[R] printing difftime summary

[Sam Steingold]

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
http://www.childpsy.net/ http://openvotingconsortium.org
http://memri.org http://camera.org http://mideasttruth.com http://pmw.org.il
MS Windows: error: the operation completed successfully.

______________________________________________
R-h...@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

[arun]

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.

[Sam Steingold]

Hi,

> * arun <fznegc...@lnubb.pbz> [2012-11-21 14:04:36 -0800]:

>
> Are you looking for some other function (difftime2string)
> ot just remove the quotes from the printed output?

I am wondering what others do when they want to see a summary of difftime.

> 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

cool, thanks.
I now think that what I want is
--8<---------------cut here---------------start------------->8---
difftime.summary <- function (v) {
s <- summary(as.numeric(v))
r <- as.data.frame(sapply(s,difftime2string),stringsAsFactors=FALSE)
names(r) <- c("string")
r[[units(v)]] <- s
r
}
> difftime.summary(infl$delay)
string secs
Min. 500.00 ms 0.5
1st Qu. 17.12 min 1027.0
Median 99.48 min 5969.0
Mean 8.30 hrs 29870.0
3rd Qu. 8.05 hrs 28970.0
Max. 6.98 days 603100.0
--8<---------------cut here---------------end--------------->8---

--
Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 11.0.11103000

http://www.childpsy.net/ http://ffii.org http://jihadwatch.org http://memri.org
http://www.memritv.org http://camera.org http://mideasttruth.com
A computer scientist is someone who fixes things that aren't broken.

[R. Michael Weylandt]

On Thu, Nov 22, 2012 at 4:01 AM, Sam Steingold <s...@gnu.org> wrote:
> Hi,
>
>> * arun <fznegc...@lnubb.pbz> [2012-11-21 14:04:36 -0800]:
>>
>> Are you looking for some other function (difftime2string)
>> ot just remove the quotes from the printed output?
>
> I am wondering what others do when they want to see a summary of difftime.
>
>> 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
>
> cool, thanks.
> I now think that what I want is
> --8<---------------cut here---------------start------------->8---
> difftime.summary <- function (v) {
> s <- summary(as.numeric(v))
> r <- as.data.frame(sapply(s,difftime2string),stringsAsFactors=FALSE)
> names(r) <- c("string")
> r[[units(v)]] <- s
> r
> }

Any reason not summary.difftime to get S3 dispatch?

MW

[Sam Steingold]

> * R. Michael Weylandt <zvpunry....@tznvy.pbz> [2012-11-22 12:11:55 +0000]:

>
>> I now think that what I want is
>> --8<---------------cut here---------------start------------->8---
>> difftime.summary <- function (v) {
>> s <- summary(as.numeric(v))
>> r <- as.data.frame(sapply(s,difftime2string),stringsAsFactors=FALSE)
>> names(r) <- c("string")
>> r[[units(v)]] <- s
>> r
>> }
>
> Any reason not summary.difftime to get S3 dispatch?

I hoped that someone will ask this :-)

1. because its argument has type "vector of difftime", not "difftime"
(coming from CLOS, I do not expect summary(vector of difftime) to
dispatch to summary.difftime, but to summary.vector.of.difftime or something)

2. because difftime.summary returns a data.frame and not a
"Classes 'summaryDefault', 'table'" as I assume summary must return.

if these are not valid issues, then I wonder why my function should not
be the system default method.

--
Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 11.0.11103000

http://www.childpsy.net/ http://memri.org http://honestreporting.com
http://jihadwatch.org http://openvotingconsortium.org http://ffii.org
Sex is like air. It's only a big deal if you can't get any.

