Re: [R] Generating Patient Data
# build off of david's suggestion
x <-
data.frame(
patient= 1:20 ,
disease =
sapply(
pmin( 2 + rpois( 20 , 2 ) , 6 ) ,
function( n ) paste0( sample( c('A','B','C','D','E','F'),
n), collapse="+" )
)
)
# break the diseases into a list, one entry per patient
y <- strsplit( as.character( x$disease ) , "\\+" )
# melt the list
library(reshape2)
z <- melt( y )
# re-name the columns in that result
names( z ) <- c( "disease" , "patient" )
# print the results to the screen
z
# compare the structure to `x` if you like
x
On Wed, Jun 25, 2014 at 2:18 AM, Abhinaba Roy
wrote:
> Hi David,
>
> I was thinking something like this:
>
> ID Disease
> 1 A
> 2 B
> 3 A
> 1C
> 2D
> 5A
> 4B
> 3D
> 2A
> ....
>
> How can this be done?
>
>
> On Wed, Jun 25, 2014 at 11:34 AM, David Winsemius
> wrote:
>
> >
> > On Jun 24, 2014, at 10:14 PM, Abhinaba Roy wrote:
> >
> > > Dear R helpers,
> > >
> > > I want to generate data for say 1000 patients (i.e., 1000 unique IDs)
> > > having suffered from various diseases in the past (say diseases
> > > A,B,C,D,E,F). The only condition imposed is that each patient should've
> > > suffered from *atleast* two diseases. So my data frame will have two
> > > columns 'ID' and 'Disease'.
> > >
> > > I want to do a basket analysis with this data, where ID will be the
> > > identifier and we will establish rules based on the 'Disease' column.
> > >
> > > How can I generate this type of data in R?
> > >
> >
> > Perhaps something along these lines for 20 cases:
> >
> > > data.frame(patient=1:20, disease = sapply(pmin(2+rpois(20, 2), 6),
> > function(n) paste0( sample( c('A','B','C','D','E','F'), n), collapse="+"
> ) )
> > + )
> >patient disease
> > 11 F+D
> > 22 F+A+D+E
> > 33 F+D+C+E
> > 44 B+D+C+A
> > 55 D+A+F+C
> > 66 E+A+D
> > 77 E+F+B+C+A+D
> > 88 A+B+C+D+E
> > 99 B+E+C+F
> > 10 10 C+A
> > 11 11 B+A+D+E+C+F
> > 12 12 B+C
> > 13 13 A+D+B+E
> > 14 14 D+C+E+F+B+A
> > 15 15 C+F+D+E+A
> > 16 16 A+C+B
> > 17 17 C+D+B+E
> > 18 18 A+B
> > 19 19 C+B+D+E+F
> > 20 20 D+C+F
> >
> > > --
> > > Regards
> > > Abhinaba Roy
> > >
> > > [[alternative HTML version deleted]]
> >
> > You should read the Posting Guide and learn to post in HTML.
> > >
> > > PLEASE do read the posting guide
> > http://www.R-project.org/posting-guide.html
> > > and provide commented, minimal, self-contained, reproducible code.
> >
> >
> > --
> > David Winsemius
> > Alameda, CA, USA
> >
> >
>
>
> --
> Regards
> Abhinaba Roy
> Statistician
> Radix Analytics Pvt. Ltd
> Ahmedabad
>
> [[alternative HTML version deleted]]
>
> __
> 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.
>
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
[R] SD of Residuals by group
Dear Community, I emphasize the use of graphical methods to examine residuals for a Panel model and want to plot the Standard Deviation of residuals by group. I used the following for plotting residuals by group: >library(lattice) >xyplot((residuals(fixed.reg1.1))~countrynr,data=data.plm) But I have no idea how to adjust this for the SD of Residuals by group ? Thanks for attention and help! Katie [[alternative HTML version deleted]] __ 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] matlab serial date to r
Hi, I have a matlab variable as serial date (class double) in the form 'dd-mmm- HH:MM:SS'. format long disp( tx(40:60,1) ) 1.0e+05 * 7.35600813091 7.35600956856 7.35601305921 7.35601654985 7.35602004049 7.35602353113 7.35602702178 7.35603397179 7.35604092182 7.35604787183 7.35605482185 7.35606177187 7.35606940080 7.35607702975 7.35608465869 7.35609228763 7.35609991657 7.35610754551 7.35611517445 7.35612280339 7.35613085329 It should be the same as datestr( tx(40:60,1), 0) 01-Jan-2014 00:00:07 01-Jan-2014 00:00:08 01-Jan-2014 00:00:11 01-Jan-2014 00:00:14 01-Jan-2014 00:00:17 01-Jan-2014 00:00:20 01-Jan-2014 00:00:23 01-Jan-2014 00:00:29 01-Jan-2014 00:00:35 01-Jan-2014 00:00:41 01-Jan-2014 00:00:47 01-Jan-2014 00:00:53 01-Jan-2014 00:00:59 01-Jan-2014 00:01:06 01-Jan-2014 00:01:13 01-Jan-2014 00:01:19 01-Jan-2014 00:01:26 01-Jan-2014 00:01:32 01-Jan-2014 00:01:39 01-Jan-2014 00:01:46 01-Jan-2014 00:01:53 I can easily convert it with Matlab but then I will obtain a character format which is useless. I can also make use of cellstr(datestr(tx(:,1), 0)) but then I can't save it in ASCII file. The origin of the Matlab format is supposed to be "-00-00". This is the only origin which results in "2014-01-01" which is my actual start date. Can somebody please tell me how I can simply convert serial datetime to datetime in R. Thanks in advance. All the best, Christoph [[alternative HTML version deleted]] __ 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] saving a 'get' object in R
Greg - I appreciate your taking the time to explain. This is very
helpful. My case was a bit unusual as I'm helping a colleague with code
to use on a regular but individual basis. I want them to name a data set
once at the top of the script and have that name propagate through
several objects that end up being saved at the end with file names tied
to the data set name. Then tomorrow, they'll do the same thing on a
different data set so each session would be pretty simple, with only 2-3
named objects at the end, though the data sets are big and cumbersome.
I'll dig into this more and apologize to those who thought this problem
was too trivial for the r-help forum.
David
On 6/24/2014 4:00 PM, Greg Snow wrote:
The main reason to avoid assign/get is that there are better ways.
You can use a list or environment more directly without using
assign/get. Also the main reason to use assign/get is to work with
global variables which are frowned on in general programming (and
prone to hard to find bugs).
Consider the following code snippet:
x <- 1:10
functionThatUsesAssign(x)
what is the value of x after running this code? If you use global
variables then the answer is "I don't know", and if x was something
that took several hours to compute and I just accidentally overwrote
it, then I am probably not happy.
Using assign often leads to the temptation to try code like:
assign( "x[5]", 25 )
which does something (no warnings, no errors) but generally not what
was being attempted.
Another common (mis)use of assign/get is to create a sequence of
variables like "data01", "data02", "data03", ... and then do the same
thing for each of the data objects. This is much better done by
putting all the objects into a single list, then you can easily
iterate over the list using lapply/sapply or still access the
individual pieces. Then when it comes time to save these objects, you
only need to save the list, not all the individual objects, same for
deleting, copying, moving, etc. And you don't have a bunch of
different objects cluttering your workspace, just a single list.
On Tue, Jun 24, 2014 at 2:57 PM, David Stevens wrote:
Thanks to all for the replies. I tried all three and they work great. I was
misinterpreting the list = parameter in save(...) and I get your point about
overwriting existing objects. I've heard about not using assign/get before.
Can anyone point me to why and what alternatives there are?
Regards
David
On 6/24/2014 2:50 PM, Henrik Bengtsson wrote:
I recommend to use saveRDS()/readRDS() instead. More convenient and
avoids the risk that load() has of overwriting existing variables with
the same name.
/Henrik
On Tue, Jun 24, 2014 at 1:45 PM, Greg Snow <538...@gmail.com> wrote:
I think that you are looking for the `list` argument in `save`.
save( list=foo, file=paste0(foo, '.Rdata') )
In general it is best to avoid using the assign function (and get when
possible). Usually there are better alternatives.
On Tue, Jun 24, 2014 at 2:35 PM, David Stevens
wrote:
R community,
Apologies if this has been answered. The concept I'm looking for is to
save() an object retrieved using get() for an object
that resulted from using assign. Something like
save(get(foo),file=paste(foo,'rData',sep=''))
where assign(foo,obj) creates an object named foo with the contents of
obj
assigned. For example, if
x <- data.frame(v1=c(1,2,3,4),v2=c('1','2','3','4'))
foo = 'my.x'
assign(foo,x)
# (... then modify foo as needed)
save(get(foo),file=paste(foo,'.rData',sep=''))
# though this generates " in save(get(foo), file = paste(foo, ".rData",
sep
= "")) :
object ‘get(foo)’ not found", whereas
get(foo)
at the command prompt yields the contents of my.x
There's a concept I'm missing here. Can anyone help?
Regards
David Stevens
--
David K Stevens, P.E., Ph.D.
