Re: [R] printing difftime summary
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$delay is a difftime vector, prints
...
delay
string:c(492.00 ms, 18.08 min, 1.77 hrs, 8.20 hrs, 8.13 hrs, 6.98
days)
secs :c( 0.5, 1085.1, 6370.2, 29534.4, 29254.0,
602949.7)
instead of something like
delay
Min.:492 ms
1st Qu.: 18.08 min
c
so, how do I arrange for a proper printing of difftime summary as a part
of the data frame summary?
* David Winsemius qjvafrz...@pbzpnfg.arg [2012-11-25 00:50:51 -0800]:
On Nov 24, 2012, at 7:48 PM, Sam Steingold wrote:
* David Winsemius qjvafrz...@pbzpnfg.arg [2012-11-23 13:14:17
-0800]:
See
http://cran.r-project.org/doc/FAQ/R-FAQ.html#How-should-I-write-summary-methods_003f
--8---cut here---start-8---
summary.difftime - function (v) {
s - summary(as.numeric(v))
r - as.data.frame(sapply(s,difftime2string),stringsAsFactors=FALSE)
names(r) - c(string)
r[[units(v)]] - s
class(r) - c(data.frame,summary.difftime)
r
}
print.summary.difftime - function (sd) print.data.frame(sd)
--8---cut here---end---8---
it appears to work for a single vector:
--8---cut here---start-8---
r1 - summary(infl$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()?)
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.
--
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.
__
R-help@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.
Re: [R] printing difftime summary
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)
r
}
print.summary.difftime - function (sd) print.data.frame(sd)
--8---cut here---end---8---
summary(infl), where infl$delay is a difftime vector, prints
...
delay
string:c(492.00 ms, 18.08 min, 1.77 hrs, 8.20 hrs, 8.13
hrs, 6.98 days)
secs :c( 0.5, 1085.1, 6370.2, 29534.4, 29254.0,
602949.7)
instead of something like
delay
Min.:492 ms
1st Qu.: 18.08 min
c
so, how do I arrange for a proper printing of difftime summary as a
part
of the data frame summary?
If you like a particular format from an existing print method then why
not look it up and copy the code?
methods(print)
--
David.
* David Winsemius qjvafrz...@pbzpnfg.arg [2012-11-25 00:50:51
-0800]:
On Nov 24, 2012, at 7:48 PM, Sam Steingold wrote:
* David Winsemius qjvafrz...@pbzpnfg.arg [2012-11-23 13:14:17
-0800]:
See
http://cran.r-project.org/doc/FAQ/R-FAQ.html#How-should-I-write-summary-methods_003f
--8---cut here---start-8---
summary.difftime - function (v) {
s - summary(as.numeric(v))
r - as.data.frame(sapply(s,difftime2string),stringsAsFactors=FALSE)
names(r) - c(string)
r[[units(v)]] - s
class(r) - c(data.frame,summary.difftime)
r
}
print.summary.difftime - function (sd) print.data.frame(sd)
--8---cut here---end---8---
it appears to work for a single vector:
--8---cut here---start-8---
r1 - summary(infl$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()?)
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.
--
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, MD
Alameda, CA, USA
__
R-help@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.
Re: [R] printing difftime summary
On Mon, Nov 26, 2012 at 4:46 PM, David Winsemius dwinsem...@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
--8---cut here---end---8---
summary(infl), where infl$delay is a difftime vector, prints
...
delay
string:c(492.00 ms, 18.08 min, 1.77 hrs, 8.20 hrs, 8.13 hrs,
6.98 days)
secs :c( 0.5, 1085.1, 6370.2, 29534.4, 29254.0,
602949.7)
instead of something like
delay
Min.:492 ms
1st Qu.: 18.08 min
c
so, how do I arrange for a proper printing of difftime summary as a part
of the data frame summary?
If you like a particular format from an existing print method then why not
look it up and copy the code?
methods(print)
--
David.
* David Winsemius qjvafrz...@pbzpnfg.arg [2012-11-25 00:50:51 -0800]:
On Nov 24, 2012, at 7:48 PM, Sam Steingold wrote:
* David Winsemius qjvafrz...@pbzpnfg.arg [2012-11-23 13:14:17
-0800]:
See
http://cran.r-project.org/doc/FAQ/R-FAQ.html#How-should-I-write-summary-methods_003f
--8---cut here---start-8---
summary.difftime - function (v) {
s - summary(as.numeric(v))
r - as.data.frame(sapply(s,difftime2string),stringsAsFactors=FALSE)
names(r) - c(string)
r[[units(v)]] - s
class(r) - c(data.frame,summary.difftime)
r
}
print.summary.difftime - function (sd) print.data.frame(sd)
--8---cut here---end---8---
it appears to work for a single vector:
--8---cut here---start-8---
r1 - summary(infl$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()?)
