lots of thanks
Ragia
> Date: Wed, 2 Dec 2015 14:12:46 -0800
> Subject: Re: [R] extract rows based on column value in a data frame
> From: bgunter.4...@gmail.com
> To: dwinsem...@comcast.net
> CC: ragi...@hotmail.com; r-help@r-project.org
>
> ...
> Perhaps
...
Perhaps also worth mentioning -- David's solution works even if there
are less than 3 rows per group, whereas mine will fail.
Cheers,
Bert
Bert Gunter
"Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom."
-- Clifford Stoll
On Wed, Dec 2, 2015 a
... or perhaps using rep() to do the indexing directly instead of matching:
dfrm[ ave(dfrm$v1, dfrm$v1, FUN =
function(x)rep(c(TRUE,FALSE),c(3,length(x)-3))), ]
There's probably another 6 dozen ways to do it, especially if you
access packages like data.table, plyr, etc.
Cheers,
Bert
Bert Gunte
On Wed, 2 Dec 2015 at 23:10 Adrian Dușa wrote:
> Dear All,
>
> I know how to fill a polygon, using a basic R graphics device:
>
> par(mai=c(0, 0, 0, 0))
> plot(1:100, type="n")
> polygon(c(20, 80, 80, 20), c(20, 20, 80, 80), col="lightblue")
>
>
> But let's say I have an outer polygon like this:
Adrian,
Draw the polygon once without the border and the whole in it, then go
back and draw the border around the outer polygon without any fill.
On Wed, Dec 2, 2015 at 9:31 AM, Adrian Dușa wrote:
> On Wed, Dec 2, 2015 at 5:19 PM, David L Carlson wrote:
>>
>> Using only base graphics, one solut
> On Dec 2, 2015, at 10:09 AM, Ragia Ibrahim wrote:
>
> Dear Group,
> I have a data frame that such as
>
> v1 v2v3v4
> 1 1 3 6
> 1 1 5 6
> 1 1 8 0
> 1 2 6 1
> 1 2 4 0
> 1 3 4 4
> 1 3
In your
> ordinalmodel <- polr(eduattain ~ dadedu, data = workdataset, weights =
> "SPFWT0", Hess = TRUE)
take the quotation marks off of SPFWT0. Like the subset argument, weights
is a literal expression, evaluated in the context of the data argument, not a
character string naming a column in the
First, a couple posting tips. It's helpful to provide some example data
people can work with. Also, please post in plain text (not html).
If you have a single standard for comparison, you might find an approach
like this helpful.
# example data
mylist <- c("AAEBCC", "AABDCC", "AABBCD")
list.2 <
Hello everyone,
I'm running an ordinal logistic and I keep getting this error:
Error in model.frame.default(formula = eduattain ~ dadedu, data =
workdataset, :
variable lengths differ (found for '(weights)')
I looked at several similar questions on the internet and ended up deleting
all the N
> On 02 Dec 2015, at 16:09, Brant Inman wrote:
>
> Thank you for your response. Here is the problem that I find with your code
> (which I had tried). When you pass a value to the subset argument of the
> function, it will not hold the quotes on the subsetting variable’s value.
>
> For examp
On Wed, Dec 2, 2015 at 5:19 PM, David L Carlson wrote:
>
> Using only base graphics, one solution would be to embed the inner
polygon in the outer one and turn off the border:
>
> par(mai=c(0, 0, 0, 0))
> plot(1:100, type="n")
> polygon(c(0, 100, 100, 0, 0, 20, 80, 80, 20, 20, 0),
> c(0, 0, 1
Hi everyone,
I have a question with reference to "glm.cluster" from the package "miceadds"
and hope that someone can help me. I am trying to calculate cluster-robust
standard errors for a glm-model with multiply imputed datasets. Everything
works just fine with glm.cluster but in the end I just
I think I am making this problem harder than it has to be and so I keep getting
stuck on what might be a trivial problem.
I have used the seqinr package to load a protein sequence alignment containing
15 protein sequences;
> library(seqinr) > x =
read.alignment("proteins.fasta",format="f
Using only base graphics, one solution would be to embed the inner polygon in
the outer one and turn off the border:
par(mai=c(0, 0, 0, 0))
plot(1:100, type="n")
polygon(c(0, 100, 100, 0, 0, 20, 80, 80, 20, 20, 0),
c(0, 0, 100, 100, 0, 20, 20, 80, 80, 20, 0), col="lightblue", border=NA)
--
Thank you for your response. Here is the problem that I find with your code
(which I had tried). When you pass a value to the subset argument of the
function, it will not hold the quotes on the subsetting variable’s value.
For example, if I want the function to do Y[Z==‘skinny’] so that we us
Your example and explanation are not complete, but I have the gut feeling that
you could do all this both more efficiently *and* more R-ish.
First of all, why would you pass Y and X separately, to ultimately build the Y
~ X formula within the body of your function?
Secondly, it seems to me that
Dear All,
I know how to fill a polygon, using a basic R graphics device:
par(mai=c(0, 0, 0, 0))
plot(1:100, type="n")
polygon(c(20, 80, 80, 20), c(20, 20, 80, 80), col="lightblue")
But let's say I have an outer polygon like this:
polygon(c(0,100,100,0), c(0,0,100,100))
Is it possible to fill t
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