On Aug 4, 2010, at 3:40 PM, Dimitri Liakhovitski wrote:
Thanks a lot, David.
It works perfectly. Of course, lapply is also a loop!
So, your method is:
z<-
data
.frame
(nam1
=
c("bbb..aba","ccc..abb","ddd..abc","eee..abd"),stringsAsFactors=FALSE)
z$nam2<-unlist(lapply( strsplit(z[[1]],split="\\.."), "[", 1))
z$nam3<-unlist(lapply( strsplit(z[[1]],split="\\.."), "[", 2))
Unless you want to use the gsub method I later offered.
And using the new package "stringr" (thank you for sharing!):
y<-data.frame(nam1=c("aaa..aba","bbb..abb","ccc..abc","ddd..abd"),
stringsAsFactors=FALSE)
library(stringr)
y$nam2<-as.data.frame(str_split_fixed(y$nam1, "\\..", 2))[[1]]
y$nam3<-as.data.frame(str_split_fixed(y$nam1, "\\..", 2))[[2]]
(y)
One question - what exactly does the square bracket in your lapply
code mean? Looks like a shortcut - I've not seen it before.
lapply( strsplit(z[[1]],split="\\.."), "[", 1)
It is just the Extract function applied with an argument of 1 to each
successive member of the list, so it is simply the series:
> strsplit(x[[1]],split="\\..")[[1]][1]
[1] "bbb"
> strsplit(x[[1]],split="\\..")[[2]][1]
[1] "ccc"
> strsplit(x[[1]],split="\\..")[[3]][1]
[1] "ddd"
Thank you!
Dimitri
On Wed, Aug 4, 2010 at 3:31 PM, David Winsemius <dwinsem...@comcast.net
> wrote:
On Aug 4, 2010, at 3:03 PM, Dimitri Liakhovitski wrote:
I am sorry, someone said that strsplit automatically works on a
column. How exactly does it work?
For example, if I want to grab just the first (or the second) part
of
the string in nam1 that should be split based on ".."
x<-data.frame(nam1=c("bbb..aba","ccc..abb","ddd..abc","eee..abd"),
stringsAsFactors=FALSE)
str(x)
strsplit(x[[1]],split="\\..")
str(strsplit(x[[1]],split="\\.."))
I am getting a list - hence, it looks like I have to go in a
loop...?
lapply( strsplit(x[[1]],split="\\.."), "[", 1)
[[1]]
[1] "bbb"
[[2]]
[1] "ccc"
[[3]]
[1] "ddd"
[[4]]
[1] "eee"
lapply( strsplit(x[[1]],split="\\.."), "[", 2)
[[1]]
[1] "aba"
[[2]]
[1] "abb"
[[3]]
[1] "abc"
[[4]]
[1] "abd"
unlist(lapply( strsplit(x[[1]],split="\\.."), "[", 2) )
[1] "aba" "abb" "abc" "abd"
unlist(lapply( strsplit(x[[1]],split="\\.."), "[", 1) )
[1] "bbb" "ccc" "ddd" "eee"
Thank you!
Dimitri
On Wed, Aug 4, 2010 at 2:39 PM, Dimitri Liakhovitski
<dimitri.liakhovit...@gmail.com> wrote:
Thank you very much, everyone!
Dimitri
On Wed, Aug 4, 2010 at 2:10 PM, David Winsemius <dwinsem...@comcast.net
>
wrote:
On Aug 4, 2010, at 1:42 PM, Dimitri Liakhovitski wrote:
I am sorry, I'd like to split my column ("names") such that all
the
beginning of a string ("X..") is gone and only the rest of the
text is
left.
I could not tell whether it was the string "X.." or the pattern
"X.."
that
was your goal for matching and removal.
x<-data.frame(names=c("X..aba","X..abb","X..abc","X..abd"))
x$names<-as.character(x$names)
a) Instead of "names" which is heavily used function name, use
something
more specific. Otherwise you get:
names(x)
"names" # and thereby avoid list comments about canines.
b) Instead of coercing a character vector back to a character
vector,
use
stringsAsFactors = FALSE.
x<-data.frame(nam1=c("X..aba","X..abb","X..abc","X..abd"),
stringsAsFactors=FALSE)
#Thus is the pattern version:
x$nam1 <- gsub("X..",'', x$nam1)
x
nam1
1 aba
2 abb
3 abc
4 abd
This is the string version:
x<-data.frame(nam1=c("X......aba","X.y.abb","X..abc","X..abd"),
stringsAsFactors=FALSE)
x$nam1 <- gsub("X\\.+",'', x$nam1)
x
nam1
1 aba
2 y.abb
3 abc
4 abd
(x)
str(x)
Can't figure out how to apply strsplit in this situation -
without
using a loop. I hope it's possible to do it without a loop - is
it?
--
David Winsemius, MD
West Hartford, CT
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