[Rd] [ subscripting sometimes loses names (PR#8192)

[atp@piskorski.com]

Here is an example of the losing names problem in stock R 2.2.0.  Note
that below, only stock R packages are loaded, and then I manually
source in just my dtk.test.brace.names() testing function, nothing
else.

Since the list-of-lists output of dtk.test.brace.names() is very
lengthy, I've manually cut-and-pasted it into a tabular format to save
space and make inspection easier.  As you can see, out of its 15 test
cases, stock R 2.2.0 fails 4 of them while the other 12 are Ok.

Too see what these simple subscripting tests actually DO, please refer
to the body of dtk.test.brace.names() from my previous emails above.


R : Copyright 2005, The R Foundation for Statistical Computing
Version 2.2.0  (2005-10-06 r35749)
> search()
[1] ".GlobalEnv"        "package:methods"   "package:graphics" 
[4] "package:grDevices" "package:datasets"  "package:utils"    
[7] "package:stats"     "Autoloads"         "package:base"     

> dtk.test.brace.names(return.results.p=T ,only="all")

Ok?  Actual Result         Desired Result
---  ------------------    ------------------
     $vec.1
BAD  $vec.1[[1]]           $vec.1[[2]]
        a    c <NA>         a  c no
        1    3   NA         1  3 NA

     $diag.1
Ok   $diag.1[[1]]          $diag.1[[2]]
     [1]  1  7 13 19 25    [1]  1  7 13 19 25

     $diag.2
Ok   $diag.2[[1]]          $diag.2[[2]]
     [1]  1  7 13 19 25    [1]  1  7 13 19 25

     $df.a.1
Ok   $df.a.1[[1]]          $df.a.1[[2]]
     a b                   a b
     4 5                   4 5

     $df.b.1
BAD  $df.b.1[[1]]          $df.b.1[[2]]
     [1] 4 5               a b
                           4 5

     $df.a.2
Ok   $df.a.2[[1]]          $df.a.2[[2]]
     c b a                 c b a
     6 5 4                 6 5 4

     $df.b.2
BAD  $df.b.2[[1]]          $df.b.2[[2]]
     [1] 6 5 4             c b a
                           6 5 4

     $df.a.3
Ok   $df.a.3[[1]]          $df.a.3[[2]]
     a b                   a b
     3 4                   3 4

     $df.b.3
BAD  $df.b.3[[1]]          $df.b.3[[2]]
     [1] 3 4               a b
                           3 4

     $df.a.4
Ok   $df.a.4[[1]]          $df.a.4[[2]]
     col1 col2             col1 col2
        2    4                2    4

     $df.b.4
Ok   $df.b.4[[1]]          $df.b.4[[2]]
       col1 col2             col1 col2
     b    2    4           b    2    4

     $df.a.5
Ok   $df.a.5[[1]]          $df.a.5[[2]]
     col1 col2             col1 col2
        2    4                2    4

     $df.b.5
Ok   $df.b.5[[1]]          $df.b.5[[2]]
     $df.b.5[[1]]$col1     $df.b.5[[2]]$col1
     [1] 2                 [1] 2
     $df.b.5[[1]]$col2     $df.b.5[[2]]$col2
     [1] 4                 [1] 4

     $df.a.6
Ok   $df.a.6[[1]]          $df.a.6[[2]]
       col1 col2             col1 col2
     b    2    4           b    2    4

     $df.b.6
Ok   $df.b.6[[1]]          $df.b.6[[2]]
       col1 col2             col1 col2
     b    2    4           b    2    4

-- 
Andrew Piskorski <atp at piskorski.com>
http://www.piskorski.com/