On Jan 4, 2012, at 3:21 PM, Rich Shepard wrote:
On Wed, 4 Jan 2012, David Winsemius wrote:
Nothing attached. I don't know what you entitled teh "compressed
dput output" but it did not pass the filters of the mailserver and
you did not copy me.
David,
It must have been stripped off as t
On Wed, 4 Jan 2012, David Winsemius wrote:
Nothing attached. I don't know what you entitled teh "compressed dput output"
but it did not pass the filters of the mailserver and you did not copy me.
David,
It must have been stripped off as too large (14K).
Regardless, I solved the problem:
Nothing attached. I don't know what you entitled teh "compressed dput
output" but it did not pass the filters of the mailserver and you did
not copy me. If chemdata is available as a text file, hten make sure
its extension is .txt and then attach it.
--
David.
On Jan 4, 2012, at 1:31 PM, R
On Wed, 4 Jan 2012, David Winsemius wrote:
You didn't ask for what was duplicated, but rather what was NOT duplicated
with that code. In the case of a dataframe it is the entire row that is
tested.
My original question was what was duplicated, but ... I changed the
function by dropping the '
On Jan 4, 2012, at 12:21 PM, Rich Shepard wrote:
On Tue, 3 Jan 2012, David Winsemius wrote:
burns.tds[ !duplicated(burns.tds) , ]
Apparently it does not matter if the site column in the data frame
is a
factor or a character, read.zoo() generates the same error. Applying
the
above pro
On Tue, 3 Jan 2012, David Winsemius wrote:
burns.tds[ !duplicated(burns.tds) , ]
Apparently it does not matter if the site column in the data frame is a
factor or a character, read.zoo() generates the same error. Applying the
above produces a long list starting with:
burns.tds[!duplicated(
On Jan 3, 2012, at 5:53 PM, Rich Shepard wrote:
On Tue, 3 Jan 2012, David Winsemius wrote:
Maybe we need to backtrack a bit.
Yes. I've been trying to do this but still have too little
experience with
R to be successful on my own.
You originally were complaining about an error that sai
On Tue, 3 Jan 2012, David Winsemius wrote:
Maybe we need to backtrack a bit.
Yes. I've been trying to do this but still have too little experience with
R to be successful on my own.
You originally were complaining about an error that said you had
duplicated index entries as you attempted t
Maybe we need to backtrack a bit. You originally were complaining
about an error that said you had duplicated index entries as you
attempted to make a zoo object. I assumed, incorrectly it now
appears, that you understood that an index in a zoo object was a
vector. You now seem to be admit
On Tue, 3 Jan 2012, David Winsemius wrote:
That's rather unconvincing. What does this show:
burns.tds[ !duplicated(burns.tds) ]
I saw that in the help page but assumed it was the opposite of duplicated;
apparently not.
burns.tds[ !duplicated(burns.tds) ]
Error in .data.frame(burns.tds, !du
On Jan 3, 2012, at 12:48 PM, Rich Shepard wrote:
On Tue, 3 Jan 2012, Rich Shepard wrote:
I _think_ the problem comes from a duplicated factor column in the
data
frame. Now I need to figure out how subset() generated that
additional
column.
Nope. That's not it.
Running 'duplicated(bur
On Jan 3, 2012, at 12:26 PM, Rich Shepard wrote:
On Tue, 3 Jan 2012, David Winsemius wrote:
How can I identify the non-unique index entries within R?
?duplicated
Thank you, David.
I _think_ the problem comes from a duplated factor column in the data
frame. Now I need to figure out how
On Tue, 3 Jan 2012, Rich Shepard wrote:
I _think_ the problem comes from a duplicated factor column in the data
frame. Now I need to figure out how subset() generated that additional
column.
Nope. That's not it.
Running 'duplicated(burns.tds, incomparables = FALSE)' produces a listing
of
On Tue, 3 Jan 2012, David Winsemius wrote:
How can I identify the non-unique index entries within R?
?duplicated
Thank you, David.
I _think_ the problem comes from a duplated factor column in the data
frame. Now I need to figure out how subset() generated that additional
column.
Regard
On Jan 3, 2012, at 11:45 AM, Rich Shepard wrote:
I have a situation that I cannot resolve by myself. When I try to
create a
zoo object (with read.zoo() ) I get this error:
Error in merge.zoo(`BC-0.5 = c(" 0.000", " 0.010", " 0.010", "
0.060", :
series cannot be merged with non-uniqu
I have a situation that I cannot resolve by myself. When I try to create a
zoo object (with read.zoo() ) I get this error:
Error in merge.zoo(`BC-0.5 = c(" 0.000", " 0.010", " 0.010", "
0.060", :
series cannot be merged with non-unique index entries in a series
This suggests that th
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