R help - Feb 2005 - reshape without timevar argument?

Hi

I have a data.frame with 2 columns. The first column is an ID column. 
The other columns are description of the ids. There is more than one 
description for each Id.
Want I want to get as a value is a data.frame where each row 
corresponds to one ID and has as many columns as different 
descriptions.

I have used a very convoluted step, but I'm very convinced there is 
an easier way to do this.

Basically, I used the reshape function, but even in this convulated 
way there is a step that I can't solve. I used a "fake" timevar
using
the table function.

df <- data.frame(id=c(rep('IDa',3), rep('IDb', 5),
rep('IDc', 2),
                    rep('IDd',5)), let=letters[1:5])
#add Freq to each id
xFreqdf <- merge(df, table(df['id']), by.x='id',
by.y='Var1')
xFreq <- xFreqdf[,'Freq']
#general way of transforming xFreq into:
xFreq2 <- c(3:1, 5:1, 2:1, 5:1)

#substitute Freq by xFreq2
xFreqdf['Freq'] <- xFreq2

#get final df
xReshape <- reshape(xFreqdf, idvar='id', v.names='let',
                     timevar='Freq', direction='wide')

The final data.frame doesn't order the columns in the way I would 
think the more obvious: let.1 is actually the 4th column instead of 
the second one.

I again thank the kindness of people on this list

Tiago R Magalhaes <tiago17 <at> socrates.Berkeley.EDU> writes:

: I have a data.frame with 2 columns. The first column is an ID column. 
: The other columns are description of the ids. There is more than one 
: description for each Id.
: Want I want to get as a value is a data.frame where each row 
: corresponds to one ID and has as many columns as different 
: descriptions.
: 
: I have used a very convoluted step, but I'm very convinced there is 
: an easier way to do this.
: 
: Basically, I used the reshape function, but even in this convulated 
: way there is a step that I can't solve. I used a "fake" timevar
using
: the table function.
: 
: df <- data.frame(id=c(rep('IDa',3), rep('IDb', 5),
rep('IDc', 2),
:                     rep('IDd',5)), let=letters[1:5])

table(df) gives a table of zeros and ones corresponding to the 
incidence matrix of IDs vs. descriptions.  Is that sufficient?