array indexing and which

Hi R friends!

I am stuck with a stupid question: I can circumvent it
but I would like to 
understand why it is wrong. It would be nice if you
could give me a hint...

Having a reproducible example, as per the posting guide, would be
helpful here. We'll use a contrived example that hopefully explains what
I can only presume you are seeing.

I have an 2D array d and do the following:
ids <- which(d[,1]>0)

Here ids contains the indices of the values in the vector d[, 1] that
are > 0.

For example:

d <- matrix(sample(0:1, 12, replace = TRUE), ncol = 2)
d

[,1] [,2]
[1,]    1    1
[2,]    1    1
[3,]    0    1
[4,]    0    0
[5,]    0    1
[6,]    1    0

ids <- which(d[, 1] > 0)
ids

[1] 1 2 6

Note that c(1, 2, 6) are the indices into the vector:

d[, 1]

[1] 1 1 0 0 0 1

of the values that are > 0.

then I have a vector gk with same column size as d and
do:
ids2 <- which(gk[ids]==1)

Here ids2 contains the indices of the values in gk[ids] that equal 1.

gk <- sample(0:1, 6, replace = TRUE)
gk

[1] 1 1 1 0 1 1

gk[ids]  # same as gk[c(1, 2, 6)]

[1] 1 1 1

ids2 <- which(gk[ids] == 1)
ids2

[1] 1 2 3

All three of the values in gk[ids] == 1.

but I can't interprete the result I get in ids2.

I get the expected result when I use:
which(gk==1 & d[,1]>0)

Here you are getting the result of logically comparing the two vectors:

gk == 1

[1]  TRUE  TRUE  TRUE FALSE  TRUE  TRUE

AND

d[, 1] > 0

[1]  TRUE  TRUE FALSE FALSE FALSE  TRUE


where the result of the comparison is the index value of each pair in
the two vectors where both values are TRUE.

Thus:

which(gk == 1 & d[, 1] > 0)

[1] 1 2 6

versus:

ids2

[1] 1 2 3

Why is the first version wrong?

It's not wrong. It is giving you what you asked for.

Your question was wrong.  :-)

The reason why I try to use the ids vectors is that I
want to avoid recomputation.

Thanks for your help!
   Werner

HTH,

Marc Schwartz