Re: [R] Passing variable names in quotes to a function

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

Re: [R] fill the area outside a polygon

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)

[R] Significance level for glm.cluster

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

[R] Looping through multiple sub elements of a list to compare to multiple components of a vector

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 =

Re: [R] fill the area outside a polygon

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,

Re: [R] Looping through multiple sub elements of a list to compare to multiple components of a vector

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

Re: [R] Passing variable names in quotes to a function

> 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

Re: [R-es] help

Hola: Nunca he utilizado el RcmdrPlugin.EZR, pero lo único que he entendido: - accrual time during which subjects are recruited to the study: tiempo que ja durado el reclutamiento de los pacientes. - Total (accural + follow-up) duration: Duración total del estudio: reclutamiento + seguimiento.

[R] Error in model.frame.default… variable lengths differ (Without NA's)

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

Re: [R] Error in model.frame.default… variable lengths differ (Without NA's)

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

Re: [R] fill the area outside a polygon

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

Re: [R] fill the area outside a polygon

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

[R-es] Interpretación coeficiente en modelo

Hola, Tenemos este modelo: Phealth = mu + 11.79 poly(weight, 3) -5.15 sex.F +0.02 height -10.67 poly(weight, 3): sex.F + height:sex.F + epsilon Donde Phealth es una respuesta binaria que representa la percepción de los encuestados sobre su estado de salud (1=excelente, 0=pobre). Tenemos el

Re: [R] extract rows based on column value in a data frame

> 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

Re: [R] extract rows based on column value in a data frame

... 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

Re: [R] extract rows based on column value in a data frame

... 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

Re: [R] extract rows based on column value in a data frame

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

Re: [R-es] igraph o network

Hola, ¿qué tal? Yo siempre he usado igraph. Debería serte suficiente para casi todos los fines habituales. Tu red es pequeña, además. El problema que tienes con ese algoritmo está indicado en el mensaje de error: "detection works for undirected graph only". Tu grafo está dirigido (¿tiene que

[R-es] Elección del nuevo logo de la Comunidad R Hispano

Hola, ¿qué tal? Este es un mensaje de interés, sobre todo, para los socios y simpatizantes de la Comunidad R HIspano, la sociación española de usuarios de R. Durante las últimas jornadas en Salamanca se consideró la necesidad de cambiar el logo de la asociación por diversos motivos: 1)

[R] fill the area outside a polygon

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

[R-es] igraph o network

Estimados, estoy “jugando” con unos datos clínicos donde realizo un análisis de redes, utilicé igraph como nerwork, pero finalmente uso igraph (creo que tiene más opciónes). Son algo de 200 nodos y algo más de 2000 relaciones. Puedo graficarlos. Los datos están de la siguiente forma, por

Re: [R] Passing variable names in quotes to a function

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