[Rd] Bugs with partial name matching during partial replacement (PR#9299)

[hin-tak.leung at cimr.cam.ac.uk]

This is a rather interesting, but I don't think it is a bug - it is
just things that "you are not supposed to do"... you are assuming
a certain evaluation order of the 4 "$" operators in
" D$ABC[D$M] = D$V[D$M] " as in:

temp1 <- D$M                 # 2nd and 4th
temp2 <- D$V[temp1]          # 3rd
D$ABC[temp1] = temp2         # 1st

What R did was this:

temp4 <- D$ABC         # make reference, expand to D$ABCD , 1st
temp1 <- D$M           # 2nd, and 4th
temp2 <- D$V[temp1]    # 3rd

temp4[temp1] <- temp2  # oh dear, it looks as if we are
D$ABC <- temp4         # trying to write to a reference,
                        # better make a copy instead

R is doing the 4 $'s roughly from left to right, if you have some ideas 
how R works inside. (I am not saying this behavior is a "good" thing,
but at least it is consistent). Basically it is a very bad habit to 
write code that depends on evaluation order of operators at the same
precendence.

The difference in behavior in the two case is probably due to
coercion, (and also how lazy R does make-a-reference versus "oops, you 
seems to try to write to a reference so I better copy it") but
I'll leave you to think about what order R is doing the combination of
the 4 $'s and coercing between types... Basically writing code that 
depends on evaluation order is a bad idea.

c.f. this bit of C code:

i =0;
++i = ++i + ++i;

what value do you think "i" should be?

amaliy1 at uic.edu wrote:
> Hello,
> 
> First the version info:
> platform       powerpc-apple-darwin8.6.0
> arch           powerpc
> os             darwin8.6.0
> system         powerpc, darwin8.6.0
> status
> major          2
> minor          3.1
> year           2006
> month          06
> day            01
> svn rev        38247
> language       R
> version.string Version 2.3.1 (2006-06-01)
> 
> I have encountered some unusual behavior when trying to create new  
> columns in a data frame that have names that would generate a partial  
> match with an existing column with a longer name.  It is my  
> understanding that replacement operations shouldn't have partial  
> matching, but it is not clear to me whether this applies only when  
> the named column exists and not for new assignments.
> 
> The first example:
> 
>  > D = data.frame(M=c(T,T,F,F,F,T,F,T,F,F,T,T,T),V=I(sprintf("ZZ%02d", 
> 1:13)),ABCD=13:1)
>  > D
>         M    V ABCD
> 1   TRUE ZZ01   13
> 2   TRUE ZZ02   12
> 3  FALSE ZZ03   11
> 4  FALSE ZZ04   10
> 5  FALSE ZZ05    9
> 6   TRUE ZZ06    8
> 7  FALSE ZZ07    7
> 8   TRUE ZZ08    6
> 9  FALSE ZZ09    5
> 10 FALSE ZZ10    4
> 11  TRUE ZZ11    3
> 12  TRUE ZZ12    2
> 13  TRUE ZZ13    1
>  > D$CBA[D$M] = D$V[D$M]
>  > D
>         M    V ABCD  CBA
> 1   TRUE ZZ01   13 ZZ01
> 2   TRUE ZZ02   12 ZZ02
> 3  FALSE ZZ03   11 <NA>
> 4  FALSE ZZ04   10 <NA>
> 5  FALSE ZZ05    9 <NA>
> 6   TRUE ZZ06    8 ZZ06
> 7  FALSE ZZ07    7 <NA>
> 8   TRUE ZZ08    6 ZZ08
> 9  FALSE ZZ09    5 <NA>
> 10 FALSE ZZ10    4 <NA>
> 11  TRUE ZZ11    3 ZZ11
> 12  TRUE ZZ12    2 ZZ12
> 13  TRUE ZZ13    1 ZZ13
>  > D$ABC[D$M] = D$V[D$M]
>  > D
>         M    V ABCD  CBA  ABC
> 1   TRUE ZZ01   13 ZZ01 ZZ01
> 2   TRUE ZZ02   12 ZZ02 ZZ02
> 3  FALSE ZZ03   11 <NA>   11
> 4  FALSE ZZ04   10 <NA>   10
> 5  FALSE ZZ05    9 <NA>    9
> 6   TRUE ZZ06    8 ZZ06 ZZ06
> 7  FALSE ZZ07    7 <NA>    7
> 8   TRUE ZZ08    6 ZZ08 ZZ08
> 9  FALSE ZZ09    5 <NA>    5
> 10 FALSE ZZ10    4 <NA>    4
> 11  TRUE ZZ11    3 ZZ11 ZZ11
> 12  TRUE ZZ12    2 ZZ12 ZZ12
> 13  TRUE ZZ13    1 ZZ13 ZZ13
> 
> I expected ABC to equal CBA with NA values in rows not assigned, but  
> instead it appears that an extraction from D$ABCD and coercion to  
> string is being performed in the process of creating D$ABC.
> 
> Here is something I believe is definitely a bug:
> 
>  > D = data.frame(M=c(T,T,F,F,F,T,F,T,F,F,T,T,T),V=1:13,ABCD=13:1)
>  > D
>         M  V ABCD
> 1   TRUE  1   13
> 2   TRUE  2   12
> 3  FALSE  3   11
> 4  FALSE  4   10
> 5  FALSE  5    9
> 6   TRUE  6    8
> 7  FALSE  7    7
> 8   TRUE  8    6
> 9  FALSE  9    5
> 10 FALSE 10    4
> 11  TRUE 11    3
> 12  TRUE 12    2
> 13  TRUE 13    1
>  > D$CBA[D$M] = D$V[D$M]
>  > D
>         M  V ABCD CBA
> 1   TRUE  1   13   1
> 2   TRUE  2   12   2
> 3  FALSE  3   11  NA
> 4  FALSE  4   10  NA
> 5  FALSE  5    9  NA
> 6   TRUE  6    8   6
> 7  FALSE  7    7  NA
> 8   TRUE  8    6   8
> 9  FALSE  9    5  NA
> 10 FALSE 10    4  NA
> 11  TRUE 11    3  11
> 12  TRUE 12    2  12
> 13  TRUE 13    1  13
>  > D$ABC[D$M] = D$V[D$M]
>  > D
>         M  V ABCD CBA ABC
> 1   TRUE  1    1   1   1
> 2   TRUE  2    2   2   2
> 3  FALSE  3   11  NA  11
> 4  FALSE  4   10  NA  10
> 5  FALSE  5    9  NA   9
> 6   TRUE  6    6   6   6
> 7  FALSE  7    7  NA   7
> 8   TRUE  8    8   8   8
> 9  FALSE  9    5  NA   5
> 10 FALSE 10    4  NA   4
> 11  TRUE 11   11  11  11
> 12  TRUE 12   12  12  12
> 13  TRUE 13   13  13  13
> 
> ABC is created as before with valued from ABCD in the unassigned  
> rows, but ABCD is being modified as well.  The only difference form  
> the previous example is that V is now just a numeric column.
> 
> Anil Maliyekkel
> 
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