Professor and Head, Environmental Engineering
Civil and Environmental Engineering
Utah Water Research Laboratory
8200 Old Main Hill
Logan, UT 84322-8200
435 797 3229 - voice
435 797 1363 - fax
david.stev...@usu.edu
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Gregory (Greg) L. Snow Ph.D.
538...@gmail.com
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
David K Stevens, P.E., Ph.D.
Professor and Head, Environmental Engineering
Civil and Environmental Engineering
Utah Water Research Laboratory
8200 Old Main Hill
Logan, UT 84322-8200
435 797 3229 - voice
435 797 1363 - fax
david.stev...@usu.edu
__
R-help@r-project.org mailing list
https://stat.ethz.c
Re: [R] saving a 'get' object in R
Seems to me you are creating your own troubles in your "requirements". If the
analysis is the same from case to case, it makes more sense to use a single set
of object names within the analysis and change only the names of the input and
output files.
---
Jeff NewmillerThe . . Go Live...
DCN:Basics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer (Solar/BatteriesO.O#. #.O#. with
/Software/Embedded Controllers) .OO#. .OO#. rocks...1k
---
Sent from my phone. Please excuse my brevity.
On June 25, 2014 7:53:33 AM PDT, David Stevens wrote:
>Greg - I appreciate your taking the time to explain. This is very
>helpful. My case was a bit unusual as I'm helping a colleague with code
>
>to use on a regular but individual basis. I want them to name a data
>set
>once at the top of the script and have that name propagate through
>several objects that end up being saved at the end with file names tied
>
>to the data set name. Then tomorrow, they'll do the same thing on a
>different data set so each session would be pretty simple, with only
>2-3
>named objects at the end, though the data sets are big and cumbersome.
>
>I'll dig into this more and apologize to those who thought this problem
>
>was too trivial for the r-help forum.
>
>David
>
>On 6/24/2014 4:00 PM, Greg Snow wrote:
>> The main reason to avoid assign/get is that there are better ways.
>> You can use a list or environment more directly without using
>> assign/get. Also the main reason to use assign/get is to work with
>> global variables which are frowned on in general programming (and
>> prone to hard to find bugs).
>>
>> Consider the following code snippet:
>>
>> x <- 1:10
>> functionThatUsesAssign(x)
>>
>> what is the value of x after running this code? If you use global
>> variables then the answer is "I don't know", and if x was something
>> that took several hours to compute and I just accidentally overwrote
>> it, then I am probably not happy.
>>
>> Using assign often leads to the temptation to try code like:
>>
>> assign( "x[5]", 25 )
>>
>> which does something (no warnings, no errors) but generally not what
>> was being attempted.
>>
>> Another common (mis)use of assign/get is to create a sequence of
>> variables like "data01", "data02", "data03", ... and then do the
>same
>> thing for each of the data objects. This is much better done by
>> putting all the objects into a single list, then you can easily
>> iterate over the list using lapply/sapply or still access the
>> individual pieces. Then when it comes time to save these objects,
>you
>> only need to save the list, not all the individual objects, same for
>> deleting, copying, moving, etc. And you don't have a bunch of
>> different objects cluttering your workspace, just a single list.
>>
>> On Tue, Jun 24, 2014 at 2:57 PM, David Stevens
> wrote:
>>> Thanks to all for the replies. I tried all three and they work
>great. I was
>>> misinterpreting the list = parameter in save(...) and I get your
>point about
>>> overwriting existing objects. I've heard about not using assign/get
>before.
>>> Can anyone point me to why and what alternatives there are?
>>>
>>> Regards
>>>
>>> David
>>>
>>>
>>> On 6/24/2014 2:50 PM, Henrik Bengtsson wrote:
I recommend to use saveRDS()/readRDS() instead. More convenient
>and
avoids the risk that load() has of overwriting existing variables
>with
the same name.
/Henrik
On Tue, Jun 24, 2014 at 1:45 PM, Greg Snow <538...@gmail.com>
>wrote:
> I think that you are looking for the `list` argument in `save`.
>
> save( list=foo, file=paste0(foo, '.Rdata') )
>
> In general it is best to avoid using the assign function (and get
>when
> possible). Usually there are better alternatives.
>
> On Tue, Jun 24, 2014 at 2:35 PM, David Stevens
>
> wrote:
>> R community,
>>
>> Apologies if this has been answered. The concept I'm looking for
>is to
>> save() an object retrieved using get() for an object
>> that resulted from using assign. Something like
>>
>> save(get(foo),file=paste(foo,'rData',sep=''))
>>
>> where assign(foo,obj) creates an object named foo with the
>contents of
>> obj
>> assigned. For example, if
>>
>> x <- data.frame(v1=c(1,2,3,4),v2=c('1','2','3','4'))
>> foo = 'my.x'
>> assign(foo,x)
>> # (... then modify foo as needed)
>> save(get(foo),file=paste(foo,'.rData',sep=''))
>>
>> # though this generates " in save(get(foo), file = paste(foo,
>".rData",
>> sep
>> = "")) :
>> object ‘get(foo)’ not found", whereas
>>
>> get(foo)
>>
>> at the command prompt yields the contents of my.
Re: [R] saving a 'get' object in R
... and wrap it all up into a single function call that could even have the user interactively supply the input data file names. Cheers, Bert Bert Gunter Genentech Nonclinical Biostatistics (650) 467-7374 "Data is not information. Information is not knowledge. And knowledge is certainly not wisdom." Clifford Stoll On Wed, Jun 25, 2014 at 8:23 AM, Jeff Newmiller wrote: > Seems to me you are creating your own troubles in your "requirements". If the > analysis is the same from case to case, it makes more sense to use a single > set of object names within the analysis and change only the names of the > input and output files. > --- > Jeff NewmillerThe . . Go Live... > DCN:Basics: ##.#. ##.#. Live Go... > Live: OO#.. Dead: OO#.. Playing > Research Engineer (Solar/BatteriesO.O#. #.O#. with > /Software/Embedded Controllers) .OO#. .OO#. rocks...1k > --- > Sent from my phone. Please excuse my brevity. > > On June 25, 2014 7:53:33 AM PDT, David Stevens wrote: >>Greg - I appreciate your taking the time to explain. This is very >>helpful. My case was a bit unusual as I'm helping a colleague with code >> >>to use on a regular but individual basis. I want them to name a data >>set >>once at the top of the script and have that name propagate through >>several objects that end up being saved at the end with file names tied >> >>to the data set name. Then tomorrow, they'll do the same thing on a >>different data set so each session would be pretty simple, with only >>2-3 >>named objects at the end, though the data sets are big and cumbersome. >> >>I'll dig into this more and apologize to those who thought this problem >> >>was too trivial for the r-help forum. >> >>David >> >>On 6/24/2014 4:00 PM, Greg Snow wrote: >>> The main reason to avoid assign/get is that there are better ways. >>> You can use a list or environment more directly without using >>> assign/get. Also the main reason to use assign/get is to work with >>> global variables which are frowned on in general programming (and >>> prone to hard to find bugs). >>> >>> Consider the following code snippet: >>> >>> x <- 1:10 >>> functionThatUsesAssign(x) >>> >>> what is the value of x after running this code? If you use global >>> variables then the answer is "I don't know", and if x was something >>> that took several hours to compute and I just accidentally overwrote >>> it, then I am probably not happy. >>> >>> Using assign often leads to the temptation to try code like: >>> >>> assign( "x[5]", 25 ) >>> >>> which does something (no warnings, no errors) but generally not what >>> was being attempted. >>> >>> Another common (mis)use of assign/get is to create a sequence of >>> variables like "data01", "data02", "data03", ... and then do the >>same >>> thing for each of the data objects. This is much better done by >>> putting all the objects into a single list, then you can easily >>> iterate over the list using lapply/sapply or still access the >>> individual pieces. Then when it comes time to save these objects, >>you >>> only need to save the list, not all the individual objects, same for >>> deleting, copying, moving, etc. And you don't have a bunch of >>> different objects cluttering your workspace, just a single list. >>> >>> On Tue, Jun 24, 2014 at 2:57 PM, David Stevens >> wrote: Thanks to all for the replies. I tried all three and they work >>great. I was misinterpreting the list = parameter in save(...) and I get your >>point about overwriting existing objects. I've heard about not using assign/get >>before. Can anyone point me to why and what alternatives there are? Regards David On 6/24/2014 2:50 PM, Henrik Bengtsson wrote: > I recommend to use saveRDS()/readRDS() instead. More convenient >>and > avoids the risk that load() has of overwriting existing variables >>with > the same name. > > /Henrik > > On Tue, Jun 24, 2014 at 1:45 PM, Greg Snow <538...@gmail.com> >>wrote: >> I think that you are looking for the `list` argument in `save`. >> >> save( list=foo, file=paste0(foo, '.Rdata') ) >> >> In general it is best to avoid using the assign function (and get >>when >> possible). Usually there are better alternatives. >> >> On Tue, Jun 24, 2014 at 2:35 PM, David Stevens >> >> wrote: >>> R community, >>> >>> Apologies if this has been answered. The concept I'm looking for >>is to >>> save() an object retrieved using get() for an object >>> that resulted from using assign. Something like >>> >>> save(get(foo),file=paste(foo,'rData',sep='')) >>> >>> where assign(foo,obj) creates an object named foo with the >>
