[19167] in s-news-athena
[S] follow-up: converting a list of dataframes into a single dataframe
daemon@ATHENA.MIT.EDU (John.Gavin@wdr.com)
Thu Sep 9 07:55:18 1999
From: John.Gavin@wdr.com
Date: Thu, 9 Sep 1999 12:49:59 +0100
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Hi,
There is a follow-up to this posting
about using 'do.call(rbind, ...)' to convert
a list of similar dataframes into single dataframe that
may be relevant to some (highlighted by Bill Dunlap).
In particular, the 'do.call(rbind, ...)' solution
does not do exactly what I claimed.
It transpires that 'rbind' demotes variables of class
'ordered' to class 'factor'.
They have to be manually converted back.
e.g. 'class(variable.name) <- c("ordered","factor")' promotes them
back
from 'factor' to 'ordered' as required.
If there are a lot of variables in the dataframe then something like
the following might work (I haven't tested it extensively)
# from the original dataframe 'data' get the list of variable names
# and decide if their class is 'ordered'
temp <- unlist(sapply(data, function(x) class(x)[1] )) == "ordered"
# remember those variables that should be of class 'ordered'
temp <- names(temp)[temp]
# reassign the class of those variables from factor to ordered
data.spread.df2 <- convert.col.type(data.spread.df, temp , "ordered")
Thanks again for the many responses to this question.
Regards,
John.
john.gavin@ubs.com
John Gavin
Quantitative Risk Models and Statistics
Warburg Dillon Read
1 Finsbury Ave (4th floor)
London EC2M 2PP, UK
> -----Original Message-----
> From: bill
> Sent: 08 September 1999 21:43
> To: Gavin, John
> Cc: bill
> Subject: Re: [S] summary: converting a list of dataframes
> into a single
> dataframe
>
>
> On Wed, 8 Sep 1999 John.Gavin@wdr.com wrote:
>
> > There were two solutions.
> > The first did exactly what I wanted
> >
> > 1) do.call("rbind",input)
> >
> > 2) output <- NULL
> > while(length(input) > 0)
> > {
> > output <- rbind(output, input[[1]])
> > input[[1]] <- NULL
> > }
>
> I think rbind() still has a bug in it such that when it combines
> columns of class "ordered","factor" it produces on of class
"factor".
> E.g.,
>
> > lapply(input, function(i)lapply(i,data.class))
> $a:
> $a$x:
> [1] "numeric"
> $a$y:
> [1] "factor"
> $a$z:
> [1] "ordered"
> $b:
> $b$x:
> [1] "numeric"
> $b$y:
> [1] "factor"
> $b$z:
> [1] "ordered"
> > lapply(do.call("rbind",input), data.class)
> $x:
> [1] "numeric"
> $y:
> [1] "factor"
> $z:
> [1] "factor"
>
> Note the z component of the output should be "ordered". As a
> workaround, you could process the output to assign the class
> c("ordered","factor") to the factors that you know to be ordered.
>
> > Also each data frame may contain 'factor' and 'ordered' columns
> > so I want to preserve the original labels.
>
> --------------------------------------------------------------
> --------------
> Bill Dunlap 22461 Mt
> Vernon-Big Lake Rd
> Data Analysis Products Div. of MathSoft, Inc. Mount
> Vernon, WA 98274
> bill@statsci.com 360-428-8146
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