On 3/08/2009, at 3:43 PM, David Winsemius wrote:
On Aug 2, 2009, at 10:46 PM, Rolf Turner wrote:
On 3/08/2009, at 1:48 PM, David Winsemius wrote:
On Aug 2, 2009, at 7:29 PM, Rolf Turner wrote:
On 3/08/2009, at 11:14 AM, David Winsemius wrote:
On Aug 2, 2009, at 7:02 PM, Noah Silverman wrote:
Hi,
It seems as if the problem was caused by an odd quirk of the
"scale"
function.
Some of my data have NA entries.
So, I substitute 0 for any NA with:
rawdata[is.na(rawdata)] <- 0
Perhaps this would have done what you intended:
rawdata[is.na(rawdata), ] <- 0
I don't think this works at all. E.g.:
set.seed(42)
junk <- matrix(rnorm(60),12,5)
junk[sample(1:60,14)] <- NA
junk[is.na(junk),] <- 0 # Throws an error.
junk[is.na(junk)] <- 0 # Gives the desired result.\\
Then how about? ....
junk2 <- replace(junk, is.na(junk), 0)
That certainly works --- and is a potentially useful piece of syntax
of which I was
previously unaware (thank you) --- but the OP's syntax also worked
(and gives identical results
to yours). I.e. the OP's problem was *not* induced by the way he
went about replacing
NA's by zeroes, irrespective of the fact that that's a silly thing
to do.
Go back to the original posting. The OP was complaining about the fact
that rawdata[is.na(rawdata)] <- 0 threw an error before he ran that R
object through scale. He did not realize that rawdata was a dataframe
in the first instance and a matrix in the second. The syntax of the
"[" operator would not support the construction that he and you were
using when it was applied to a dataframe. The code I offered does not
seem to suffer from that deficiency.
OK. I get it now.
cheers,
Rolf Turner
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