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.
--
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, MD
Alameda, CA, USA
__
R-help@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.
__
R-help@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.
Re: [R] printing difftime summary
* David Winsemius qjvafrz...@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.
__
R-help@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.
Re: [R] printing difftime summary
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 DateDelta
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-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf
Of Sam Steingold
Sent: Monday, November 26, 2012 2:20 PM
To: r-help@r-project.org; David Winsemius
Subject: Re: [R] printing difftime summary
* David Winsemius qjvafrz...@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.
__
R-help@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.
__
R-help@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.
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 minNULL:21.80 min
12d2665c906bb232: 1 NULL:20.32 min NULL:3.26 hrs
12d2802f1435b4cd: 1 NULL:5.52 hrsNULL:13.78 hrs
12d292988f5f8422: 1 NULL:2.81 hrsNULL: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$string)
rownames(t) - rownames(sd)
t
[,1]
Min.492.00 ms
1st Qu. 3.70 min
Median 20.32 min
Mean5.52 hrs
3rd Qu. 2.81 hrs
Max.6.95 days
as.table(t)
A
Min.492.00 ms
1st Qu. 3.70 min
Median 20.32 min
Mean5.52 hrs
3rd Qu. 2.81 hrs
Max.6.95 days
format(as.table(t))
A
Min.492.00 ms
1st Qu. 3.70 min
Median 20.32 min
Mean5.52 hrs
3rd Qu. 2.81 hrs
Max.6.95 days
--8---cut here---end---8---
* William Dunlap jqha...@gvopb.pbz [2012-11-26 23:02:48 +]:
It looks like summary.data.frame(d) calls format(d[[i]]) for i in
seq_len(ncol(d))
and pastes the results together into a table object for printing. Hence,
write
a format.summary.difftime if you want objects of class summary.difftime
(which
I assume summary.difftime produces) to be formatted as you wish when a
difftime object is in a data.frame. Once you've written it, have your
print.summary.difftime
call it too.
E.g., with the following methods
summary.difftime - function(x, ...) {
ret - quantile(x, p=(0:2)/2, na.rm=TRUE)
class(ret) - c(summary.difftime, class(ret))
ret
}
format.summary.difftime - function(x, ...) c(Min.Med.Max =
paste(collapse=..., NextMethod(format)))
print.summary.difftime - function(x, ...){ print(format(x), quote=FALSE) ;
invisible(x) }
I get
d - data.frame(Num=1:5, Date=as.Date(2012-11-26)+(0:4),
Delta=diff(as.Date(2012-11-26)+2^(0:5)))
summary(d)
Num DateDelta
Min. :1 Min. :2012-11-26 Min.Med.Max: 1 days... 4 days...16 days
1st Qu.:2 1st Qu.:2012-11-27
Median :3 Median :2012-11-28
Mean :3 Mean :2012-11-28
3rd Qu.:4 3rd Qu.:2012-11-29
Max. :5 Max. :2012-11-30
summary(d$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-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf
Of Sam Steingold
Sent: Monday, November 26, 2012 2:20 PM
To: r-help@r-project.org; David Winsemius
Subject: Re: [R] printing difftime summary
* David Winsemius qjvafrz...@pbzpnfg.arg [2012-11-26 08:46:35 -0800]:
On Nov 26, 2012, at 7:14 AM, Sam Steingold wrote:
summary(infl), where infl$delay is a difftime vector, prints
...
delay
string:c(492.00 ms, 18.08 min, 1.77 hrs, 8.20 hrs, 8.13 hrs,
6.98 days)
secs :c( 0.5, 1085.1, 6370.2, 29534.4, 29254.0,
602949.7)
instead of something like
delay
Min.:492 ms
1st Qu.: 18.08 min
c
so, how do I arrange for a proper printing of difftime summary as a
part
of the data frame summary?