[R. Michael Weylandt]

On Thu, Nov 22, 2012 at 5:49 PM, Sam Steingold <s...@gnu.org> wrote:
>> * R. Michael Weylandt <zvpunry....@tznvy.pbz> [2012-11-22 12:11:55 +0000]:
>>
>>> I now think that what I want is
>>> --8<---------------cut here---------------start------------->8---
>>> difftime.summary <- function (v) {
>>> s <- summary(as.numeric(v))
>>> r <- as.data.frame(sapply(s,difftime2string),stringsAsFactors=FALSE)
>>> names(r) <- c("string")
>>> r[[units(v)]] <- s
>>> r
>>> }
>>
>> Any reason not summary.difftime to get S3 dispatch?
>
> I hoped that someone will ask this :-)
>
> 1. because its argument has type "vector of difftime", not "difftime"
> (coming from CLOS, I do not expect summary(vector of difftime) to
> dispatch to summary.difftime, but to summary.vector.of.difftime or something)

I'm not sure that's a suitable distinction in R. (Almost) All objects
are vectors (either generic or atomic) and all that....

>
> 2. because difftime.summary returns a data.frame and not a
> "Classes 'summaryDefault', 'table'" as I assume summary must return.

See http://cran.r-project.org/doc/FAQ/R-FAQ.html#How-should-I-write-summary-methods_003f

[Sam Steingold]

> * R. Michael Weylandt <zvpunry....@tznvy.pbz> [2012-11-23 09:13:36 +0000]:

>
>> 2. because difftime.summary returns a data.frame and not a
>> "Classes 'summaryDefault', 'table'" as I assume summary must return.
>
> See http://cran.r-project.org/doc/FAQ/R-FAQ.html#How-should-I-write-summary-methods_003f

what are the requirements on the class summary.foo?
does it have to inherit from some other class?
how do I define a class?

--
Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 11.0.11103000

http://www.childpsy.net/ http://dhimmi.com http://honestreporting.com
http://thereligionofpeace.com http://iris.org.il http://americancensorship.org
In the race between idiot-proof software and idiots, the idiots are winning.

[David Winsemius]

On Nov 23, 2012, at 12:35 PM, Sam Steingold wrote:

>> * R. Michael Weylandt <zvpunry....@tznvy.pbz> [2012-11-23
>> 09:13:36 +0000]:
>>
>>> 2. because difftime.summary returns a data.frame and not a
>>> "Classes 'summaryDefault', 'table'" as I assume summary must return.
>>
>> See http://cran.r-project.org/doc/FAQ/R-FAQ.html#How-should-I-write-summary-methods_003f
>
> what are the requirements on the class summary.foo?

I'm not sure it makes sense to frame the question this way.
summary.foo would nt be a class but rather a 'summary' method/function
that applied to items of class 'foo.

>
> does it have to inherit from some other class?

There is implicit inheritance from the vector "class" and sometimes
default methods will assume a numeric vector. But if you defin an
object to be of a particular class, it would not need to have an
explicit inheritance defined/

> how do I define a class?

That's pretty easy. Read:

?class # and the pages to which it links.

(And also read ?methods, and the pages to which it links. Then read
Sect 5 Object-oriented programming in the "R Language Definition".)

--

David Winsemius, MD
Alameda, CA, USA

[Sam Steingold]

> * David Winsemius <qjvaf...@pbzpnfg.arg> [2012-11-23 13:14:17 -0800]:
>
>>> See http://cran.r-project.org/doc/FAQ/R-FAQ.html#How-should-I-write-summary-methods_003f

--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("data.frame","summary.difftime")
r
}
print.summary.difftime <- function (sd) print.data.frame(sd)
--8<---------------cut here---------------end--------------->8---

it appears to work for a single vector:

--8<---------------cut here---------------start------------->8---
> r1 <- summary(infl$delay)
> r1
string secs
Min. 492.00 ms 0.5
1st Qu. 18.08 min 1085.0
Median 1.77 hrs 6370.0
Mean 8.20 hrs 29530.0
3rd Qu. 8.12 hrs 29250.0
Max. 6.98 days 602900.0
> str(r1)
Classes 'summary.difftime' and 'data.frame': 6 obs. of 2 variables:
$ string: chr "492.00 ms" "18.08 min" "1.77 hrs" "8.20 hrs" ...
$ secs :Classes 'summaryDefault', 'table' num [1:6] 4.92e-01 1.08e+03 6.37e+03 2.95e+04 2.92e+04 ...
--8<---------------cut here---------------end--------------->8---

but not as a part of data frame:

