Hi Michael,
Thanks a lot, your solution is exactly what I needed!
-M
On Wed, Jun 20, 2012 at 9:44 PM, R. Michael Weylandt <
michael.weyla...@gmail.com> <michael.weyla...@gmail.com> wrote:
> Something like:
>
> X <- as.character(data.name)
> X <- as.double(gsub("%","", X))/100
>
> mean(X)
> quantile(X)
>
> etc.
>
> Untested [I'm not at my computer so the args for gsub() might be out of
> order] but one possible solution pattern.
>
> Michael
>
> On Jun 20, 2012, at 8:26 PM, C W <tmrs...@gmail.com> wrote:
>
> > Hi R list,
> > I imported values from Excel, there is a column with numbers like 45%,
> 65%,
> > 12%.
> >
> > I want to find its mean. What should I use?
> >
> > strisplit()
> > split()
> > parse()
> >
> > Data from dput(),
> >
> > structure(c(78L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = c("",
> >
> > "-0.15%", "-0.34%", "-1.3%", "-10.77%", "-100.00%", "-11.45%",
> >
> > "-12.53%", "-13.06%", "-15.36%", "-15.82%", "-16.96%", "-18.71%",
> >
> > "-2.02%", "-2.94%", "-21.23%", "-25.00%", "-26.20%", "-29.79%",
> >
> > "-3.16%", "-3.67%", "-30.52%", "-33.44%", "-37.48%", "-37.89%",
> >
> > "-39.42%", "-45.88%", "-5.09%", "-51.64%", "-61.58%", "-62.87%",
> >
> > "-63.51%", "-7.00%", "-7.90%", "-8.33%", "-8.58%", "-8.88%",
> >
> > "-91.10%", "-94.08%", "-96.01%", "0.98%", "10.00%", "10.04%",
> >
> > "10.64%", "11.11%", "114.32%", "12.09%", "12.68%", "13.77%",
> >
> > "14.10%", "15.51%", "16.25%", "16.93%", "16.94%", "18.57%", "18.88%",
> >
> > "2.46%", "2.55%", "2.79%", "2.93%", "20.00%", "22.67%", "24.50%",
> >
> > "25.76%", "28.18%", "3.26%", "3.80%", "3.83%", "36.05%", "37.22%",
> >
> > "40.63%", "5.53%", "5.70%", "6.19%", "6.62%", "6.72%", "63.33%",
> >
> > "7.14%", "7.21%", "7.39%", "9.15%", "9.99%", "95.00%"), class = "factor")
> >
> > Thanks,
> >
> > Mike
> >
> > [[alternative HTML version deleted]]
> >
> > ______________________________________________
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> > and provide commented, minimal, self-contained, reproducible code.
>
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