Displaying 5 results from an estimated 5 matches for "nocols".
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2009 Aug 17
1
image() generates many border lines in pdf, not on screen (quartz) - R 2.9.1 GUI 1.28 Tiger build 32-bit (5444) - OS X 10.5.8
...read on this issue, including R mailing lists (e.g.
http://finzi.psych.upenn.edu/R/Rhelp02/archive/30588.html), e.g. if
someone claims those lines would not matter and go away when printing.
In my case they don't.
BTW, my image() statement looks like this:
image(rfc, y, -t(zVals), zlim=c(-noCols-0.5,0.5), col =
heat.colors(noCols), axes=F, add=T)
where z is a 127 x 3 matrix.
Since I need high resolution variants for final publication I need to
know whether it is possible to force image() to give up drawing these
borders (I tried already several low level par parameters, all to no
ava...
2009 Feb 09
2
R equivalent of SAS Cochran-Mantel-Haenszel tests?
In SAS, for a two-way (or 3-way, stratified) table, the CMH option in
SAS PROC FREQ gives
3 tests that take ordinality of the factors into account, for both
variables, just the column variable
or neither. Is there an equivalent in R?
The mantelhaen.test in stats gives something quite different (a test of
conditional independence for
*nominal* factors in a 3-way table).
e.g. I'd like to
2008 Mar 10
2
write.table with row.names=FALSE unnecessarily slow?
...e[quote > 0]
}
}
- d <- dimnames(x)
- if(is.null(d[[1]])) d[[1]] <- seq_len(nrow(x))
+ d <- list(if (makeRownames==TRUE) row.names(x) else NULL,
+ if (makeColnames==TRUE) names(x) else NULL)
p <- ncol(x)
}
nocols <- p==0
improves performance at least in proportion to nrow(x):
> system.time({
+ write.table(df, '/tmp/dftest1.txt', row.names=FALSE)
+ }, gcFirst=TRUE)
user system elapsed
8.132 0.608 8.899
Martin
--
Martin Morgan
Computational Biology / Fred Hutchinson Cancer Re...
2018 Jul 05
0
write.table with quote=TRUE fails on nested data.frames
...se {
if (qset)
quote <- if (length(x))
which(unlist(lapply(x, function(x) is.character(x) ||
is.factor(x))))
else numeric()
d <- list(if (makeRownames) row.names(x), if (makeColnames) names(x))
}
p <- ncol(x)
}
nocols <- p == 0L
if (is.logical(quote))
quote <- NULL
else if (is.numeric(quote)) {
if (any(quote < 1L | quote > p))
stop("invalid numbers in 'quote'")
}
else stop("invalid 'quote' specification")
rn <- FALSE
rnames <- NULL...
2007 Nov 22
5
testing independence of categorical variables
hi,
is there a way of calculating of measuring dependence between two
categorical variables. i tried using the chi square test to test for
independence but i got error saying that the lengths of the two
vectors don't match. Suppose X and Y are two factors. X has 5 levels
and Y has 7 levels. This is what i tried doing
>temp<-chisq.test(x,y)
but got error "the lengths of the two