Displaying 2 results from an estimated 2 matches for "zz05".
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zz01
2006 Sep 05
2
Bugs with partial name matching during partial replacement (PR#9202)
...ly 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...
2006 Oct 16
1
Bugs with partial name matching during partial replacement (PR#9299)
...s.
>
> 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 ZZ...