp-Woodcutter Complex - 15 to 60% slope (60/25/15) > Budlamp-Woodcutter Complex - 15 to 60% slope (60/25/15) > Terrarossa-Blacktail-Pyeatt Complex - 1 to 40% slope (40/35/15/10) > Terrarossa-Blacktail-Pyeatt Complex - 1 to 40% slope (40/35/15/10) > > How can I split the single column into three columns like name > (Budlamp-Woodcutter Complex), slope (15 to 60% slope) andpercentage(60/25/15)> > thanks# example data frame mydat <- data.frame(Name = c("Budlamp-Woodcutter Complex - 15 to 60% slope (60/25/15)", "Budlamp-Woodcutter Complex - 15 to 60% slope (60/25/15)", "Terrarossa-Blacktail-Pyeatt Complex - 1 to 40% slope (40/35/15/10)", "Terrarossa-Blacktail-Pyeatt Complex - 1 to 40% slope (40/35/15/10)"), x=1:4) # find where the three splitting strings are dash <- regexpr(" - ", mydat$Name) leftp <- regexpr(" \\(", mydat$Name) rightp <- regexpr("\\)", mydat$Name) # define three new columns based on the location of the splitting strings mydat$name <- substring(mydat$Name, 1, dash-1) mydat$slope <- substring(mydat$Name, dash + attr(dash, "match.length"), leftp-1) mydat$percentage <- substring(mydat$Name, leftp + attr(leftp, "match.length"), rightp-1) mydat [[alternative HTML version deleted]]
?strsplit -- Bert On Thu, Aug 16, 2012 at 11:41 AM, Sapana Lohani <lohani.sapana at ymail.com> wrote:> Hi, > > > I have a table in which one column has the name of the objects as shown below. > > > Name > > Budlamp-Woodcutter Complex - 15 to 60% slope (60/25/15) > Budlamp-Woodcutter Complex - 15 to 60% slope (60/25/15) > Terrarossa-Blacktail-Pyeatt Complex - 1 to 40% slope (40/35/15/10) > Terrarossa-Blacktail-Pyeatt Complex - 1 to 40% slope (40/35/15/10) > > How can I split the single column into three columns like name (Budlamp-Woodcutter Complex), slope (15 to 60% slope) and percentage (60/25/15) > > thanks > > [[alternative HTML version deleted]] > > ______________________________________________ > R-help at r-project.org mailing list > 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.-- Bert Gunter Genentech Nonclinical Biostatistics Internal Contact Info: Phone: 467-7374 Website: http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm
Hello,
Try the following.
x <- c( "Budlamp-Woodcutter Complex - 15 to 60% slope (60/25/15)",
"Budlamp-Woodcutter Complex - 15 to 60% slope (60/25/15)",
"Terrarossa-Blacktail-Pyeatt Complex - 1 to 40% slope (40/35/15/10)",
"Terrarossa-Blacktail-Pyeatt Complex - 1 to 40% slope (40/35/15/10)" )
lapply(strsplit(x, " - "), function(.x){
s1 <- .x[1]
s2 <- gsub(" \\(.*\\)$", "", .x[2])
s3 <- gsub("^.*(\\(.*\\)$)", "\\1", .x[2])
c(s1, s2, s3)
})
If you want to have the output in the form of matrix or data.frame,
change the above to:
xx <- lapply(...etc...)
mat <- do.call(rbind, xx) # matrix
dat <- data.frame(mat) # data.frame
Hope this helps,
Rui Barradas
Em 16-08-2012 19:41, Sapana Lohani escreveu:> Hi,
>
>
> I have a table in which one column has the name of the objects as shown
below.
>
>
> Name
>
> Budlamp-Woodcutter Complex - 15 to 60% slope (60/25/15)
> Budlamp-Woodcutter Complex - 15 to 60% slope (60/25/15)
> Terrarossa-Blacktail-Pyeatt Complex - 1 to 40% slope (40/35/15/10)
> Terrarossa-Blacktail-Pyeatt Complex - 1 to 40% slope (40/35/15/10)
>
> How can I split the single column into three columns like name
(Budlamp-Woodcutter Complex), slope (15 to 60% slope) and percentage (60/25/15)
>
> thanks
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list
> 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.
HI,
Slightly different way to do this:
dat1<-readLines(textConnection("Name
Budlamp-Woodcutter Complex - 15 to 60% slope (60/25/15)
Budlamp-Woodcutter Complex - 15 to 60% slope (60/25/15)
Terrarossa-Blacktail-Pyeatt Complex - 1 to 40% slope (40/35/15/10)
Terrarossa-Blacktail-Pyeatt Complex - 1 to 40% slope (40/35/15/10)
"))
dat2<-data.frame(dat1)
dat3<-data.frame(Name=dat2[2:5,],slope=dat2[2:5,],percentage=dat2[2:5,])
dat3$Name<-gsub("(.*Complex)\\s-\\s(.*slope).*(\\(.*\\))","\\1",dat3$Name)
dat3$slope<-gsub("(.*Complex)\\s-\\s(.*slope).*(\\(.*\\))","\\2",dat3$slope)
dat3$percentage<-gsub("(.*Complex)\\s-\\s(.*slope).*(\\(.*\\))","\\3",dat3$percentage)
?
