Dimitris Rizopoulos wrote:
On 3/9/2010 1:36 PM, Paul Hiemstra wrote:
Dimitris Rizopoulos wrote:
one approach is the following:
mat <- matrix(rnorm(100*45), 100, 45)
mat[sample(100*45, 50)] <- 0
index <- rowMeans(mat == 0) == 0
mat[index, ]
Dimitris,
You use quite a complicated syntax to get the index. I think the
following syntax using apply is more easy to understand:
well, this way is much more efficient to compute, especially if you
have many rows. Compare the following to see the difference:
mat <- matrix(rnorm(2*1e06), 1e06, 2)
mat[sample(2*1e06, 50)] <- 0
system.time(index1 <- !apply(mat == 0, MARGIN = 1, any))
system.time(index2 <- rowMeans(mat == 0) == 0 )
all.equal(index1, index2)
Point taken, I totally agree that for large matrices you should use your
approach. Didn't know that it made such a difference, great to learn
something new :). But I still like the other syntax more ;).
cheers,
Paul
Best,
Dimitris
# Note, MARGIN equal to 1 means loop over rows
# If any member of a row is zero
index = apply(mat == 0, MARGIN = 1, any)
# If all members of a row are zero
index = apply(mat == 0, MARGIN = 1, all)
cheers and hope it helps,
Paul
I hope it helps.
Best,
Dimitris
On 3/9/2010 11:05 AM, ogbos okike wrote:
Hi Everybody,
I have a matrix of about 45 columns. Some of the rows contain zeros.
Using
data1<-data[complete.cases(data),], I can remove the "NA" rows. But
I am
unable to tackle that of zeros.
Can anybody give me an idea of how to remove rows containing zeros
in a
matrix.
Thanks so much
Best
Ogbos
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