Re: [R] matlab serial date to r
See ?as.Date. I am guessing these are days and fractional days. Try x <- 7.35600813091e5 as.POSIXct((x - 719529)*86400, origin = "1970-01-01") On 25/06/2014 14:56, Christoph Schlächter wrote: Hi, I have a matlab variable as serial date (class double) in the form 'dd-mmm- HH:MM:SS'. format long disp( tx(40:60,1) ) 1.0e+05 * 7.35600813091 7.35600956856 7.35601305921 7.35601654985 7.35602004049 7.35602353113 7.35602702178 7.35603397179 7.35604092182 7.35604787183 7.35605482185 7.35606177187 7.35606940080 7.35607702975 7.35608465869 7.35609228763 7.35609991657 7.35610754551 7.35611517445 7.35612280339 7.35613085329 It should be the same as datestr( tx(40:60,1), 0) 01-Jan-2014 00:00:07 01-Jan-2014 00:00:08 01-Jan-2014 00:00:11 01-Jan-2014 00:00:14 01-Jan-2014 00:00:17 01-Jan-2014 00:00:20 01-Jan-2014 00:00:23 01-Jan-2014 00:00:29 01-Jan-2014 00:00:35 01-Jan-2014 00:00:41 01-Jan-2014 00:00:47 01-Jan-2014 00:00:53 01-Jan-2014 00:00:59 01-Jan-2014 00:01:06 01-Jan-2014 00:01:13 01-Jan-2014 00:01:19 01-Jan-2014 00:01:26 01-Jan-2014 00:01:32 01-Jan-2014 00:01:39 01-Jan-2014 00:01:46 01-Jan-2014 00:01:53 I can easily convert it with Matlab but then I will obtain a character format which is useless. I can also make use of cellstr(datestr(tx(:,1), 0)) but then I can't save it in ASCII file. The origin of the Matlab format is supposed to be "-00-00". This is the only origin which results in "2014-01-01" which is my actual start date. Can somebody please tell me how I can simply convert serial datetime to datetime in R. Thanks in advance. All the best, Christoph [[alternative HTML version deleted]] __ 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. -- Brian D. Ripley, rip...@stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ 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] matlab serial date to r
I think the character format for this data is the most versatile and clear option. You do have to prevent the R input function (read.csv? read.table?) from converting it to factor when you read it in, but then you can use as.POSIXct with a format argument (see ?strptime) to obtain useful timestamp values in R. You also need to be clear about time zones, but that is true regardless of the software you use. --- Jeff NewmillerThe . . Go Live... DCN:Basics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/BatteriesO.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k --- Sent from my phone. Please excuse my brevity. On June 25, 2014 6:56:38 AM PDT, "Christoph Schlächter" wrote: >Hi, > >I have a matlab variable as serial date (class double) in the form >'dd-mmm- HH:MM:SS'. > >format long >disp( tx(40:60,1) ) > >1.0e+05 * > > 7.35600813091 > 7.35600956856 > 7.35601305921 > 7.35601654985 > 7.35602004049 > 7.35602353113 > 7.35602702178 > 7.35603397179 > 7.35604092182 > 7.35604787183 > 7.35605482185 > 7.35606177187 > 7.35606940080 > 7.35607702975 > 7.35608465869 > 7.35609228763 > 7.35609991657 > 7.35610754551 > 7.35611517445 > 7.35612280339 > 7.35613085329 > >It should be the same as > >datestr( tx(40:60,1), 0) > >01-Jan-2014 00:00:07 >01-Jan-2014 00:00:08 >01-Jan-2014 00:00:11 >01-Jan-2014 00:00:14 >01-Jan-2014 00:00:17 >01-Jan-2014 00:00:20 >01-Jan-2014 00:00:23 >01-Jan-2014 00:00:29 >01-Jan-2014 00:00:35 >01-Jan-2014 00:00:41 >01-Jan-2014 00:00:47 >01-Jan-2014 00:00:53 >01-Jan-2014 00:00:59 >01-Jan-2014 00:01:06 >01-Jan-2014 00:01:13 >01-Jan-2014 00:01:19 >01-Jan-2014 00:01:26 >01-Jan-2014 00:01:32 >01-Jan-2014 00:01:39 >01-Jan-2014 00:01:46 >01-Jan-2014 00:01:53 > >I can easily convert it with Matlab but then I will obtain a character >format which is useless. I can also make use of >cellstr(datestr(tx(:,1), >0)) but then I can't save it in ASCII file. > >The origin of the Matlab format is supposed to be "-00-00". This is >the >only origin which results in "2014-01-01" which is my actual start >date. > >Can somebody please tell me how I can simply convert serial datetime to >datetime in R. > >Thanks in advance. > >All the best, > >Christoph > > [[alternative HTML version deleted]] > >__ >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] Generating Patient Data
Hi, Check if this works: set.seed(495) dat <- data.frame(ID=sample(1:10,20,replace=TRUE), Disease=sample(LETTERS[1:6], 20, replace=TRUE) ) subset(melt(table(dat)[rowSums(!!table(dat))>1,]), !!value,select=1:2) ID Disease 1 2 A 3 4 A 4 6 A 6 10 A 8 3 B 15 4 C 16 6 C 20 3 D 22 6 D 24 10 D 26 3 E 27 4 E 29 7 E 31 2 F 33 4 F 35 7 F A.K. On Wednesday, June 25, 2014 1:17 AM, Abhinaba Roy wrote: Dear R helpers, I want to generate data for say 1000 patients (i.e., 1000 unique IDs) having suffered from various diseases in the past (say diseases A,B,C,D,E,F). The only condition imposed is that each patient should've suffered from *atleast* two diseases. So my data frame will have two columns 'ID' and 'Disease'. I want to do a basket analysis with this data, where ID will be the identifier and we will establish rules based on the 'Disease' column. How can I generate this type of data in R? -- Regards Abhinaba Roy [[alternative HTML version deleted]] __ 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] matlab serial date to r
Hi, May be this helps: dat <- read.table(text="7.35600813091 7.35600956856 7.35601305921 7.35601654985 7.35602004049 7.35602353113 7.35602702178 7.35603397179 7.35604092182 7.35604787183 7.35605482185 7.35606177187 7.35606940080 7.35607702975 7.35608465869 7.35609228763 7.35609991657 7.35610754551 7.35611517445 7.35612280339 7.35613085329",sep="",header=F, colClasses="character") library(chron) t1 <- chron(1.0e+05 *as.numeric(dat[,1])) -719529 format(as.POSIXct(paste(as.Date(dates(t1)), times(t1)%%1)),"%m-%b-%Y %H:%M:%S") A.K. On Wednesday, June 25, 2014 10:00 AM, Christoph Schlächter wrote: Hi, I have a matlab variable as serial date (class double) in the form 'dd-mmm- HH:MM:SS'. format long disp( tx(40:60,1) ) 1.0e+05 * 7.35600813091 7.35600956856 7.35601305921 7.35601654985 7.35602004049 7.35602353113 7.35602702178 7.35603397179 7.35604092182 7.35604787183 7.35605482185 7.35606177187 7.35606940080 7.35607702975 7.35608465869 7.35609228763 7.35609991657 7.35610754551 7.35611517445 7.35612280339 7.35613085329 It should be the same as datestr( tx(40:60,1), 0) 01-Jan-2014 00:00:07 01-Jan-2014 00:00:08 01-Jan-2014 00:00:11 01-Jan-2014 00:00:14 01-Jan-2014 00:00:17 01-Jan-2014 00:00:20 01-Jan-2014 00:00:23 01-Jan-2014 00:00:29 01-Jan-2014 00:00:35 01-Jan-2014 00:00:41 01-Jan-2014 00:00:47 01-Jan-2014 00:00:53 01-Jan-2014 00:00:59 01-Jan-2014 00:01:06 01-Jan-2014 00:01:13 01-Jan-2014 00:01:19 01-Jan-2014 00:01:26 01-Jan-2014 00:01:32 01-Jan-2014 00:01:39 01-Jan-2014 00:01:46 01-Jan-2014 00:01:53 I can easily convert it with Matlab but then I will obtain a character format which is useless. I can also make use of cellstr(datestr(tx(:,1), 0)) but then I can't save it in ASCII file. The origin of the Matlab format is supposed to be "-00-00". This is the only origin which results in "2014-01-01" which is my actual start date. Can somebody please tell me how I can simply convert serial datetime to datetime in R. Thanks in advance. All the best, Christoph [[alternative HTML version deleted]] __ 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.
[R] Duncan test: 2-way ANOVA without repetition, but with multiple subjects
Hello
I have a question on how to perform a Duncan test after I set a model like this
(see below).
The data consist of a dependent variable (PHt) and two dependent variables
(REGION (3 levels) AND MANAGEMENT (3 levels)).
