R help - Jul 2017 - select from data frame

Dear All,

wonder if you could please assist with the following 

df<-data.frame(ID=c(1,1,1,2,2,3,3,4,4,5,5),samples=c("A","B","C","A","C","A","D","C","B","A","C"))

from this data frame the goal is to extract the value of 3 from the ID column
based on the logic that the ID=3 in the data frame has NO row that would pair 3
with either "B", AND/OR "C" in the samples column...


much appreciate your help...

thanks,
 Andras

> On Jul 15, 2017, at 4:01 AM, Andras Farkas via R-help <r-help at
r-project.org> wrote:
> 
> Dear All,
> 
> wonder if you could please assist with the following 
> 
>
df<-data.frame(ID=c(1,1,1,2,2,3,3,4,4,5,5),samples=c("A","B","C","A","C","A","D","C","B","A","C"))
> 
> from this data frame the goal is to extract the value of 3 from the ID
column based on the logic that the ID=3 in the data frame has NO row that would
pair 3 with either "B", AND/OR "C" in the samples column...
> 
This returns a vector that determines if either of those characters are in the
character values of that factor column you created. Coercing to character is
needed because leaving samples as a factor generated an invalid factor level
warning and gave useless results.

 with( df, ave( as.character(samples), ID, FUN=function(x) {!any(x %in%
c("B","C"))}))
 [1] "FALSE" "FALSE" "FALSE" "FALSE"
"FALSE" "TRUE"  "TRUE"  "FALSE"
"FALSE"
[10] "FALSE" "FALSE"

You can then use it to extract and consolidate to a single value (although
wrapping with as.logical was needed because `ave` returned character class
values):

 unique( df$ID[ as.logical(   # fails without this since "FALSE" !=
FALSE
                    with( df, 
                       ave( as.character(samples), ID, FUN=function(x) {!any(x
%in% c("B","C"))})))
              ] )
#[1] 3

The same sort of logic could also be constructed with a for-loop:
> for (x in unique(df$ID) ) { if ( !any( df$samples[df$ID==x] %in%
c("b","C")) ) print(x) }
[1] 3

Although you are warned that for-loops do not return values and you might need
to make an assignment rather than just printing.

-- 

David Winsemius
Alameda, CA, USA

If I understand correctly, no looping (ave(), for()) or type casting
(as.character()) is needed -- indexing and matching suffice:
> with(df, ID[!ID %in% unique(ID[samples %in% c("B","C")
])])
[1] 3 3



Cheers,

Bert


Bert Gunter

"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )


On Sat, Jul 15, 2017 at 8:54 AM, David Winsemius <dwinsemius at
comcast.net> wrote:
>
>> On Jul 15, 2017, at 4:01 AM, Andras Farkas via R-help <r-help at
r-project.org> wrote:
>>
>> Dear All,
>>
>> wonder if you could please assist with the following
>>
>>
df<-data.frame(ID=c(1,1,1,2,2,3,3,4,4,5,5),samples=c("A","B","C","A","C","A","D","C","B","A","C"))
>>
>> from this data frame the goal is to extract the value of 3 from the ID
column based on the logic that the ID=3 in the data frame has NO row that would
pair 3 with either "B", AND/OR "C" in the samples column...
>>
>
> This returns a vector that determines if either of those characters are in
the character values of that factor column you created. Coercing to character is
needed because leaving samples as a factor generated an invalid factor level
warning and gave useless results.
>
>  with( df, ave( as.character(samples), ID, FUN=function(x) {!any(x %in%
c("B","C"))}))
>  [1] "FALSE" "FALSE" "FALSE"
"FALSE" "FALSE" "TRUE"  "TRUE" 
"FALSE" "FALSE"
> [10] "FALSE" "FALSE"
>
> You can then use it to extract and consolidate to a single value (although
wrapping with as.logical was needed because `ave` returned character class
values):
>
>  unique( df$ID[ as.logical(   # fails without this since "FALSE"
!= FALSE
>                     with( df,
>                        ave( as.character(samples), ID, FUN=function(x)
{!any(x %in% c("B","C"))})))
>               ] )
> #[1] 3
>
> The same sort of logic could also be constructed with a for-loop:
>
>> for (x in unique(df$ID) ) { if ( !any( df$samples[df$ID==x] %in%
c("b","C")) ) print(x) }
> [1] 3
>
> Although you are warned that for-loops do not return values and you might
need to make an assignment rather than just printing.
>
> --
>
> David Winsemius
> Alameda, CA, USA
>
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