--8<---------------cut here---------------start------------->8---
> a <- summary(infl)
Error in summary.difftime(X[[22L]], ...) :
unused argument(s) (maxsum = 7, digits = 12)
--8<---------------cut here---------------end--------------->8---

I guess I should somehow accept a list of options in summary.difftime()
and pass them on to the inner call to summary() (or should it be
explicitly summary.numeric()?)

how do I do that?

--
Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 11.0.11103000

http://www.childpsy.net/ http://camera.org http://jihadwatch.org
http://americancensorship.org http://truepeace.org http://memri.org
Why do you never call me back after I scream that I will never talk to you again?!

[David Winsemius]

In the usual way. If you know that the function will be called with
arguments from the summary.data.frame function then you should allow
the argument list to accept them. You can ignore them or provide
provisions for them. You just can't define your function to have only
one argument if you expect (as you should since you passes summary a
dataframe object) that it might be called within summary.data.frame.

This is the argument list for summary.data.frame:

> summary.data.frame
function (object, maxsum = 7, digits = max(3, getOption("digits") -
3), ...)

> how do I do that?

summary.difftime <- function (v, ... ) { ................

There are many asked and answered questions on rhelp about how to deal
with the "dots" arguments.

--
David Winsemius, MD
Alameda, CA, USA

[Sam Steingold]

this overcomes the summary generation, but not printing:

--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("data.frame","summary.difftime")
r
}
print.summary.difftime <- function (sd) print.data.frame(sd)
--8<---------------cut here---------------end--------------->8---

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?

> * David Winsemius <qjvaf...@pbzpnfg.arg> [2012-11-25 00:50:51 -0800]:

>> 1.08e+03 6.37e+03 2.95e+04 2.92e+04 ...

Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 11.0.11103000

http://www.childpsy.net/ http://www.memritv.org http://memri.org
http://honestreporting.com http://dhimmi.com http://openvotingconsortium.org
People with a good taste are especially appreciated by cannibals.

[David Winsemius]

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

methods(print)

--
David.

David Winsemius, MD
Alameda, CA, USA

[R. Michael Weylandt]

On Mon, Nov 26, 2012 at 4:46 PM, David Winsemius <dwins...@comcast.net> wrote:
>
> On Nov 26, 2012, at 7:14 AM, Sam Steingold wrote:
>
>> this overcomes the summary generation, but not printing:
>>
>> --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("data.frame","summary.difftime")

Surely reversed no? summary.difftime inherits from data.frame I would
have assumed.

>> r
>> }
>> print.summary.difftime <- function (sd) print.data.frame(sd)

What is this supposed to do exactly? If you have inheritance why have
the subclass method do nothing other than call the parent method?

Michael

[Sam Steingold]

> * David Winsemius <qjvaf...@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

http://www.childpsy.net/ http://palestinefacts.org http://think-israel.org
http://www.memritv.org http://openvotingconsortium.org http://mideasttruth.com
The force of gravity doubles when acting on a body on a couch.

[William Dunlap]

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

[Sam Steingold]

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 <jqh...@gvopb.pbz> [2012-11-26 23:02:48 +0000]:

http://www.childpsy.net/ http://openvotingconsortium.org
http://ffii.org http://www.memritv.org http://americancensorship.org
There are two ways to write error-free programs; only the third one works.

[William Dunlap]

> 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.

[Sam Steingold]

Looks like
format.summary.difftime <- function(sd, ...) structure(sd$string,
names=rownames(sd))
does the job.
any reason not to use it?

--
Sam Steingold <http://sds.podval.org> <http://www.childpsy.net/>