MANAGEMENT
N PS
REGION A 196 196 196
V 196 196 196
H 196 196 196
196 is the number of trees on which PHt was estimated, but we cannot consider
them as repetitions, they are subjects. Therefore, only one repetition per
REGION*MANAGEMENT factor.
I set a two way ANOVA without repetition. Although within
model <-
aov(PHt~REGION*MANAGEMENT+Error(subject_f/(REGION*MANAGEMENT)),data=obsHETf)
Then I try to run a Duncan test for REGION and also for MANAGEMENT. I get this
error message.
Error in as.data.frame.default(x[[i]], optional = TRUE, stringsAsFactors =
stringsAsFactors) :
cannot coerce class "c("aovlist", "listof")" to a data.frame
Is there anyone who could give me a clue on what it is wrong? Maybe it isnot
correct to call for a Duncan.test() for such type of model?
If I fit only the mean (mean of the 196 observations) within each
REGION*MANAGEMENT factor, then Duncan.test() works as expected. The model then
looks as simple as this.
model <- aov(PHt~REGION+MANAGEMENT,data=obsHET)
Thanks in advance.
R.
[[alternative HTML version deleted]]
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Re: [R] Generating Patient Data
Also, you can do: library(dplyr) dat%>%group_by(ID)%>%filter(length(unique(Disease))>1)%>%arrange(Disease,ID) A.K. On Wednesday, June 25, 2014 3:45 AM, arun wrote: Forgot about: library(reshape2) On , arun wrote: Hi, Check if this works: set.seed(495) dat <- data.frame(ID=sample(1:10,20,replace=TRUE), Disease=sample(LETTERS[1:6], 20, replace=TRUE) ) subset(melt(table(dat)[rowSums(!!table(dat))>1,]), !!value,select=1:2) ID Disease 1 2 A 3 4 A 4 6 A 6 10 A 8 3 B 15 4 C 16 6 C 20 3 D 22 6 D 24 10 D 26 3 E 27 4 E 29 7 E 31 2 F 33 4 F 35 7 F A.K. On Wednesday, June 25, 2014 1:17 AM, Abhinaba Roy wrote: Dear R helpers, I want to generate data for say 1000 patients (i.e., 1000 unique IDs) having suffered from various diseases in the past (say diseases A,B,C,D,E,F). The only condition imposed is that each patient should've suffered from *atleast* two diseases. So my data frame will have two columns 'ID' and 'Disease'. I want to do a basket analysis with this data, where ID will be the identifier and we will establish rules based on the 'Disease' column. How can I generate this type of data in R? -- Regards Abhinaba Roy [[alternative HTML version deleted]] __ 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] partykit ctree: minbucket and case weights
Dear Amber, your data contains missing values and you don't use surrogate splits to deal with them. So, the observations are passed down the tree randomly (there is no "majority" argument to "ctree_control"!) and thus it might happen that too small terminal nodes are created. Simply use surrogate split and the tree will be deterministic with correct-sized terminal nodes (maxsurrogate = 3, for example). Best, Torsten On Mon, 9 Jun 2014, Amber Dawn Nolder wrote: I have attached the data set (cavl) and R code used when I got the results I posted about. I included the code I used at the top of the document. Below that is the version of R used and some of the results I obtained. Many thanks! Amber On Wed, 4 Jun 2014 09:12:15 +0200 (CEST) Torsten Hothorn wrote: On Tue, 3 Jun 2014, Amber Dawn Nolder wrote: I apologize for my lack of knowledge with R. I usually load my data as a csv file. May I send that to you? I was not sure if I could do so on the list. yes, and the R code you used. Thanks, Torsten Thank you? On Fri, 30 May 2014 09:37:23 +0200 (CEST) Torsten Hothorn wrote: Amber, this looks like an error -- could you pls send me a reproducible example so that I can track the problem down? Best, Torsten Prof. Dr. Torsten Hothorn = \\ Universitaet Zuerich \\ Institut fuer Epidemiologie, Biostatistik und \\ Praevention, Abteilung Biostatistik // Hirschengraben 84// CH-8001 Zuerich // Schweiz// == Telephon: +41 44 634 48 17 Fax: +41 44 634 43 86 Web: http://tiny.uzh.ch/6p On Wed, 28 May 2014, Achim Zeileis wrote: Falls Du es nicht eh gesehen hast... lg, Z -- Forwarded message -- Date: Wed, 28 May 2014 17:16:12 -0400 From: Amber Dawn Nolder To: r-help@r-project.org Subject: [R] partykit ctree: minbucket and case weights Hello, I am an R novice, and I am using the "partykit" package to create regression trees. I used the following to generate the trees: ctree(y~x1+x2+x3+x4,data=my_data,control=ctree_control(testtype = "Bonferroni", mincriterion = 0.90, minsplit = 12, minbucket = 4, majority = TRUE) I thought that "minbucket" set the minimum value for the sum of weights in each terminal node, and that each case weight is 1, unless otherwise specified. In which case, the sum of case weights in a node should equal the number of cases (n) in that node. However, I sometimes obtain a tree with a terminal node that contains fewer than 4 cases. My data set has a total of 36 cases. The dependent and all independent variables are continuous data. Variables x1 and x2 contain missing (NA) values. Could someone please explain why I am getting these results? Am I mistaken about the value of case weights or about the use of minbucket to restrict the size of a terminal node? This is an example of the output: Model formula: y ~ x1 + x2 + x3 + x4 Fitted party: [1] root | [2] x4 <= 30: 0.927 (n = 17, err = 1.1) | [3] x4 > 30 | | [4] x2 <= 43: 0.472 (n = 8, err = 0.4) | | [5] x2 > 43 | | | [6] x3 <= 0.4: 0.282 (n = 3, err = 0.0) | | | [7] x3 > 0.4: 0.020 (n = 8, err = 0.0) Number of inner nodes:3 Number of terminal nodes: 4 Many thanks! Amber Nolder Graduate Student Indiana University of Pennsylvania __ 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] Duncan test: 2-way ANOVA without repetition, but with multiple subjects
You didn't say which package Duncan.test is in. glht has the ability.
glht, and probably any other package's test, cannot work with aovlist objects.
They require aov objects. That means you must rewrite your Error() statement
into the main model. Please see the entire maiz example in ?mmc to see how
to rewrite your model.
## install.packages("HH") ## if necessary
require(HH)
?mmc
On Wed, Jun 25, 2014 at 6:04 AM, Rosario Garcia Gil
wrote:
> Hello
>
> I have a question on how to perform a Duncan test after I set a model like
> this (see below).
>
> The data consist of a dependent variable (PHt) and two dependent variables
> (REGION (3 levels) AND MANAGEMENT (3 levels)).
>
>
>MANAGEMENT
> N PS
> REGION A 196 196 196
> V 196 196 196
> H 196 196 196
>
> 196 is the number of trees on which PHt was estimated, but we cannot consider
> them as repetitions, they are subjects. Therefore, only one repetition per
> REGION*MANAGEMENT factor.
>
> I set a two way ANOVA without repetition. Although within
>
> model <-
> aov(PHt~REGION*MANAGEMENT+Error(subject_f/(REGION*MANAGEMENT)),data=obsHETf)
>
> Then I try to run a Duncan test for REGION and also for MANAGEMENT. I get
> this error message.
>
> Error in as.data.frame.default(x[[i]], optional = TRUE, stringsAsFactors =
> stringsAsFactors) :
> cannot coerce class "c("aovlist", "listof")" to a data.frame
>
> Is there anyone who could give me a clue on what it is wrong? Maybe it isnot
> correct to call for a Duncan.test() for such type of model?
>
> If I fit only the mean (mean of the 196 observations) within each
> REGION*MANAGEMENT factor, then Duncan.test() works as expected. The model
> then looks as simple as this.
>
> model <- aov(PHt~REGION+MANAGEMENT,data=obsHET)
>
> Thanks in advance.
> R.
>
>
> [[alternative HTML version deleted]]
>
> __
> 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.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Simple permutation question
So my company has hired a few young McKinsey guys from overseas for a
couple of weeks to help us with a production line optimization. They
probably charge what I make in a year, but that's OK because I just
never have the time to really dive into one particular time, and I have
to hand it to the consultants that they came up with one or two really
clever ideas to model the production line. Of course it's up to me to
feed them the real data which they then churn through their Excel
models that they cook up during the nights in their hotel rooms, and
which I then implement back into my experimental system using live data.
Anyway, whenever they need something or come up with something I skip
out of the room, hack it into R, export the CSV and come back in about
half the time it takes Excel to even read in the data, let alone
process it. Of course that gor them curious, and I showed off a couple
of scripts that condense their abysmal Excel convolutions in a few
lean and mean lines of R code.
Anyway, I'm in my office with this really attractive, clever young
McKinsey girl (I'm in my mid-forties, married with kids and all, but I
still enjoyed impressing a woman with computer stuff, of all things!),
and one of her models involves a simple permutation of five letters --
"A" through "E".
