[R] Factor contingency tables

[David Winsemius]

On Jan 21, 2010, at 12:38 PM, b k wrote:

> Hello,
>
> I know there must be a simple soluton to this problem but it eludes me
> currently.
>
> My data is partitioned into two subsets, each subset has a common  
> column
> factor but with varying levels:
>
> levels(fdf_ghc$AgeDemo)
> [1] "26TO35" "36TO45" "46TO55" "56TO65" "66TO75" "76TO85"
> levels(fdf_ghcnull$AgeDemo)
> [1] "26TO35"  "36TO45"  "46TO55"  "56TO65"  "66TO75"  "76TO85"   
> "86TO100"
> table(fdf_ghc$AgeDemo)
> 26TO35 36TO45 46TO55 56TO65 66TO75 76TO85
>     6     14     21     31     19     14
> table(fdf_ghcnull$AgeDemo)
> 26TO35  36TO45  46TO55  56TO65  66TO75  76TO85 86TO100
>      5       5      10       7       8       4       1
> I need to construct a common contingency table from the two lists,  
> but rbind
> recycles values due to the differing levels:
>
> rbind(table(fdf_ghc$AgeDemo), table(fdf_ghcnull$AgeDemo))
>     26TO35 36TO45 46TO55 56TO65 66TO75 76TO85 86TO100
> [1,]      6     14     21     31     19     14       6
> [2,]      5      5     10      7      8      4       1
> Warning message:
> In rbind(table(fdf_ghc$AgeDemo), table(fdf_ghcnull$AgeDemo)) :
>  number of columns of result is not a multiple of vector length (arg  
> 1)

  fdf_ghc<- data.frame(AgeDemo=factor( c("26TO35", "36TO45", "46TO55",  
"56TO65", "66TO75", "76TO85") ))
  fdf_ghcnull <- data.frame(AgeDemo= factor( c( "26TO35",  "36TO45",   
"46TO55",  "56TO65",  "66TO75"  ,"76TO85",  "86TO100")  ))

table( rbind(subset(fdf_ghcnull, select=AgeDemo), subset(fdf_ghc,  
select=AgeDemo)) )
# You need a strategy that prevents wiping out the factor level labels.
# in my test case:

  26TO35  36TO45  46TO55  56TO65  66TO75  76TO85 86TO100
       2       2       2       2       2       2       1

Whereas concatenation seemed to remove the factor labels although it  
keeps the counts:
 > factor(c(fdf_ghcnull$AgeDemo, fdf_ghc$AgeDemo))
  [1] 1 2 3 4 5 6 7 1 2 3 4 5 6
Levels: 1 2 3 4 5 6 7

>
> I need something I can pass to fisher.test() or chisq.test().

I'm not sure I see how those would accept a single factor or one row  
table for a test.

-- 

David Winsemius, MD
Heritage Laboratories
West Hartford, CT