[R] Table Construction from calculations

Christos Hatzis christos at nuverabio.com
Sat Mar 10 03:41:10 CET 2007


Your "data table" basis is actually a dataframe, whose first column is
non-numeric.  That's what is causing the problem.
 
Try removing the first column of the dataframe before adding the row to your
matrix:

test <- latpoints + basis[2, -1]

-Christos

Christos Hatzis, Ph.D.
Nuvera Biosciences, Inc.
400 West Cummings Park
Suite 5350
Woburn, MA 01801
Tel: 781-938-3830
www.nuverabio.com

> -----Original Message-----
> From: r-help-bounces at stat.math.ethz.ch 
> [mailto:r-help-bounces at stat.math.ethz.ch] On Behalf Of Seth Imhoff
> Sent: Friday, March 09, 2007 9:23 PM
> To: r-help at stat.math.ethz.ch
> Subject: [R] Table Construction from calculations
> 
> Hi-
> 
> I am trying to create a table of values by adding  pairs of 
> vectors, but am running into some problems.  The problem is 
> best expressed by a simple example.
> 
> Starting with a data table "basis":
>   atom   x   y   z
> 1   Cu 0.0 0.0 0.0
> 2   Cu 0.5 0.5 0.5
> 
> I want to add 0.5 0.5 0.5  (and also the 0 0 0 but it 
> wouldn't change the values below so I won't refer to it in 
> the rest of the example) to a list of vectors in the form of:
>  > latpoints
>    V1 V2 V3
> 1   0  0  0
> 2   0  0  1
> 3   0  0  2
> 4   0  0  3
> 5   0  1  1
> 
> so that I end up with a table such as:
> V1  V2  V3
> 0.5 0.5 0.5
> 0.5 0.5 1.5
> 0.5 0.5 2.5
> 0.5 0.5 3.5
> 0.5 1.5 1.5
> 
> I've tried many variations on the following: (not just cat, 
> but most of the data/data.table options)
> 
>  test = for(i in 1:5) {cat(basis[1,2:4] + 
> latticemultipliers[i,], append=TRUE)}
> 
> However, I either end up with an error telling me that cat 
> doesn't handle "type 'list' " or with a table with length of 
> 1 such as:
>      x    y    z
> 2  0.5 1.5 1.5
> 
> Which is simply the last value that the loop calculates.
> 
> Does anyone know what function handles lists of the form I am 
> using, or have a better suggestion on how to get the form that I want.
> 
> Thanks in advance,
> Seth Imhoff
> 
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