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john.gavin@ubs.com

2006-Feb-27 19:33 UTC

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

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}}

Patrick Burns

2006-Feb-27 19:55 UTC

head link

[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}}
>
>______________________________________________
>R-help at stat.math.ethz.ch mailing list
>https://stat.ethz.ch/mailman/listinfo/r-help
>PLEASE do read the posting guide!
http://www.R-project.org/posting-guide.html
>
>
>
>  
>

john.gavin@ubs.com

2006-Feb-28 09:31 UTC

head link

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

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}}
> >
> >______________________________________________
> >R-help at stat.math.ethz.ch mailing list
> >https://stat.ethz.ch/mailman/listinfo/r-help
> >PLEASE do read the posting guide! 
http://www.R-project.org/posting-guide.html>
>
>
>  
>

Visit our website at http://www.ubs.com

This message contains confidential information and is intend...{{dropped}}

Petr Pikal

2006-Feb-28 10:24 UTC

head link

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

Hi

not a complete solution but

z.f<-matrix(z%in%1, ncol=3)

gives you a matrix of logicals

and

apply(apply(z.f*1, 1,cumsum),2,function(x) sum(x==0))
[1] 0 0 1 2 1 3

shall give you number of values to drop from each row.

Then you maybe could use it to manipulate your z matrix.

HTH
Petr




On 28 Feb 2006 at 9:31, john.gavin at ubs.com wrote:

Date sent:      	Tue, 28 Feb 2006 09:31:51 -0000
From:           	<john.gavin at ubs.com>
To:             	<pburns at pburns.seanet.com>, <Jeff.Laake at
noaa.gov>
Copies to:      	r-help at stat.math.ethz.ch
Subject:        	Re: [R] elements in each row of a matrix to the left.
> 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}}
> > >
> > >______________________________________________
> > >R-help at stat.math.ethz.ch mailing list
> > >https://stat.ethz.ch/mailman/listinfo/r-help
> > >PLEASE do read the posting guide! 
> http://www.R-project.org/posting-guide.html
> >
> >
> >
> >  
> >
> 
> 
> Visit our website at http://www.ubs.com
> 
> This message contains confidential information and is\ > i...{{dropped}}

Patrick Burns

2006-Feb-28 10:27 UTC

head link

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

Okay, I think this does the requested operation:

 > single.shift
function (x)
{
        r <- rle(is.na(x))
        if(!r$values[1]) return(x)
        num <- r$length[1]
        c(x[-1:-num], rep(NA, num))
}
 > t(apply(z, 1, single.shift))


john.gavin at ubs.com wrote:
>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}}
>>    
>>
>>>______________________________________________
>>>R-help at stat.math.ethz.ch mailing list
>>>https://stat.ethz.ch/mailman/listinfo/r-help
>>>PLEASE do read the posting guide! 
>>>      
>>>
>http://www.R-project.org/posting-guide.html
>  
>
>>
>> 
>>
>>    
>>
>
>
>Visit our website at http://www.ubs.com
>
>This message contains confidential information and is intended only 
>for the individual named.  If you are not the named addressee you 
>should not disseminate, distribute or copy this e-mail.  Please 
>notify the sender immediately by e-mail if you have received this 
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>
>
>  
>

john.gavin@ubs.com

2006-Feb-28 11:10 UTC

head link

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

Hi Patrick,

Yes, that works. Thanks for your time.

Regards,

John.
> -----Original Message-----
> From: Patrick Burns [mailto:pburns at pburns.seanet.com]
> Sent: 28 February 2006 10:28
> To: Gavin, John
> Cc: Jeff.Laake at noaa.gov; r-help at stat.math.ethz.ch
> Subject: Re: [R] elements in each row of a matrix to the left.
> 
> 
> Okay, I think this does the requested operation:
> 
>  > single.shift
> function (x)
> {
>         r <- rle(is.na(x))
>         if(!r$values[1]) return(x)
>         num <- r$length[1]
>         c(x[-1:-num], rep(NA, num))
> }
>  > t(apply(z, 1, single.shift))
> 
> 
> john.gavin at ubs.com wrote:
> 
> >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}}
> >>    
> >>
> >>>______________________________________________
> >>>R-help at stat.math.ethz.ch mailing list
> >>>https://stat.ethz.ch/mailman/listinfo/r-help
> >>>PLEASE do read the posting guide! 
> >>>      
> >>>
> >http://www.R-project.org/posting-guide.html
> >  
> >
> >>
> >> 
> >>
> >>    
> >>
> >
> >
> >Visit our website at http://www.ubs.com
> >
> >This message contains confidential information and is intended only 
> >for the individual named.  If you are not the named addressee you 
> >should not disseminate, distribute or copy this e-mail.  Please 
> >notify the sender immediately by e-mail if you have received this 
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R help - Feb 2006 - elements in each row of a matrix to the left.

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

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

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

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

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

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