On Jun 22, 2011, at 9:46 AM, Marius Hofert wrote:
Hi David,
thanks for the quick response. That's nice. Is there also a way
without loading an additional package? I'd prefer loading less
packages if possible.
> xtb <- xtabs(value ~ year + block, data = df)
> xtb
block
year a b c
2000 1 0 5
2001 2 4 6
2002 3 0 0
> as.data.frame(xtb)
year block Freq
1 2000 a 1
2 2001 a 2
3 2002 a 3
4 2000 b 0
5 2001 b 4
6 2002 b 0
7 2000 c 5
8 2001 c 6
9 2002 c 0
> xt.df <- as.data.frame(xtb)
> xt.df[xt.df[,3] != 0 , ]
year block Freq
1 2000 a 1
2 2001 a 2
3 2002 a 3
5 2001 b 4
7 2000 c 5
8 2001 c 6
As Hadley pointed out yesterday this does coerce the margin names to
factors,
> str(xt.df)
'data.frame': 9 obs. of 3 variables:
$ year : Factor w/ 3 levels "2000","2001",..: 1 2 3 1 2 3 1 2 3
$ block: Factor w/ 3 levels "a","b","c": 1 1 1 2 2 2 3 3 3
$ Freq : num 1 2 3 0 4 0 5 6 0
--
David.
Cheers,
Marius
On 2011-06-22, at 15:38 , David Winsemius wrote:
On Jun 22, 2011, at 9:19 AM, Marius Hofert wrote:
Hi,
and what's the simplest way to obtain a *data.frame* with all years?
The matching seems more difficult here because the years can/will
show up several times...
(df <- data.frame(year=c(2000, 2001, 2002, 2001, 2000, 2001),
block=c("a","a","a","b","c","c"), value=1:6))
(df. <- data.frame(year=rep(2000:2002, 3), block=rep(c("a", "b",
"c"), each=3), value=0))
# how to fill in the given values?
These days I think most people would reach for melt() in either
reshape or reshape2 packages:
require(reshape2)
Loading required package: reshape2
melt(xtb)
year block value
1 2000 a 1
2 2001 a 2
3 2002 a 3
4 2000 b 0
5 2001 b 4
6 2002 b 0
7 2000 c 5
8 2001 c 6
9 2002 c 0
It seems to do a good job of guessing what you want whereas the
reshape function in my hands is very failure prone (... yes, the
failings are mine.)
--
David
Cheers,
Marius
On 2011-06-22, at 14:40 , Dennis Murphy wrote:
I saw it as an xtabs object - I didn't think to check whether it
was
also a matrix object. Thanks for the clarification, David.
Dennis
On Wed, Jun 22, 2011 at 4:59 AM, David Winsemius <dwinsem...@comcast.net
> wrote:
On Jun 21, 2011, at 6:51 PM, Dennis Murphy wrote:
Ahhh...you want a matrix. xtabs() doesn't easily allow coercion
to a
matrix object, so try this instead:
What am I missing? A contingency table already inherits from
matrix-class
and if you insisted on coercion it appears simple:
xtb <- xtabs(value ~ year + block, data = df)
is.matrix(xtb)
[1] TRUE
as.matrix(xtb)
block
year a b c
2000 1 0 5
2001 2 4 6
2002 3 0 0
--
David.
library(reshape)
as.matrix(cast(df, year ~ block, fill = 0))
a b c
2000 1 0 5
2001 2 4 6
2002 3 0 0
Hopefully this is more helpful...
Dennis
On Tue, Jun 21, 2011 at 3:35 PM, Dennis Murphy
<djmu...@gmail.com> wrote:
Hi:
xtabs(value ~ year + block, data = df)
block
year a b c
2000 1 0 5
2001 2 4 6
2002 3 0 0
HTH,
Dennis
On Tue, Jun 21, 2011 at 3:13 PM, Marius Hofert
<m_hof...@web.de> wrote:
Dear expeRts,
In the minimal example below, I have a data.frame containing
three
"blocks" of years
(the years are subsets of 2000 to 2002). For each year and
block a
certain "value" is given.
I would like to create a matrix that has row names given by
all years
("2000", "2001", "2002"),
and column names given by all blocks ("a", "b", "c"); the
entries are
then given by the
corresponding value or zero if not year-block combination
exists.
What's a short way to achieve this?
Of course one can setup a matrix and use for loops (see
below)... but
that's not nice.
The problem is that the years are not running from 2000 to
2002 for all
three "blocks"
(the second block only has year 2001, the third one has only
2000 and
2001).
In principle, table() nicely solves such a problem (see
below) and fills
in zeros.
This is what I would like in the end, but all non-zero
entries should be
given by df$value,
not (as table() does) by their counts.
Cheers,
Marius
(df <- data.frame(year=c(2000, 2001, 2002, 2001, 2000, 2001),
block=c("a","a","a","b","c","c"), value=1:6))
table(df[,1:2]) # complements the years and fills in 0
year <- c(2000, 2001, 2002)
block <- c("a", "b", "c")
res <- matrix(0, nrow=3, ncol=3, dimnames=list(year, block))
for(i in 1:3){ # year
for(j in 1:3){ # block
for(k in 1:nrow(df)){
if(df[k,"year"]==year[i] && df[k,"block"]==block[j])
res[i,j]
<- df[k,"value"]
}
}
}
res # does the job; but seems complicated
David Winsemius, MD
West Hartford, CT
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David Winsemius, MD
West Hartford, CT
David Winsemius, MD
West Hartford, CT
______________________________________________
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