And that's when I find out that R doesn't have a permutation function.
How is that possible? R has EVERYTHING, but not that? I'm
flabbergasted. Stumped. And now it's up to me to spend the evening at
home coding that model, and the only thing I really need is that
permutation.
So this is my first attempt:
perm.broken <- function(x) {
if (length(x) == 1) return(x)
sapply(1:length(x), function(i) {
cbind(x[i], perm(x[-i]))
})
}
But it doesn't work:
> perm.broken(c("A", "B", "C"))
[,1] [,2] [,3]
[1,] "A" "B" "C"
[2,] "A" "B" "C"
[3,] "B" "A" "A"
[4,] "C" "C" "B"
[5,] "C" "C" "B"
[6,] "B" "A" "A"
>
And I can't figure out for the life of me why. It should work because I
go through the elements of x in order, use that in the leftmost column,
and slap the permutation of the remaining elements to the right. What
strikes me as particularly odd is that there doesn't even seem to be a
systematic sequence of letters in any of the columns. OK, since I
really need that function I wrote this piece of crap:
perm.stupid <- function(x) {
b <- as.matrix(expand.grid(rep(list(x), length(x
b[!sapply(1:nrow(b), function(r) any(duplicated(b[r,]))),]
}
It works, but words cannot describe its ugliness. And it gets really
slow really fast with growing x.
So, anyway. My two questions are:
1. Does R really, really, seriously lack a permutation function?
2. OK, stop kidding me. So what's it called?
3. Why doesn't my recursive function work, and what would a
working version look like?
Thanks,
robert
__
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Simple permutation question
It is called sample(,replace=F), where the default argument is sampling
without replacement.
Try
x <- c("A","B","C","D","E")
sample(x)
Brian
Brian S. Cade, PhD
U. S. Geological Survey
Fort Collins Science Center
2150 Centre Ave., Bldg. C
Fort Collins, CO 80526-8818
email: ca...@usgs.gov
tel: 970 226-9326
On Wed, Jun 25, 2014 at 2:22 PM, Robert Latest wrote:
> So my company has hired a few young McKinsey guys from overseas for a
> couple of weeks to help us with a production line optimization. They
> probably charge what I make in a year, but that's OK because I just
> never have the time to really dive into one particular time, and I have
> to hand it to the consultants that they came up with one or two really
> clever ideas to model the production line. Of course it's up to me to
> feed them the real data which they then churn through their Excel
> models that they cook up during the nights in their hotel rooms, and
> which I then implement back into my experimental system using live data.
>
> Anyway, whenever they need something or come up with something I skip
> out of the room, hack it into R, export the CSV and come back in about
> half the time it takes Excel to even read in the data, let alone
> process it. Of course that gor them curious, and I showed off a couple
> of scripts that condense their abysmal Excel convolutions in a few
> lean and mean lines of R code.
>
> Anyway, I'm in my office with this really attractive, clever young
> McKinsey girl (I'm in my mid-forties, married with kids and all, but I
> still enjoyed impressing a woman with computer stuff, of all things!),
> and one of her models involves a simple permutation of five letters --
> "A" through "E".
>
> And that's when I find out that R doesn't have a permutation function.
> How is that possible? R has EVERYTHING, but not that? I'm
> flabbergasted. Stumped. And now it's up to me to spend the evening at
> home coding that model, and the only thing I really need is that
> permutation.
>
> So this is my first attempt:
>
> perm.broken <- function(x) {
> if (length(x) == 1) return(x)
> sapply(1:length(x), function(i) {
> cbind(x[i], perm(x[-i]))
> })
> }
>
> But it doesn't work:
> > perm.broken(c("A", "B", "C"))
> [,1] [,2] [,3]
> [1,] "A" "B" "C"
> [2,] "A" "B" "C"
> [3,] "B" "A" "A"
> [4,] "C" "C" "B"
> [5,] "C" "C" "B"
> [6,] "B" "A" "A"
> >
>
> And I can't figure out for the life of me why. It should work because I
> go through the elements of x in order, use that in the leftmost column,
> and slap the permutation of the remaining elements to the right. What
> strikes me as particularly odd is that there doesn't even seem to be a
> systematic sequence of letters in any of the columns. OK, since I
> really need that function I wrote this piece of crap:
>
> perm.stupid <- function(x) {
> b <- as.matrix(expand.grid(rep(list(x), length(x
> b[!sapply(1:nrow(b), function(r) any(duplicated(b[r,]))),]
> }
>
> It works, but words cannot describe its ugliness. And it gets really
> slow really fast with growing x.
>
> So, anyway. My two questions are:
> 1. Does R really, really, seriously lack a permutation function?
> 2. OK, stop kidding me. So what's it called?
> 3. Why doesn't my recursive function work, and what would a
>working version look like?
>
> Thanks,
> robert
>
> __
> 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.
>
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Generating Patient Data
On Jun 24, 2014, at 11:18 PM, Abhinaba Roy wrote:
> Hi David,
>
> I was thinking something like this:
>
> ID Disease
> 1 A
> 2 B
> 3 A
> 1C
> 2D
> 5A
> 4B
> 3D
> 2A
> ....
>
> How can this be done?
do.call(rbind, lapply( 1:20, function(pt) {
data.frame( patient=pt,
disease= sample( c('A','B','C','D','E','F'),
pmin(2+rpois(1, 2), 6)) )}) )
--
David.
>
>
> On Wed, Jun 25, 2014 at 11:34 AM, David Winsemius
> wrote:
>
> On Jun 24, 2014, at 10:14 PM, Abhinaba Roy wrote:
>
> > Dear R helpers,
> >
> > I want to generate data for say 1000 patients (i.e., 1000 unique IDs)
> > having suffered from various diseases in the past (say diseases
> > A,B,C,D,E,F). The only condition imposed is that each patient should've
> > suffered from *atleast* two diseases. So my data frame will have two
> > columns 'ID' and 'Disease'.
> >
> > I want to do a basket analysis with this data, where ID will be the
> > identifier and we will establish rules based on the 'Disease' column.
> >
> > How can I generate this type of data in R?
> >
>
> Perhaps something along these lines for 20 cases:
>
> > data.frame(patient=1:20, disease = sapply(pmin(2+rpois(20, 2), 6),
> > function(n) paste0( sample( c('A','B','C','D','E','F'), n), collapse="+" ) )
> + )
>patient disease
> 11 F+D
> 22 F+A+D+E
> 33 F+D+C+E
> 44 B+D+C+A
> 55 D+A+F+C
> 66 E+A+D
> 77 E+F+B+C+A+D
> 88 A+B+C+D+E
> 99 B+E+C+F
> 10 10 C+A
> 11 11 B+A+D+E+C+F
> 12 12 B+C
> 13 13 A+D+B+E
> 14 14 D+C+E+F+B+A
> 15 15 C+F+D+E+A
> 16 16 A+C+B
> 17 17 C+D+B+E
> 18 18 A+B
> 19 19 C+B+D+E+F
> 20 20 D+C+F
>
> > --
> > Regards
> > Abhinaba Roy
> >
> > [[alternative HTML version deleted]]
>
> You should read the Posting Guide and learn to post in HTML.
> >
> > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
>
>
> --
> David Winsemius
> Alameda, CA, USA
>
>
>
>
> --
> Regards
> Abhinaba Roy
> Statistician
> Radix Analytics Pvt. Ltd
> Ahmedabad
>
David Winsemius
Alameda, CA, USA
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Simple permutation question
I think Robert wants deterministic permutations. In the e1071
package -- load with library(e1071) -- there is a function
permutations():
Description:
Returns a matrix containing all permutations of the integers
'1:n' (one permutation per row).
Usage:
permutations(n)
Arguments:
n: Number of element to permute.
so, starting with
x <- c("A","B","C","D","E")
library(e1071)
P <- permutations(length(x))
then, for say the 27th of these 120 permutations of x,
x[P[27,]]
will return it.
Ted.
On 25-Jun-2014 20:38:45 Cade, Brian wrote:
> It is called sample(,replace=F), where the default argument is sampling
> without replacement.
> Try
> x <- c("A","B","C","D","E")
> sample(x)
>
> Brian
>
> Brian S. Cade, PhD
>
> U. S. Geological Survey
> Fort Collins Science Center
> 2150 Centre Ave., Bldg. C
> Fort Collins, CO 80526-8818
>
> email: ca...@usgs.gov
> tel: 970 226-9326
>
>
>
> On Wed, Jun 25, 2014 at 2:22 PM, Robert Latest wrote:
>
>> So my company has hired a few young McKinsey guys from overseas for a
>> couple of weeks to help us with a production line optimization. They
>> probably charge what I make in a year, but that's OK because I just
>> never have the time to really dive into one particular time, and I have
>> to hand it to the consultants that they came up with one or two really
>> clever ideas to model the production line. Of course it's up to me to
>> feed them the real data which they then churn through their Excel
>> models that they cook up during the nights in their hotel rooms, and
>> which I then implement back into my experimental system using live data.
