Forget I asked. There was a typo in my example (stringsAsFactor
instead of stringAsFactors) which explained the difference. My
apologies.
My second question however still stands: How does on create a
data.frame with given column types and given dimensions? Thanks.
Regards,
Jan
Quoting Jan van der Laan <rh...@eoos.dds.nl>:
I use the following code to create two data.frames d1 and d2 from a list:
types <- c("integer", "character", "double")
nlines <- 10
d1 <- as.data.frame(lapply(types, do.call, list(nlines)),
stringsAsFactor=FALSE)
l2 <- lapply(types, do.call, list(nlines))
d2 <- as.data.frame(l2, stringsAsFactors=FALSE)
I would expect d1 and d2 to be the same, however, in d1 the second
column is a factor while in d2 it is a character (which I would expect):
str(d1)
'data.frame': 10 obs. of 3 variables:
$ c.0L..0L..0L..0L..0L..0L..0L..0L..0L..0L.: int 0 0 0 0 0 0 0 0 0 0
$ c........................................: Factor w/ 1 level "": 1 1
1 1 1 1 1 1 1 1
$ c.0..0..0..0..0..0..0..0..0..0. : num 0 0 0 0 0 0 0 0 0 0
str(d2)
'data.frame': 10 obs. of 3 variables:
$ c.0L..0L..0L..0L..0L..0L..0L..0L..0L..0L.: int 0 0 0 0 0 0 0 0 0 0
$ c........................................: chr "" "" "" "" ...
$ c.0..0..0..0..0..0..0..0..0..0. : num 0 0 0 0 0 0 0 0 0 0
As different but related question: I use the commands above to create
an 'empty' data.frame with specified column types and dimensions. I
need this data.frame to pass on to my c++ routines. Is there a more
simple/elegant way of creating this data.frame?
Regards,
Jan
PS:
I am running R on 64 bit Ubuntu 11.04:
sessionInfo()
R version 2.12.1 (2010-12-16)
Platform: x86_64-pc-linux-gnu (64-bit)
locale:
[1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C
[3] LC_TIME=en_US.UTF-8 LC_COLLATE=en_US.UTF-8
[5] LC_MONETARY=C LC_MESSAGES=en_US.UTF-8
[7] LC_PAPER=en_US.UTF-8 LC_NAME=C
[9] LC_ADDRESS=C LC_TELEPHONE=C
[11] LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C
attached base packages:
[1] stats graphics grDevices utils datasets methods base
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