discussed
> within
> > > R's
> > > > documentation, but
> > > > you can find info on it in V&R's "S
> Programming",
> > > esp. p.24 and 4.3,
> > > > "Extracting or replacing coefficients".
> > > >
&
"S Programming",
> > esp. p.24 and 4.3,
> > > "Extracting or replacing coefficients".
> > >
> > > No doubt, other S/R books explain it also.
> > >
> > > Cheers,
> > >
> > > Bert Gunter
> > >
u can find info on it in V&R's "S Programming",
> esp. p.24 and 4.3,
> > "Extracting or replacing coefficients".
> >
> > No doubt, other S/R books explain it also.
> >
> > Cheers,
> >
> > Bert Gunter
> > Genente
xplain it also.
>
> Cheers,
>
> Bert Gunter
> Genentech Nonclinical Statistics
> 47374
>
>
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On
> Behalf Of John Kane
> Sent: Thursday, March 13, 2008 11:53 AM
> To: Henrique
cal Statistics
47374
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of John Kane
Sent: Thursday, March 13, 2008 11:53 AM
To: Henrique Dallazuanna
Cc: R R-help
Subject: Re: [R] Splitting a set of vectors in a list (Solved )
My thanks to Henrique Dallazuann
Blood ell ! Thanks very much, it does help give a
feel for what's happening. I'll have to do some
experimenting.
--- Henrique Dallazuanna <[EMAIL PROTECTED]> wrote:
> A example:
>
> x <- rnorm(5)
>
> x[3]
> `[`(x, 3)
> `[[`(x, 3)
>
> x[3:4]
> `[`(x, 3:4)
> `[[`(x, 3:4) # Error
>
>
> On 13/
A example:
x <- rnorm(5)
x[3]
`[`(x, 3)
`[[`(x, 3)
x[3:4]
`[`(x, 3:4)
`[[`(x, 3:4) # Error
On 13/03/2008, John Kane <[EMAIL PROTECTED]> wrote:
> Ah ?Extract. Thanks
>
> Unless I'm missing something there is nothing in the
> Usage or Examples to suggest to a naive reader like me
> that one
Ah ?Extract. Thanks
Unless I'm missing something there is nothing in the
Usage or Examples to suggest to a naive reader like me
that one can use an unbalanced [ or [[, that is
withoug a corresponding ] or ]].
I probably am just not understanding the details.
--- Henrique Dallazuanna <[EMAIL
or the suggestive :)
?"["
b
On Mar 13, 2008, at 2:58 PM, Henrique Dallazuanna wrote:
On 13/03/2008, John Kane <[EMAIL PROTECTED]> wrote:
Can you point me to some documentation that discusses
these usages. I have seen them before but I have never
actually figured out how to use them.?
See ?
On 13/03/2008, John Kane <[EMAIL PROTECTED]> wrote:
> My thanks to Henrique Dallazuanna and Phil Spector.
> Both solutions worked well.
> Phil suggested that an alterative to my function would
> be
> vect1 = sapply(mylist,'[[',1)
> and I see that Henrique used `[` in his solution.
>
> Can you
My thanks to Henrique Dallazuanna and Phil Spector.
Both solutions worked well.
Phil suggested that an alterative to my function would
be
vect1 = sapply(mylist,'[[',1)
and I see that Henrique used `[` in his solution.
Can you point me to some documentation that discusses
these usages. I have se
If I understand correctly, try this:
as.data.frame(lapply(mylist, `[`, 1:max(unlist(lapply(mylist, length)
On 13/03/2008, John Kane <[EMAIL PROTECTED]> wrote:
> I have a set of character vectors of uneven length
> that I have stored in a list. I can easily enough get
> any column of them
Ideally something like this:
==
t(cbind( c("cat" , "peach" , NA, NA), bbb <- c("dog"
, "apple" ,"iron", NA),
ccb <- c("rabbit" ,"orange" ,"zinc" , "silk" )))
==
Thanks
--- Erik Iverson <[
What would you want your output matrix to look like given mylist?
John Kane wrote:
> I have a set of character vectors of uneven length
> that I have stored in a list. I can easily enough get
> any column of them using lapply but what I want is to
> be able to create a matrix of them. Other tha
I have a set of character vectors of uneven length
that I have stored in a list. I can easily enough get
any column of them using lapply but what I want is to
be able to create a matrix of them. Other than some
kind of brute force looping approach I have drawn a
blank.
Would somebody please su
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