If you like a particular format from an existing print method then why
not look it up and copy the code?
methods(print)
the problem is that I cannot figure out which function prints this:
delay
string:c(492.00 ms, 18.08 min, 1.77 hrs, 8.20 hrs, 8.13 hrs,
6.98 days)
secs :c( 0.5, 1085.1, 6370.2, 29534.4, 29254.0,
602949.7)
I added cat()s to print.summary.difftime and I do not see them, so it
appears that I have no direct control over how a summary.difftime is
printed as a part of a summary of a data.frame.
--8---cut here---start-8---
summary.difftime - function (v, ...) {
s - summary(as.numeric(v), ...)
r - as.data.frame(sapply(s,difftime2string),stringsAsFactors=FALSE)
names(r) - c(string)
r[[units(v)]] - s
class(r) - c(summary.difftime,data.frame)
invisible(r
Re: [R] printing difftime summary
why do I see NULLs?!
because
... format.difftime does a reasonable job (except that it does not copy
the input names to its output).
Replace your call of the form
format(difftimeObject)
with
structure(format(difftimeObject), names=names(difftimeObject))
to work around this.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-Original Message-
From: Sam Steingold [mailto:sam.steing...@gmail.com] On Behalf Of Sam
Steingold
Sent: Monday, November 26, 2012 4:09 PM
To: William Dunlap
Cc: r-help@r-project.org; David Winsemius
Subject: Re: [R] printing difftime summary
Thanks a lot - almost there!
--8---cut here---start-8---
format.summary.difftime - function(sd, ...) {
t - matrix(sd$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 minNULL:21.80 min
12d2665c906bb232: 1 NULL:20.32 min NULL:3.26 hrs
12d2802f1435b4cd: 1 NULL:5.52 hrsNULL:13.78 hrs
12d292988f5f8422: 1 NULL:2.81 hrsNULL: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$string)
rownames(t) - rownames(sd)
t
[,1]
Min.492.00 ms
1st Qu. 3.70 min
Median 20.32 min
Mean5.52 hrs
3rd Qu. 2.81 hrs
Max.6.95 days
as.table(t)
A
Min.492.00 ms
1st Qu. 3.70 min
Median 20.32 min
Mean5.52 hrs
3rd Qu. 2.81 hrs
Max.6.95 days
format(as.table(t))
A
Min.492.00 ms
1st Qu. 3.70 min
Median 20.32 min
Mean5.52 hrs
3rd Qu. 2.81 hrs
Max.6.95 days
--8---cut here---end---8---
* William Dunlap jqha...@gvopb.pbz [2012-11-26 23:02:48 +]:
It looks like summary.data.frame(d) calls format(d[[i]]) for i in
seq_len(ncol(d))
and pastes the results together into a table object for printing. Hence,
write
a format.summary.difftime if you want objects of class summary.difftime
(which
I assume summary.difftime produces) to be formatted as you wish when a
difftime object is in a data.frame. Once you've written it, have your
print.summary.difftime
call it too.
E.g., with the following methods
summary.difftime - function(x, ...) {
ret - quantile(x, p=(0:2)/2, na.rm=TRUE)
class(ret) - c(summary.difftime, class(ret))
ret
}
format.summary.difftime - function(x, ...) c(Min.Med.Max =
paste(collapse=..., NextMethod(format)))
print.summary.difftime - function(x, ...){ print(format(x), quote=FALSE) ;
invisible(x) }
I get
d - data.frame(Num=1:5, Date=as.Date(2012-11-26)+(0:4),
Delta=diff(as.Date(2012-11-26)+2^(0:5)))
summary(d)
Num DateDelta
Min. :1 Min. :2012-11-26 Min.Med.Max: 1 days... 4 days...16 days
1st Qu.:2 1st Qu.:2012-11-27
Median :3 Median :2012-11-28
Mean :3 Mean :2012-11-28
3rd Qu.:4 3rd Qu.:2012-11-29
Max. :5 Max. :2012-11-30
summary(d$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-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf
Of Sam Steingold
Sent: Monday, November 26, 2012 2:20 PM
To: r-help@r-project.org; David Winsemius
Subject: Re: [R] printing difftime summary
* David Winsemius qjvafrz...@pbzpnfg.arg [2012-11-26 08:46:35 -0800]:
On Nov 26, 2012, at 7:14 AM, Sam Steingold wrote:
summary(infl), where infl$delay is a difftime vector, prints
...
delay
string:c(492.00 ms, 18.08 min, 1.77 hrs, 8.20 hrs, 8.13 hrs,
6.98 days)
secs :c( 0.5, 1085.1, 6370.2, 29534.4, 29254.0,
602949.7)
instead of something like
delay
Min.:492 ms
1st Qu.: 18.08 min
c
so, how do I arrange for a proper printing of difftime summary as a
part
of the data frame summary?
