Re: [R] elements in each row of a matrix to the left.

From: <john.gavin_at_ubs.com>
Date: Tue 28 Feb 2006 - 20:31:51 EST


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@pburns.seanet.com]
> Sent: 27 February 2006 19:55
> To: Gavin, John
> Cc: r-help@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@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@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@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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> >https://stat.ethz.ch/mailman/listinfo/r-help
> >PLEASE do read the posting guide!
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>
>
>

>

>

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https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html Received on Tue Feb 28 20:51:30 2006

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