Hello,
Just a doubt. Are you looking for some other function (difftime2string) ot
just remove the quotes from the printed output?
If it is the latter, then this should do it.
res<-do.call(data.frame,lapply(s,difftime2string))
names(res)<-names(s)
res
# Min. 1st Qu. Median Mean 3rd Qu. Max.
#1 500.00 ms 17.12 min 99.48 min 8.30 hrs 8.05 hrs 6.98 days
A.K.
----- Original Message -----
From: Sam Steingold <s...@gnu.org>
To: r-help@r-project.org
Cc:
Sent: Wednesday, November 21, 2012 2:22 PM
Subject: [R] printing difftime summary
Hi,
I have a vector of difftime objects and I want to see its summary.
Alas:
--8<---------------cut here---------------start------------->8---
> summary(infl$delay)
Length Class Mode
9008386 difftime numeric
--8<---------------cut here---------------end--------------->8---
this is almost completely useless.
I can use as.numeric:
--8<---------------cut here---------------start------------->8---
> s <- summary(as.numeric(infl$delay))
> dput(s)
structure(c(0.5, 1027, 5969, 29870, 28970, 603100), .Names = c("Min.",
"1st Qu.", "Median", "Mean", "3rd Qu.", "Max."), class = c("summaryDefault",
"table"))
> s
Min. 1st Qu. Median Mean 3rd Qu. Max.
0.5 1027.0 5969.0 29870.0 28970.0 603100.0
--8<---------------cut here---------------end--------------->8---
but the printed representation is very unreadable: the fact that
603100.0 is almost exactly 7 days is not obvious.
Okay, maybe as.difftime will help?
--8<---------------cut here---------------start------------->8---
> as.difftime(s,units="secs")
Time differences in secs
Min. 1st Qu. Median Mean 3rd Qu. Max.
0.5 1027.0 5969.0 29870.0 28970.0 603100.0
> as.difftime(s/3600,units="hours")
Time differences in hours
Min. 1st Qu. Median Mean 3rd Qu. Max.
1.388889e-04 2.852778e-01 1.658056e+00 8.297222e+00 8.047222e+00 1.675278e+02
--8<---------------cut here---------------end--------------->8---
nope; still unreadable.
What I really want to see _printed_ is something likes this:
--8<---------------cut here---------------start------------->8---
> sapply(s,difftime2string)
Min. 1st Qu. Median Mean 3rd Qu. Max.
"500.00 ms" "17.12 min" "99.48 min" "8.30 hrs" "8.05 hrs" "6.98 days"
--8<---------------cut here---------------end--------------->8---
except that the quotes are not needed in the printed output.
Here I wrote:
--8<---------------cut here---------------start------------->8---
difftime2string <- function (x) {
if (x < 1) return(sprintf("%.2f ms",x*1000))
if (x < 100) return(sprintf("%.2f sec",x))
if (x < 6000) return(sprintf("%.2f min",x/60))
if (x < 108000) return(sprintf("%.2f hrs",x/3600))
if (x < 400*24*3600) return(sprintf("%.2f days",x/(24*3600)))
sprintf("%.2f years",x/(365.25*24*3600))
}
--8<---------------cut here---------------end--------------->8---
So, what is "The Right R Way" to print a summary of difftime objects?
Thanks!
--
Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 11.0.11103000
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