[R] OK - I got the data - now what? :-)

[Mark Knecht]

On Sun, Jul 5, 2009 at 12:30 PM, Henrique Dallazuanna<wwwhsd at gmail.com> wrote:
> Try this:
>
> subset(test[3,], select=C1:C6)[,subset(test[3,], select = C1:C6) > 0]
>
> subset(test[6,], select=C1:C6)[,subset(test[6,], select = C1:C6) > 0]
>
>

I must admit I like this one. Pleasing to look at. It seems
approachable. Thanks!

If I understand this the second subset gets evaluated first producing
either TRUE or FALSE, and then the first subset gets evaluated but
only for the entries that are TRUE? Is that the process?

Thanks,
Mark

>
> On Sun, Jul 5, 2009 at 1:19 PM, Mark Knecht <markknecht at gmail.com> wrote:
>>
>> On Sun, Jul 5, 2009 at 8:18 AM, David Winsemius<dwinsemius at comcast.net>
>> wrote:
>> >
>> > On Jul 5, 2009, at 10:50 AM, Uwe Ligges wrote:
>> >
>> >>
>> >>
>> >> David Winsemius wrote:
>> >>>
>> >>> So if your values are calculated from other values then consider using
>> >>> all.equal()
>> >>> And repeated applications of the testing criteria process are
>> >>> effective:
>> >>> test[3,][which(names(test)=="C1"):(which(test[3,] == 0.0)-1)]
>> >>>   C1   C2   C3
>> >>> 3 0.52 0.66 0.51
>> >>> (and a warning that does not seem accurate to me.)
>> >>> In which(names(test) == "C1"):(which(test[3, ] == 0) - 1) :
>> >>>  numerical expression has 3 elements: only the first used
>> >>
>> >>
>> >> David,
>> >>
>> >> # which(test[3,] == 0.0)
>> >> [1] 6 7 8
>> >>
>> >> and in a:b a and b must be length 1 vectors (scalars) otherwise just
>> >> the
>> >> first element (in this case 6) is used.
>> >>
>> >> That leads us to the conclusion that writing the line above is not
>> >> really
>> >> the cleanest way or you intended something different ....
>> >
>> > Thanks, Uwe. I see my confusion. I did want 6 to be used  and it looks
>> > as
>> > though I would not be getting in truouble this way, but a cleaner method
>> > would be to access only the first element of which(test[3, ] == 0):
>> >
>> > test[3,][ which(names(test) == "C1") : (which(test[3,] == 0.0)[1]-1) ]
>> >
>> >>
>> >> David
>> >
>> >>> Seems to me that all of the element were used. I cannot explain that
>> >>> warning but am pretty sure it can be ignored.
>> >>>
>> >
>> > David Winsemius, MD
>> > Heritage Laboratories
>> > West Hartford, CT
>> >
>> >
>>
>> OK - making lots more headway. Thanks for your help.
>>
>> QUESTION: How do I handle the case where I'm testing for 0 and don't
>> find it? In this case I need to all of the row from C1:C6.
>>
>> test <- data.frame(A=1:10, B=100, C1=runif(10), C2=runif(10),
>> C3=runif(10), C4=runif(10), C5=runif(10), C6=runif(10))
>> test<-round(test,2)
>>
>> #Make array ragged
>> test$C3[2]<-0;test$C4[2]<-0;test$C5[2]<-0;test$C6[2]<-0
>> test$C4[3]<-0;test$C5[3]<-0;test$C6[3]<-0
>> test$C6[7]<-0
>> test$C4[8]<-0;test$C5[8]<-0;test$C6[8]<-0
>>
>> test
>>
>> #C1 always the same so calculate it only once
>> StartCol <- which(names(test)=="C1")
>>
>> #Print row 3 explicitly
>> test[3,][StartCol :(which(test[3,] == 0.0)[1]-1)]
>>
>> #Row 6 fails because 0 is not found
>> test[6,][StartCol :(which(test[6,] == 0.0)[1]-1)]
>>
>> EndCol <- which(test[6,] == 0.0)[1]-1
>> EndCol
>>
>> Thanks,
>> Mark
>>
>> ______________________________________________
>> R-help at 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.
>
>
>
> --
> Henrique Dallazuanna
> Curitiba-Paraná-Brasil
> 25° 25' 40" S 49° 16' 22" O
>