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

[Petr Pikal]

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