newbie lapply question

Hi --

I just noticed the following (R 2.6.1 on OSX)
lapply(c(as.Date('2007-01-01')), I)
[[1]]
[1] 13514

This is a bit surprising.. Why does lapply unclass the object?  Sorry for 
such a basic question, I don't seem able to produce the right google keywords.

Ranjan

Hi --

I just noticed the following (R 2.6.1 on OSX)

lapply(c(as.Date('2007-01-01')), I)
[[1]]
[1] 13514

This is a bit surprising.. Why does lapply unclass the object?  Sorry for
such a basic question, I don't seem able to produce the right google keywords.
Did you not read the help page?:

Arguments:

        X: a vector (atomic or list) or an expressions vector.  Other
           objects (including classed objects) will be coerced by
           'as.list'.

and
as.list(c(as.Date('2007-01-01')))
[[1]]
[1] 13514

BTW, the c() is redundant here: you are concatenating one item only.

As to why as.list() removes the class, read its help page which tells you.

Perhaps if you told us what you are trying to achieve we might be able to 
help you achieve it.
Brian D. Ripley,                  ripley at stats.ox.ac.uk
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford,             Tel:  +44 1865 272861 (self)
1 South Parks Road,                     +44 1865 272866 (PA)
Oxford OX1 3TG, UK                Fax:  +44 1865 272595
[...]
Thanks I'll read it more carefully.
Perhaps if you told us what you are trying to achieve we might be able to 
help you achieve it.

I have a function which takes a date as an argument.  I've tested it, and 
I'd like to run it over a range of dates.  So I'm looking at apply- or 
map- type functions.
-- 
Brian D. Ripley,                  ripley at stats.ox.ac.uk
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford,             Tel:  +44 1865 272861 (self)
1 South Parks Road,                     +44 1865 272866 (PA)
Oxford OX1 3TG, UK                Fax:  +44 1865 272595

I am not sure to understand your problem, but it seems to me that you 
can use directly the function on the range of the dates:

 > x=as.Date(c('2007-01-01','2007-01-02'))
 > fff=function(x){y=x+1;return(y)}
 > fff(x)
[1] "2007-01-02" "2007-01-03"
 > class(fff(x))
[1] "Date"

Perhaps your function use a different input (not a vector of dates but a 
dataframe)?

domenico vistocco
On Wed, 5 Dec 2007, Prof Brian Ripley wrote:

[...]

Thanks I'll read it more carefully.

Perhaps if you told us what you are trying to achieve we might be able to 
help you achieve it.

I have a function which takes a date as an argument.  I've tested it, and 
I'd like to run it over a range of dates.  So I'm looking at apply- or 
map- type functions.

-- 
Brian D. Ripley,                  ripley at stats.ox.ac.uk
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford,             Tel:  +44 1865 272861 (self)
1 South Parks Road,                     +44 1865 272866 (PA)
Oxford OX1 3TG, UK                Fax:  +44 1865 272595

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