I managed to parse more complex XML files as well. The trick was to
manually determine the position of the child nodes of interest, after
which they can be parsed in a loop. For example:
require(XML)
doc <- xmlTreeParse("file.xml",getDTD=T,addAttributeNamespaces=T)
r <- xmlRoot(doc)
#find the nodes of interest
r[[i]][[j]]....
#then read them
xmldata <- list(NULL)
for (i in 1:xmlSize(r[[2]][[1]])) {
xmldata[[i]] <- as.data.frame(xmlSApply(r[[2]][[1]][[i]],xmlValue))
}
--- Barry Rowlingson <B.Rowlingson at lancaster.ac.uk> wrote:
Gabor Grothendieck wrote:
You could check out the ctv package that was recently announced. It uses XML so its source would provide an example. If its a one-time operation, Excel reads XML and you could then use one of the many Excel to R possibilities.
For an xml file like this:
<?xml version="1.0"?>
<variables>
<a>100</a>
<b>23</b>
<z>666</z>
</variables>
its a one-liner with the XML package (library(XML)):
xmlReadSimple <-
function(xmlFile){
as.list(xmlSApply(xmlRoot(xmlTreeParse(xmlFile)),xmlValue))
}
add an lapply(...,as.numeric) for conversion to numbers.
sweet.
Baz
______________________________________________ R-help at stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __________________________________________________