>>
>> Anyway, whenever they need something or come up with something I skip
>> out of the room, hack it into R, export the CSV and come back in about
>> half the time it takes Excel to even read in the data, let alone
>> process it. Of course that gor them curious, and I showed off a couple
>> of scripts that condense their abysmal Excel convolutions in a few
>> lean and mean lines of R code.
>>
>> Anyway, I'm in my office with this really attractive, clever young
>> McKinsey girl (I'm in my mid-forties, married with kids and all, but I
>> still enjoyed impressing a woman with computer stuff, of all things!),
>> and one of her models involves a simple permutation of five letters --
>> "A" through "E".
>>
>> And that's when I find out that R doesn't have a permutation function.
>> How is that possible? R has EVERYTHING, but not that? I'm
>> flabbergasted. Stumped. And now it's up to me to spend the evening at
>> home coding that model, and the only thing I really need is that
>> permutation.
>>
>> So this is my first attempt:
>>
>> perm.broken <- function(x) {
>> if (length(x) == 1) return(x)
>> sapply(1:length(x), function(i) {
>> cbind(x[i], perm(x[-i]))
>> })
>> }
>>
>> But it doesn't work:
>> > perm.broken(c("A", "B", "C"))
>> [,1] [,2] [,3]
>> [1,] "A" "B" "C"
>> [2,] "A" "B" "C"
>> [3,] "B" "A" "A"
>> [4,] "C" "C" "B"
>> [5,] "C" "C" "B"
>> [6,] "B" "A" "A"
>> >
>>
>> And I can't figure out for the life of me why. It should work because I
>> go through the elements of x in order, use that in the leftmost column,
>> and slap the permutation of the remaining elements to the right. What
>> strikes me as particularly odd is that there doesn't even seem to be a
>> systematic sequence of letters in any of the columns. OK, since I
>> really need that function I wrote this piece of crap:
>>
>> perm.stupid <- function(x) {
>> b <- as.matrix(expand.grid(rep(list(x), length(x
>> b[!sapply(1:nrow(b), function(r) any(duplicated(b[r,]))),]
>> }
>>
>> It works, but words cannot describe its ugliness. And it gets really
>> slow really fast with growing x.
>>
>> So, anyway. My two questions are:
>> 1. Does R really, really, seriously lack a permutation function?
>> 2. OK, stop kidding me. So what's it called?
>> 3. Why doesn't my recursive function work, and what would a
>>working version look like?
>>
>> Thanks,
>> robert
>>
>> __
>> 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.
>>
>
> [[alternative HTML version deleted]]
>
> __
> 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.
-
E-Mail: (Ted Harding)
Date: 25-Jun-2014 Time: 21:55:42
This message was sent by XFMail
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The TERR team will be at useR 2014, June 30th to July 3rd, to share the latest enhancements and news around TIBCO Enterprise Runtime for R. In addition to providing demos at the TIBCO TERR table in the exhibition area, some of the senior members of the TERR team will be presenting at the conference: Louis Bajuk-Yorgan, Sr. Dir., Product Management:“Deploying R into Business Intelligence and Real-time Applications”. Business track, Session 5, Wednesday 16:00 Stephen Kaluzny, Lead Statistician for TERR: “Software Testing and the R Language” Business track, Session 6, Thursday 10:00 Michael Sannella, Lead Architect for TERR: "The Compatibility Challenge: Examining R and Developing TERR". Posters 1, Tuesday, 17:30 -- Lou Bajuk-Yorgan Sr. Director, Product Management Spotfire, TIBCO Software 206-802-2328 lba...@tibco.com Twitter: @LouBajuk http://spotfire.tibco.com __ 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] Simple permutation question
Assuming you want all of the permutations not just a random permutation (such
as sample() give you):
> require(e1071)
> indx <- permutations(5)
> head(indx)
[,1] [,2] [,3] [,4] [,5]
[1,]12345
[2,]21345
[3,]23145
[4,]13245
[5,]31245
[6,]32145
> tail(indx)
[,1] [,2] [,3] [,4] [,5]
[115,]54123
[116,]54213
[117,]54231
[118,]54132
[119,]54312
[120,]54321
If you want them converted to numbers try
> apply(ind, 1, function(x) paste(LETTERS[x], collapse=" "))
[1] "A B C D E" "B A C D E" "B C A D E" "A C B D E" "C A B D E" "C B A D E"
> tail(perm.ltrs)
[1] "E D A B C" "E D B A C" "E D B C A" "E D A C B" "E D C A B" "E D C B A"
This is not the only permutation function in R, but this one has the advantage
of being symmetrical. The last permutation is the reverse of the first, the
penultimate the reverse of the second, etc.
-
David L Carlson
Department of Anthropology
Texas A&M University
College Station, TX 77840-4352
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Cade, Brian
Sent: Wednesday, June 25, 2014 3:39 PM
To: Robert Latest
Cc: r-help@r-project.org
Subject: Re: [R] Simple permutation question
It is called sample(,replace=F), where the default argument is sampling
without replacement.
Try
x <- c("A","B","C","D","E")
sample(x)
Brian
Brian S. Cade, PhD
U. S. Geological Survey
Fort Collins Science Center
2150 Centre Ave., Bldg. C
Fort Collins, CO 80526-8818
email: ca...@usgs.gov
tel: 970 226-9326
On Wed, Jun 25, 2014 at 2:22 PM, Robert Latest wrote:
> So my company has hired a few young McKinsey guys from overseas for a
> couple of weeks to help us with a production line optimization. They
> probably charge what I make in a year, but that's OK because I just
> never have the time to really dive into one particular time, and I have
> to hand it to the consultants that they came up with one or two really
> clever ideas to model the production line. Of course it's up to me to
> feed them the real data which they then churn through their Excel
> models that they cook up during the nights in their hotel rooms, and
> which I then implement back into my experimental system using live data.
>
> Anyway, whenever they need something or come up with something I skip
> out of the room, hack it into R, export the CSV and come back in about
> half the time it takes Excel to even read in the data, let alone
> process it. Of course that gor them curious, and I showed off a couple
> of scripts that condense their abysmal Excel convolutions in a few
> lean and mean lines of R code.
>
> Anyway, I'm in my office with this really attractive, clever young
> McKinsey girl (I'm in my mid-forties, married with kids and all, but I
> still enjoyed impressing a woman with computer stuff, of all things!),
> and one of her models involves a simple permutation of five letters --
> "A" through "E".
>
> And that's when I find out that R doesn't have a permutation function.
> How is that possible? R has EVERYTHING, but not that? I'm
> flabbergasted. Stumped. And now it's up to me to spend the evening at
> home coding that model, and the only thing I really need is that
> permutation.
>
> So this is my first attempt:
>
> perm.broken <- function(x) {
> if (length(x) == 1) return(x)
> sapply(1:length(x), function(i) {
> cbind(x[i], perm(x[-i]))
> })
> }
>
> But it doesn't work:
> > perm.broken(c("A", "B", "C"))
> [,1] [,2] [,3]
> [1,] "A" "B" "C"
> [2,] "A" "B" "C"
> [3,] "B" "A" "A"
> [4,] "C" "C" "B"
> [5,] "C" "C" "B"
> [6,] "B" "A" "A"
> >
>
> And I can't figure out for the life of me why. It should work because I
> go through the elements of x in order, use that in the leftmost column,
> and slap the permutation of the remaining elements to the right. What
> strikes me as particularly odd is that there doesn't even seem to be a
> systematic sequence of letters in any of the columns. OK, since I
> really need that function I wrote this piece of crap:
>
> perm.stupid <- function(x) {
> b <- as.matrix(expand.grid(rep(list(x), length(x
> b[!sapply(1:nrow(b), function(r) any(duplicated(b[r,]))),]
> }
>
> It works, but words cannot describe its ugliness. And it gets really
> slow really fast with growing x.
>
> So, anyway. My two questions are:
> 1. Does R really, really, seriously lack a permutation function?
> 2. OK, stop kidding me. So what's it called?
> 3. Why doesn't my recursive function work, and what would a
>working version look like?
>
> Thanks,
> robert
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting gui
Re: [R] Simple permutation question
and further...
See ?Recall for how to do recursion in R.
However, it is my understanding that recursion is not that efficient
in R. A chain of function environments must be created, and this does
not scale well. (Comments from real experts welcome here).
Cheers,
Bert
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
"Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom."
Clifford Stoll
On Wed, Jun 25, 2014 at 1:38 PM, Cade, Brian wrote:
> It is called sample(,replace=F), where the default argument is sampling
> without replacement.
> Try
> x <- c("A","B","C","D","E")
> sample(x)
>
> Brian
>
> Brian S. Cade, PhD
>
> U. S. Geological Survey
> Fort Collins Science Center
> 2150 Centre Ave., Bldg. C
> Fort Collins, CO 80526-8818
>
> email: ca...@usgs.gov
> tel: 970 226-9326
>
>
>
> On Wed, Jun 25, 2014 at 2:22 PM, Robert Latest wrote:
>
>> So my company has hired a few young McKinsey guys from overseas for a
>> couple of weeks to help us with a production line optimization. They
>> probably charge what I make in a year, but that's OK because I just
>> never have the time to really dive into one particular time, and I have
>> to hand it to the consultants that they came up with one or two really
>> clever ideas to model the production line. Of course it's up to me to
>> feed them the real data which they then churn through their Excel
>> models that they cook up during the nights in their hotel rooms, and
>> which I then implement back into my experimental system using live data.
