[Rd] Bugs with partial name matching during partial replacement (PR#9202)
amaliy1 at uic.edu
amaliy1 at uic.edu
Tue Sep 5 03:40:02 CEST 2006
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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