Dear Marna,
If you want to extract the middle of those intervals, please find below
an improved variant of Luigi's code.
Note:
- it is more efficient to process the levels of a factor, instead of all
the individual strings;
- I envision that there are benefits in a large data frame (> 1 million
rows) - although I have not explicitly checked it;
- the code also handles better the open/closed intervals;
- the returned data structure may require some tweaking (currently
returns a data.frame);
### Middle of an Interval
mid.factor = function(x, inf.to = NULL, split.str=",") {
lvl0 = levels(x); lvl = lvl0;
lvl = sub("^[(\\[]", "", lvl);
lvl = sub("[])]$", "", lvl); # tricky;
lvl = strsplit(lvl, split.str);
lvl = lapply(lvl, function(x) as.numeric(x));
if( ! is.null(inf.to)) {
FUN = function(x) {
if(any(x == Inf)) 1
else if(any(x == - Inf)) -1
else 0;
}
whatInf = sapply(lvl, FUN);
# TODO: more advanced;
lvl[whatInf == -1] = inf.to[1];
lvl[whatInf == 1] = inf.to[2];
}
mid = sapply(lvl, mean);
lvl = data.frame(lvl=lvl0, mid=mid);
merge(data.frame(lvl=x), lvl, by="lvl");
}
# uses the daT data frame;
# requires a factor:
# - this is probably the case with the original data;
daT$group = as.factor(daT$group);
mid.factor(daT$group);
I have uploaded this code also on my GitHub list of useful data tools:
https://github.com/discoleo/R/blob/master/Stat/Tools.Data.R
Sincerely,
Leonard
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