>>
>> Anyway, whenever they need something or come up with something I skip
>> out of the room, hack it into R, export the CSV and come back in about
>> half the time it takes Excel to even read in the data, let alone
>> process it. Of course that gor them curious, and I showed off a couple
>> of scripts that condense their abysmal Excel convolutions in a few
>> lean and mean lines of R code.
>>
>> Anyway, I'm in my office with this really attractive, clever young
>> McKinsey girl (I'm in my mid-forties, married with kids and all, but I
>> still enjoyed impressing a woman with computer stuff, of all things!),
>> and one of her models involves a simple permutation of five letters --
>> "A" through "E".
>>
>> And that's when I find out that R doesn't have a permutation function.
>> How is that possible? R has EVERYTHING, but not that? I'm
>> flabbergasted. Stumped. And now it's up to me to spend the evening at
>> home coding that model, and the only thing I really need is that
>> permutation.
>>
>> So this is my first attempt:
>>
>> perm.broken <- function(x) {
>> if (length(x) == 1) return(x)
>> sapply(1:length(x), function(i) {
>> cbind(x[i], perm(x[-i]))
>> })
>> }
>>
>> But it doesn't work:
>> > perm.broken(c("A", "B", "C"))
>> [,1] [,2] [,3]
>> [1,] "A" "B" "C"
>> [2,] "A" "B" "C"
>> [3,] "B" "A" "A"
>> [4,] "C" "C" "B"
>> [5,] "C" "C" "B"
>> [6,] "B" "A" "A"
>> >
>>
>> And I can't figure out for the life of me why. It should work because I
>> go through the elements of x in order, use that in the leftmost column,
>> and slap the permutation of the remaining elements to the right. What
>> strikes me as particularly odd is that there doesn't even seem to be a
>> systematic sequence of letters in any of the columns. OK, since I
>> really need that function I wrote this piece of crap:
>>
>> perm.stupid <- function(x) {
>> b <- as.matrix(expand.grid(rep(list(x), length(x
>> b[!sapply(1:nrow(b), function(r) any(duplicated(b[r,]))),]
>> }
>>
>> It works, but words cannot describe its ugliness. And it gets really
>> slow really fast with growing x.
>>
>> So, anyway. My two questions are:
>> 1. Does R really, really, seriously lack a permutation function?
>> 2. OK, stop kidding me. So what's it called?
>> 3. Why doesn't my recursive function work, and what would a
>>working version look like?
>>
>> Thanks,
>> robert
>>
>> __
>> 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.
>>
>
> [[alternative HTML version deleted]]
>
> __
> 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] Simple permutation question
A couple of typos in the last section. It should be
> perm.ltrs <- apply(indx, 1, function(x) paste(LETTERS[x], collapse=" "))
> tail(perm.ltrs)
[1] "E D A B C" "E D B A C" "E D B C A" "E D A C B" "E D C A B" "E D C B A"
> tail(perm.ltrs)
[1] "E D A B C" "E D B A C" "E D B C A" "E D A C B" "E D C A B" "E D C B A"
>
David C
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of David L Carlson
Sent: Wednesday, June 25, 2014 4:02 PM
To: Cade, Brian; Robert Latest
Cc: r-help@r-project.org
Subject: Re: [R] Simple permutation question
Assuming you want all of the permutations not just a random permutation (such
as sample() give you):
> require(e1071)
> indx <- permutations(5)
> head(indx)
[,1] [,2] [,3] [,4] [,5]
[1,]12345
[2,]21345
[3,]23145
[4,]13245
[5,]31245
[6,]32145
> tail(indx)
[,1] [,2] [,3] [,4] [,5]
[115,]54123
[116,]54213
[117,]54231
[118,]54132
[119,]54312
[120,]54321
If you want them converted to numbers try
> apply(ind, 1, function(x) paste(LETTERS[x], collapse=" "))
[1] "A B C D E" "B A C D E" "B C A D E" "A C B D E" "C A B D E" "C B A D E"
> tail(perm.ltrs)
[1] "E D A B C" "E D B A C" "E D B C A" "E D A C B" "E D C A B" "E D C B A"
This is not the only permutation function in R, but this one has the advantage
of being symmetrical. The last permutation is the reverse of the first, the
penultimate the reverse of the second, etc.
-
David L Carlson
Department of Anthropology
Texas A&M University
College Station, TX 77840-4352
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Cade, Brian
Sent: Wednesday, June 25, 2014 3:39 PM
To: Robert Latest
Cc: r-help@r-project.org
Subject: Re: [R] Simple permutation question
It is called sample(,replace=F), where the default argument is sampling
without replacement.
Try
x <- c("A","B","C","D","E")
sample(x)
Brian
Brian S. Cade, PhD
U. S. Geological Survey
Fort Collins Science Center
2150 Centre Ave., Bldg. C
Fort Collins, CO 80526-8818
email: ca...@usgs.gov
tel: 970 226-9326
On Wed, Jun 25, 2014 at 2:22 PM, Robert Latest wrote:
> So my company has hired a few young McKinsey guys from overseas for a
> couple of weeks to help us with a production line optimization. They
> probably charge what I make in a year, but that's OK because I just
> never have the time to really dive into one particular time, and I have
> to hand it to the consultants that they came up with one or two really
> clever ideas to model the production line. Of course it's up to me to
> feed them the real data which they then churn through their Excel
> models that they cook up during the nights in their hotel rooms, and
> which I then implement back into my experimental system using live data.
>
> Anyway, whenever they need something or come up with something I skip
> out of the room, hack it into R, export the CSV and come back in about
> half the time it takes Excel to even read in the data, let alone
> process it. Of course that gor them curious, and I showed off a couple
> of scripts that condense their abysmal Excel convolutions in a few
> lean and mean lines of R code.
>
> Anyway, I'm in my office with this really attractive, clever young
> McKinsey girl (I'm in my mid-forties, married with kids and all, but I
> still enjoyed impressing a woman with computer stuff, of all things!),
> and one of her models involves a simple permutation of five letters --
> "A" through "E".
>
> And that's when I find out that R doesn't have a permutation function.
> How is that possible? R has EVERYTHING, but not that? I'm
> flabbergasted. Stumped. And now it's up to me to spend the evening at
> home coding that model, and the only thing I really need is that
> permutation.
>
> So this is my first attempt:
>
> perm.broken <- function(x) {
> if (length(x) == 1) return(x)
> sapply(1:length(x), function(i) {
> cbind(x[i], perm(x[-i]))
> })
> }
>
> But it doesn't work:
> > perm.broken(c("A", "B", "C"))
> [,1] [,2] [,3]
> [1,] "A" "B" "C"
> [2,] "A" "B" "C"
> [3,] "B" "A" "A"
> [4,] "C" "C" "B"
> [5,] "C" "C" "B"
> [6,] "B" "A" "A"
> >
>
> And I can't figure out for the life of me why. It should work because I
> go through the elements of x in order, use that in the leftmost column,
> and slap the permutation of the remaining elements to the right. What
> strikes me as particularly odd is that there doesn't even seem to be a
> systematic sequence of letters in any of the columns. OK, since I
> really need that function I wrote this piece of crap:
>
> perm.stupid <- function(x) {
> b <- as.matrix(expand.grid(rep(list(x), length(x
> b[!s
Re: [R] Simple permutation question
The brokenness of your perm.broken function arises from the attempted use
of sapply to bind matrices together, which is not something sapply does.
perm.fixed <- function( x ) {
if ( length( x ) == 1 ) return( matrix( x, nrow=1 ) )
lst <- lapply( seq_along( x )
, function( i ) {
cbind( x[ i ], perm.jdn( x[ -i ] ) )
}
)
do.call(rbind, lst)
}
On Wed, 25 Jun 2014, Robert Latest wrote:
So my company has hired a few young McKinsey guys from overseas for a
couple of weeks to help us with a production line optimization. They
probably charge what I make in a year, but that's OK because I just
never have the time to really dive into one particular time, and I have
to hand it to the consultants that they came up with one or two really
clever ideas to model the production line. Of course it's up to me to
feed them the real data which they then churn through their Excel
models that they cook up during the nights in their hotel rooms, and
which I then implement back into my experimental system using live data.
Anyway, whenever they need something or come up with something I skip
out of the room, hack it into R, export the CSV and come back in about
half the time it takes Excel to even read in the data, let alone
process it. Of course that gor them curious, and I showed off a couple
of scripts that condense their abysmal Excel convolutions in a few
lean and mean lines of R code.
Anyway, I'm in my office with this really attractive, clever young
McKinsey girl (I'm in my mid-forties, married with kids and all, but I
still enjoyed impressing a woman with computer stuff, of all things!),
and one of her models involves a simple permutation of five letters --
"A" through "E".
