printing difftime summary

why do I see NULLs?!

because

... format.difftime does a reasonable job (except that it does not copy
the input names to its output).

Replace your call of the form
  format(difftimeObject)
with
  structure(format(difftimeObject), names=names(difftimeObject))
to work around this.


Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com

-----Original Message-----
From: Sam Steingold [mailto:sam.steingold at gmail.com] On Behalf Of Sam Steingold
Sent: Monday, November 26, 2012 4:09 PM
To: William Dunlap
Cc: r-help at r-project.org; David Winsemius
Subject: Re: [R] printing difftime summary

Thanks a lot - almost there!

--8<---------------cut here---------------start------------->8---
format.summary.difftime <- function(sd, ...) {
  t <- matrix(sd$string)
  rownames(t) <- rownames(sd)
  print(t)
  format(as.table(t))
}
print.summary.difftime <- function (sd, ...) {
  print(format(sd), quote=FALSE)
  invisible(sd)
}
--8<---------------cut here---------------end--------------->8---

this almost works:

--8<---------------cut here---------------start------------->8---

summary(delays)

             share.id         min              max
 12cf12372b87cce9:      1   NULL:492.00 ms   NULL:492.00 ms
 12cf36060bdb9581:      1   NULL:3.70 min    NULL:21.80 min
 12d2665c906bb232:      1   NULL:20.32 min   NULL:3.26 hrs
 12d2802f1435b4cd:      1   NULL:5.52 hrs    NULL:13.78 hrs
 12d292988f5f8422:      1   NULL:2.81 hrs    NULL:16.20 hrs
 12d29dd2894e2790:      1   NULL:6.95 days   NULL:6.98 days
--8<---------------cut here---------------end--------------->8---

why do I see NULLs?!

--8<---------------cut here---------------start------------->8---

t <- matrix(sd$string)
rownames(t) <- rownames(sd)
t

        [,1]
Min.    "492.00 ms"
1st Qu. "3.70 min"
Median  "20.32 min"
Mean    "5.52 hrs"
3rd Qu. "2.81 hrs"
Max.    "6.95 days"

as.table(t)

        A
Min.    492.00 ms
1st Qu. 3.70 min
Median  20.32 min
Mean    5.52 hrs
3rd Qu. 2.81 hrs
Max.    6.95 days

format(as.table(t))

        A
Min.    "492.00 ms"
1st Qu. "3.70 min "
Median  "20.32 min"
Mean    "5.52 hrs "
3rd Qu. "2.81 hrs "
Max.    "6.95 days"

--8<---------------cut here---------------end--------------->8---

* William Dunlap <jqhaync at gvopb.pbz> [2012-11-26 23:02:48 +0000]:

It looks like summary.data.frame(d) calls format(d[[i]]) for i in seq_len(ncol(d))
and pastes the results together into a "table" object for printing.  Hence, write
a format.summary.difftime if you want objects of class "summary.difftime" (which
I assume summary.difftime produces) to be formatted as you wish when a
difftime object is in a data.frame.  Once you've written it, have your

print.summary.difftime

call it too.

E.g., with the following methods
summary.difftime <- function(x, ...) {
         ret <- quantile(x, p=(0:2)/2, na.rm=TRUE)
         class(ret) <- c("summary.difftime", class(ret))
         ret
}
format.summary.difftime <- function(x, ...) c(Min.Med.Max =
paste(collapse="...", NextMethod("format")))
print.summary.difftime <- function(x, ...){ print(format(x), quote=FALSE) ; invisible(x) }

I get

d <- data.frame(Num=1:5, Date=as.Date("2012-11-26")+(0:4),
Delta=diff(as.Date("2012-11-26")+2^(0:5)))
summary(d)

      Num         Date                    Delta
 Min.   :1   Min.   :2012-11-26   Min.Med.Max: 1 days... 4 days...16 days
 1st Qu.:2   1st Qu.:2012-11-27
 Median :3   Median :2012-11-28
 Mean   :3   Mean   :2012-11-28
 3rd Qu.:4   3rd Qu.:2012-11-29
 Max.   :5   Max.   :2012-11-30

summary(d$Delta)

                Min.Med.Max
 1 days... 4 days...16 days

My summary.difftime inherits from difftime so the format method is not really
needed, as format.difftime does a reasonable job (except that it does not copy
the input names to its output).  I put it in to show how it gets called.


Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com

-----Original Message-----
From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On

Behalf

Of Sam Steingold
Sent: Monday, November 26, 2012 2:20 PM
To: r-help at r-project.org; David Winsemius
Subject: Re: [R] printing difftime summary

* David Winsemius <qjvafrzvhf at pbzpnfg.arg> [2012-11-26 08:46:35 -0800]:

On Nov 26, 2012, at 7:14 AM, Sam Steingold wrote:

summary(infl), where infl$delay is a difftime vector, prints

...

   delay
string:c("492.00 ms", "18.08 min", "1.77 hrs", "8.20 hrs", "8.13 hrs",
"6.98 days")
secs  :c("     0.5", "  1085.1", "  6370.2", " 29534.4", " 29254.0",
"602949.7")



instead of something like

  delay
  Min.:    492 ms
  1st Qu.: 18.08 min

&c

so, how do I arrange for a proper printing of difftime summary as a
part
of the data frame summary?

If you like a particular format from an existing print method then why
not look it up and copy the code?

methods(print)

the problem is that I cannot figure out which function prints this:

   delay
string:c("492.00 ms", "18.08 min", "1.77 hrs", "8.20 hrs", "8.13 hrs",
"6.98 days")
secs  :c("     0.5", "  1085.1", "  6370.2", " 29534.4", " 29254.0",
"602949.7")

I added cat()s to print.summary.difftime and I do not see them, so it
appears that I have no direct control over how a summary.difftime is
printed as a part of a summary of a data.frame.


--8<---------------cut here---------------start------------->8---
summary.difftime <- function (v, ...) {
  s <- summary(as.numeric(v), ...)
  r <- as.data.frame(sapply(s,difftime2string),stringsAsFactors=FALSE)
  names(r) <- c("string")
  r[[units(v)]] <- s
  class(r) <- c("summary.difftime","data.frame")
  invisible(r)
}
print.summary.difftime <- function (sd, ...) {
  cat("[[[print.summary.difftime]]]\n")
  print(list(...))
  print.data.frame(sd, ...)
}
--8<---------------cut here---------------end--------------->8---

--
Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 11.0.11103000
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--
Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 11.0.11103000
http://www.childpsy.net/ http://openvotingconsortium.org
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There are two ways to write error-free programs; only the third one works.