If you like a particular format from an existing print method then why
not look it up and copy the code?
methods(print)
the problem is that I cannot figure out which
Re: [R] printing difftime summary
Looks like
format.summary.difftime - function(sd, ...) structure(sd$string,
names=rownames(sd))
does the job.
any reason not to use it?
On Mon, Nov 26, 2012 at 7:36 PM, William Dunlap wdun...@tibco.com wrote:
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.steing...@gmail.com] On Behalf Of Sam
Steingold
Sent: Monday, November 26, 2012 4:09 PM
To: William Dunlap
Cc: r-help@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$string)
rownames(t) - rownames(sd)
print(t)
format(as.table(t))
}
print.summary.difftime - function (sd, ...) {
print(format(sd), quote=FALSE)
invisible(sd)
}
--8---cut here---end---8---
this almost works:
--8---cut here---start-8---
summary(delays)
share.id min max
12cf12372b87cce9: 1 NULL:492.00 ms NULL:492.00 ms
12cf36060bdb9581: 1 NULL:3.70 minNULL:21.80 min
12d2665c906bb232: 1 NULL:20.32 min NULL:3.26 hrs
12d2802f1435b4cd: 1 NULL:5.52 hrsNULL:13.78 hrs
12d292988f5f8422: 1 NULL:2.81 hrsNULL:16.20 hrs
12d29dd2894e2790: 1 NULL:6.95 days NULL:6.98 days
--8---cut here---end---8---
why do I see NULLs?!
--8---cut here---start-8---
t - matrix(sd$string)
rownames(t) - rownames(sd)
t
[,1]
Min.492.00 ms
1st Qu. 3.70 min
Median 20.32 min
Mean5.52 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
Mean5.52 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
Mean5.52 hrs
3rd Qu. 2.81 hrs
Max.6.95 days
--8---cut here---end---8---
* William Dunlap jqha...@gvopb.pbz [2012-11-26 23:02:48 +]:
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 DateDelta
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$Delta)
Min.Med.Max
1 days... 4 days...16 days
My summary.difftime inherits from difftime so the format method is not
really
needed, as format.difftime does a reasonable job (except that it does not
copy
the input names to its output). I put it in to show how it gets called.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On
Behalf
Of Sam Steingold
Sent: Monday, November 26, 2012 2:20 PM
To: r-help@r-project.org; David Winsemius
Subject: Re: [R] printing difftime summary
* David Winsemius qjvafrz...@pbzpnfg.arg [2012-11-26 08:46:35 -0800]:
On Nov 26, 2012, at 7:14 AM, Sam Steingold wrote:
summary(infl), where infl$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
Re: [R] printing difftime summary
On Nov 24, 2012, at 7:48 PM, Sam Steingold wrote:
* David Winsemius qjvafrz...@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()?)
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
__
R-help@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.
Re: [R] printing difftime summary
* David Winsemius qjvafrz...@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?!
__
R-help@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.
Re: [R] printing difftime summary
On Thu, Nov 22, 2012 at 5:49 PM, Sam Steingold s...@gnu.org wrote:
* R. Michael Weylandt zvpunry.jrlyn...@tznvy.pbz [2012-11-22 12:11:55
+]:
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
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-help@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.
Re: [R] printing difftime summary
* R. Michael Weylandt zvpunry.jrlyn...@tznvy.pbz [2012-11-23 09:13:36 +]: 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. __ R-help@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.
Re: [R] printing difftime summary
On Nov 23, 2012, at 12:35 PM, Sam Steingold wrote: * R. Michael Weylandt zvpunry.jrlyn...@tznvy.pbz [2012-11-23 09:13:36 +]: 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 __ R-help@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.
Re: [R] printing difftime summary
On Thu, Nov 22, 2012 at 4:01 AM, Sam Steingold s...@gnu.org wrote:
Hi,
* arun fznegcvax...@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
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-help@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.
__
R-help@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.
Re: [R] printing difftime summary
* R. Michael Weylandt zvpunry.jrlyn...@tznvy.pbz [2012-11-22 12:11:55
+]:
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-help@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.
Re: [R] printing difftime summary
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.
1.39e-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-help@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.
__
R-help@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.
Re: [R] printing difftime summary
Hi,
* arun fznegcvax...@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-help@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.