And that's when I find out that R doesn't have a permutation function.
How is that possible? R has EVERYTHING, but not that? I'm
flabbergasted. Stumped. And now it's up to me to spend the evening at
home coding that model, and the only thing I really need is that
permutation.
So this is my first attempt:
perm.broken <- function(x) {
if (length(x) == 1) return(x)
sapply(1:length(x), function(i) {
cbind(x[i], perm(x[-i]))
})
}
But it doesn't work:
perm.broken(c("A", "B", "C"))
[,1] [,2] [,3]
[1,] "A" "B" "C"
[2,] "A" "B" "C"
[3,] "B" "A" "A"
[4,] "C" "C" "B"
[5,] "C" "C" "B"
[6,] "B" "A" "A"
And I can't figure out for the life of me why. It should work because I
go through the elements of x in order, use that in the leftmost column,
and slap the permutation of the remaining elements to the right. What
strikes me as particularly odd is that there doesn't even seem to be a
systematic sequence of letters in any of the columns. OK, since I
really need that function I wrote this piece of crap:
perm.stupid <- function(x) {
b <- as.matrix(expand.grid(rep(list(x), length(x
b[!sapply(1:nrow(b), function(r) any(duplicated(b[r,]))),]
}
It works, but words cannot describe its ugliness. And it gets really
slow really fast with growing x.
So, anyway. My two questions are:
1. Does R really, really, seriously lack a permutation function?
2. OK, stop kidding me. So what's it called?
3. Why doesn't my recursive function work, and what would a
working version look like?
Thanks,
robert
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Re: [R] Simple permutation question
sorry... editing on the fly... try:
perm.fixed <- function( x ) {
if ( length( x ) == 1 ) return( matrix( x, nrow=1 ) )
lst <- lapply( seq_along( x )
, function( i ) {
cbind( x[ i ], perm.fixed( x[ -i ] ) )
}
)
do.call( rbind, lst )
}
On Wed, 25 Jun 2014, Jeff Newmiller wrote:
The brokenness of your perm.broken function arises from the attempted use of
sapply to bind matrices together, which is not something sapply does.
perm.fixed <- function( x ) {
if ( length( x ) == 1 ) return( matrix( x, nrow=1 ) )
lst <- lapply( seq_along( x )
, function( i ) {
cbind( x[ i ], perm.jdn( x[ -i ] ) )
}
)
do.call(rbind, lst)
}
On Wed, 25 Jun 2014, Robert Latest wrote:
So my company has hired a few young McKinsey guys from overseas for a
couple of weeks to help us with a production line optimization. They
probably charge what I make in a year, but that's OK because I just
never have the time to really dive into one particular time, and I have
to hand it to the consultants that they came up with one or two really
clever ideas to model the production line. Of course it's up to me to
feed them the real data which they then churn through their Excel
models that they cook up during the nights in their hotel rooms, and
which I then implement back into my experimental system using live data.
Anyway, whenever they need something or come up with something I skip
out of the room, hack it into R, export the CSV and come back in about
half the time it takes Excel to even read in the data, let alone
process it. Of course that gor them curious, and I showed off a couple
of scripts that condense their abysmal Excel convolutions in a few
lean and mean lines of R code.
Anyway, I'm in my office with this really attractive, clever young
McKinsey girl (I'm in my mid-forties, married with kids and all, but I
still enjoyed impressing a woman with computer stuff, of all things!),
and one of her models involves a simple permutation of five letters --
"A" through "E".
And that's when I find out that R doesn't have a permutation function.
How is that possible? R has EVERYTHING, but not that? I'm
flabbergasted. Stumped. And now it's up to me to spend the evening at
home coding that model, and the only thing I really need is that
permutation.
So this is my first attempt:
perm.broken <- function(x) {
if (length(x) == 1) return(x)
sapply(1:length(x), function(i) {
cbind(x[i], perm(x[-i]))
})
}
But it doesn't work:
perm.broken(c("A", "B", "C"))
[,1] [,2] [,3]
[1,] "A" "B" "C"
[2,] "A" "B" "C"
[3,] "B" "A" "A"
[4,] "C" "C" "B"
[5,] "C" "C" "B"
[6,] "B" "A" "A"
And I can't figure out for the life of me why. It should work because I
go through the elements of x in order, use that in the leftmost column,
and slap the permutation of the remaining elements to the right. What
strikes me as particularly odd is that there doesn't even seem to be a
systematic sequence of letters in any of the columns. OK, since I
really need that function I wrote this piece of crap:
perm.stupid <- function(x) {
b <- as.matrix(expand.grid(rep(list(x), length(x
b[!sapply(1:nrow(b), function(r) any(duplicated(b[r,]))),]
}
It works, but words cannot describe its ugliness. And it gets really
slow really fast with growing x.
So, anyway. My two questions are:
1. Does R really, really, seriously lack a permutation function?
2. OK, stop kidding me. So what's it called?
3. Why doesn't my recursive function work, and what would a
working version look like?
Thanks,
robert
__
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PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
---
Jeff NewmillerThe . . Go Live...
DCN:Basics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer (Solar/BatteriesO.O#. #.O#. with
/Software/Embedded Controllers) .OO#. .OO#. rocks...1k
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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---
Jeff NewmillerThe . . Go Live...
DCN:Basics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer (Solar/BatteriesO.O#. #.O#. with
/Software/Embedded Contro
[R] plot a 3-D marked point process
Hello All I intend to plot a 3-D marked point process. Could you please help me to find a code or a related package? Thanks [[alternative HTML version deleted]] __ 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] How to troubleshoot issue with curve() function?
I am trying to plot data that I imported with read. table using a Weibull
distribution. Here's the code I used for importing the data:
data <- read.table(file =
"C:/Users/Steven/OneDrive/Unfiled/Temp/phase2.csv", header = FALSE, sep = ",",
dec = ".", col.names = c("cycle_time", "cycle_length", "red_time",
"prot_green", "perm_green", "veh_count", "vc", "max_delay", "prot_occ",
"perm_occ", "red5_occ", "gor", "ror", "arrival_time"), na.strings = "NA")
And here's my code for plotting:
curve(dexp(data(arrival_time), rate=0.06), from = 0, to = 60, main =
"Exponential distribution")
I keep getting this error message:
Error in curve(dexp(data(arrival_time), rate = 0.06), from = 0, to = 60, :
'expr' must be a function, or a call or an expression containing 'x'
How to I begin to troubleshoot this? The only thing I can think of is that my
variable of interest, "arrival_time", does contain a number of rows with
blanks, but would that be the underlying cause? If so, can I get the function
to ignore blank rows?
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Re: [R] How to troubleshoot issue with curve() function?
I’d start trying to understand the error message - it is actually pretty clear
(see ?curve).
I suspect curve is not the function you want to use, rather something like
plot(data$arrival_time, dexp(data$arrival_time, rate=0.06), xlim = c(0, 60),
type=‘l’, main = "Exponential distribution”)
assuming your arrival times lie in the specified window.
Just a wild guess though, as you didn’t provide data (see footer)
On 25 Jun 2014, at 23:02 , Lavrenz, Steven M wrote:
> I am trying to plot data that I imported with read. table using a Weibull
> distribution. Here's the code I used for importing the data:
>
> data <- read.table(file =
> "C:/Users/Steven/OneDrive/Unfiled/Temp/phase2.csv", header = FALSE, sep =
> ",", dec = ".", col.names = c("cycle_time", "cycle_length", "red_time",
> "prot_green", "perm_green", "veh_count", "vc", "max_delay", "prot_occ",
> "perm_occ", "red5_occ", "gor", "ror", "arrival_time"), na.strings = "NA")
>
> And here's my code for plotting:
>
> curve(dexp(data(arrival_time), rate=0.06), from = 0, to = 60, main =
> "Exponential distribution")
>
> I keep getting this error message:
>
> Error in curve(dexp(data(arrival_time), rate = 0.06), from = 0, to = 60, :
> 'expr' must be a function, or a call or an expression containing 'x'
>
> How to I begin to troubleshoot this? The only thing I can think of is that my
> variable of interest, "arrival_time", does contain a number of rows with
> blanks, but would that be the underlying cause? If so, can I get the function
> to ignore blank rows?
>
> __
> 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.
>
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] plot a 3-D marked point process
Ferra Xu yahoo.com> writes: > > Hello All > > I intend to plot a 3-D marked point process. > Could you please help me to find a code or a related package? How about scatterplot3d or rgl::plot3d ? Googling "r 3d point plot" gets you a lot of good starting points. (The easiest way to indicate the marks would be by modifying colour according to mark class.) __ 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] Space-time non-homogeneous Poisson process
I am wondering if I want to model a space-time non-homogeneous Poisson process that has more probability of occurrence in some areas in space, how should I do that? For example we want to model the intensity function of earthquake occurrence that is more probable to happen in some areas (around earthquake line) and therefore the intensity function is a function of space and time. Thanks, [[alternative HTML version deleted]] __ 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.
