[R] elements in each row of a matrix to the left.
john.gavin@ubs.com
john.gavin at ubs.com
Tue Feb 28 10:31:51 CET 2006
Hi Patrick/Jeff,
> Does
>
> t(apply(z, 1, sort, na.last=TRUE))
>
> do what you want?
Not quite.
t(apply(z, 1, sort, na.last=TRUE))
[,1] [,2] [,3]
[1,] 1 1 NA
[2,] 1 1 NA
[3,] 1 1 NA
[4,] 1 NA NA
[5,] 1 NA NA
[6,] NA NA NA
Row 2 is the problem.
I dont want to move all NAs to the end of each row.
I just want to move all of the NAs before the first non-NA element,
if any, to the end of each row.
So in my example, rows 1 and 2 should remain unchanged.
What I have got at the moment is ugly
shiftLeft <- function(z)
{ x <- as.data.frame(t(z)) # work with cols not rows.
ans <- lapply(x, function(xx)
{ # get indices of first and last non-NA element
ind <- which(!is.na(xx))
ind <- ind[c(1, length(ind))]
# if all NAs or if first element is non-NA do no work
if (any(is.na(ind)) || ind[1] == 1) xx else
{ temp <- numeric(length(xx)) ; temp[] <- NA
# move elements in posns ind[1] to ind[2] to the start
temp[1:(ind[2]-ind[1]+1)] <- xx[ind[1]:ind[2]]
temp
} # if
}) # lapply
ans <- as.matrix(data.frame(ans))
dimnames(ans) <- dimnames(z)
t(ans)
}
> z ; shiftLeft(z)
[,1] [,2] [,3]
[1,] 1 1 NA
[2,] 1 NA 1
[3,] NA 1 1
[4,] NA NA 1
[5,] NA 1 NA
[6,] NA NA NA
[,1] [,2] [,3]
[1,] 1 1 NA
[2,] 1 NA 1
[3,] 1 1 NA
[4,] 1 NA NA
[5,] 1 NA NA
[6,] NA NA NA
I feel that there is probably a shorter vectorised way to do this.
In general, I have matrices (z) with several thousand rows and
and few hundred columns so vectorisation would help.
Regards,
John.
> -----Original Message-----
> From: Patrick Burns [mailto:pburns at pburns.seanet.com]
> Sent: 27 February 2006 19:55
> To: Gavin, John
> Cc: r-help at stat.math.ethz.ch
> Subject: Re: [R] elements in each row of a matrix to the left.
>
> John,
>
> Does
>
> t(apply(z, 1, sort, na.last=TRUE))
>
> do what you want?
>
>
> Patrick Burns
> patrick at burns-stat.com
> +44 (0)20 8525 0696
> http://www.burns-stat.com
> (home of S Poetry and "A Guide for the Unwilling S User")
>
> john.gavin at ubs.com wrote:
>
> >Hi,
> >
> >Given a matrix like
> >
> >(z <- matrix(c(
> >1, 1, NA, NA, NA, NA,
> >1, NA, 1, NA, 1, NA,
> >NA, 1, 1, 1, NA, NA), ncol = 3))
> >
> > [,1] [,2] [,3]
> >[1,] 1 1 NA
> >[2,] 1 NA 1
> >[3,] NA 1 1
> >[4,] NA NA 1
> >[5,] NA 1 NA
> >[6,] NA NA NA
> >
> >is there a vectorised way to produce the output like
> >
> > [,1] [,2] [,3]
> >[1,] 1 1 NA
> >[2,] 1 NA 1
> >[3,] 1 1 NA
> >[4,] 1 NA NA
> >[5,] 1 NA NA
> >[6,] NA NA NA
> >
> >That is, given an n by m matrix, and going row by row,
> >if the first non-NA element is in column k
> >I want to move elements in columns from k to m
> >to columns 1 to m-k+1 with NAs filling in from
> >m-k+2 to m.
> >
> >
> >
> >>version
> >>
> >>
> > _
> >platform i386-pc-mingw32
> >arch i386
> >os mingw32
> >system i386, mingw32
> >status
> >major 2
> >minor 2.1
> >year 2005
> >month 12
> >day 20
> >svn rev 36812
> >language R
> >
> >Regards,
> >
> >John.
> >
> >John Gavin <john.gavin at ubs.com>,
> >Quantitative Risk Control,
> >UBS Investment Bank, 6th floor,
> >100 Liverpool St., London EC2M 2RH, UK.
> >Phone +44 (0) 207 567 4289
> >Fax +44 (0) 207 568 5352
> >
> >Visit our website at http://www.ubs.com
> >
> >This message contains confidential information and is
> intend...{{dropped}}
> >
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