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Subseting by more than one factor...

Fernando Henrique Ferraz Pereira da Rosa wrote:
It didn't? It does for me:

R> subset(z, p4 == 1 & p6 == 1)
  [1] p1   p2   p3   p4   p5   p6   p7   p8   p9   p10  pred
<0 rows> (or 0-length row.names)
R> subset(z, p2 == 1 & p8 == 1)
    p1 p2 p3 p4 p5 p6 p7 p8 p9 p10      pred
10  0  1  0  0  0  0  0  1  0   0 0.5862069
R> subset(z, (p2 == 1 & p3 == 0) | p5 == 1)
    p1 p2 p3 p4 p5 p6 p7 p8 p9 p10      pred
1   0  1  0  0  0  0  0  0  0   0 0.5862069
6   0  0  0  0  1  0  0  0  0   0 0.5862069
8   0  1  0  0  1  0  0  0  0   0 0.5862069
9   0  1  0  0  0  0  1  0  0   0 0.5862069
10  0  1  0  0  0  0  0  1  0   0 0.5862069
11  0  1  0  0  0  0  0  0  0   1 0.5862069
R> version
          _
platform i386-pc-mingw32
arch     i386
os       mingw32
system   i386, mingw32
status
major    1
minor    7.0
year     2003
month    04
day      16
language R
R>


[snip]

Regards,
Sundar