?dat3
#???????????????????????????????? Name?????????? slope??? percentage
#1????????? Budlamp-Woodcutter Complex 15 to 60% slope??? (60/25/15)
#2????????? Budlamp-Woodcutter Complex 15 to 60% slope??? (60/25/15)
#3 Terrarossa-Blacktail-Pyeatt Complex? 1 to 40% slope (40/35/15/10)
#4 Terrarossa-Blacktail-Pyeatt Complex? 1 to 40% slope (40/35/15/10)
A.K.
________________________________
From: Sapana Lohani <lohani.sapana at ymail.com>
To: arun <smartpink111 at yahoo.com>
Cc: R help <r-help at r-project.org>
Sent: Thursday, August 16, 2012 7:08 PM
Subject: Re: [R] How to extract from a column in a table?
Thank you :)
________________________________
From: arun <smartpink111 at yahoo.com>
To: Sapana Lohani <lohani.sapana at ymail.com>
Cc: R help <r-help at r-project.org>
Sent: Thursday, August 16, 2012 1:06 PM
Subject: Re: [R] How to extract from a column in a table?
Hi,
Forgot the last part.
colnames(dat3New)<-c("name","slope","percentage")
?dat3New
???????????????????????????????? name?????????? slope??? percentage
1????????? Budlamp-Woodcutter Complex 15 to 60% slope??? (60/25/15)
2????????? Budlamp-Woodcutter Complex 15 to 60% slope??? (60/25/15)
3 Terrarossa-Blacktail-Pyeatt Complex? 1 to 40% slope (40/35/15/10)
4 Terrarossa-Blacktail-Pyeatt Complex? 1 to 40% slope (40/35/15/10)
A.K.
----- Original Message -----
From: Sapana Lohani <lohani.sapana at ymail.com>
To: "r-help at r-project.org" <r-help at r-project.org>
Cc:
Sent: Thursday, August 16, 2012 2:41 PM
Subject: [R] How to extract from a column in a table?
Hi,
I have a table in which one column has the name of the objects as shown below.
Name
Budlamp-Woodcutter Complex - 15 to 60% slope (60/25/15)
Budlamp-Woodcutter Complex - 15 to 60% slope (60/25/15)
Terrarossa-Blacktail-Pyeatt Complex - 1 to 40% slope (40/35/15/10)
Terrarossa-Blacktail-Pyeatt Complex - 1 to 40% slope (40/35/15/10)
How can I split the single column into three columns like name
(Budlamp-Woodcutter Complex), slope (15 to 60% slope) and percentage
(60/25/15)
thanks
??? [[alternative HTML version deleted]]
______________________________________________
R-help at r-project.org mailing list
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.
Hi I am new to R so am struggling with the commands here I have one column in a table that looks like slope (60/25/15) slope (90/10) slope (40/35/15/10) slope (40/25/25/10) I want to have 4 columns with just the number inside the parenthesis. when there is no number that cell can have 0. I want the output like this 60 25 15 0 90 10 0 0 40 35 15 10 40 25 25 10 Can somebody help me?? Thanks [[alternative HTML version deleted]]
HI,
Try this:
dat1<-data.frame(slope=c("slope (60/25/15)","slope
(90/10)","slope (40/35/15/10)","slope (40/25/25/10)" ))
dat1$slope<-gsub("slope\\s+.*(\\(.*\\))","\\1",dat1$slope)
?dat1$slope<-gsub("\\((.*)\\)","\\1",dat1$slope)
dat2<-strsplit(dat1$slope,"/")
dat2[[1]][4]<-0
dat2[[2]][3:4]<-0
data.frame(do.call(rbind,dat2))
#? X1 X2 X3 X4
#1 60 25 15? 0
#2 90 10? 0? 0
#3 40 35 15 10
#4 40 25 25 10
----- Original Message -----
From: Sapana Lohani <lohani.sapana at ymail.com>
To: "r-help at r-project.org" <r-help at r-project.org>
Cc:
Sent: Friday, August 17, 2012 1:05 AM
Subject: [R] Unequal splits from a column
Hi I am new to R so am struggling with the commands here
I have one column in a table that looks like
slope (60/25/15)
slope (90/10)
slope (40/35/15/10)
slope (40/25/25/10)
I want to have 4 columns with just the number inside the parenthesis. when there
is no number that cell can have 0. I want the output like this
60 25 15 0
90 10 0 0
40 35 15 10
40 25 25 10
Can somebody help me??
Thanks
??? [[alternative HTML version deleted]]
______________________________________________
R-help at r-project.org mailing list
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.