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Hello R usuer, The code given below superimposes a pie diagram on another plot containing some points. However, I would like to center the pie diagram on the xy location on the plot, but not on the center. is there any way to re-center pic diagram. Any suggestion or better alternative are highly appreciated. Thank you in advance for your help. Regards, Bibke library(visualFields) library(car)

Hello R users, My problem is that the data I've got is in the minimum number of columns with each ward (geographic area) appearing multiple times. The first 30 terms look like this > HHum02 CASW Btype Yr CO2Group NumVeh 170597 00CCFA CARS 2002 C 2 170598 00CCFA CARS 2002 D 2 170599 00CCFA CARS 2002 E 22 170600 00CCFA CARS 2002

Dear all, Can anyone take a look at my program below? There are two functions: f1 (lambda,z,p1) and f2(p1,cl, cu). I fixed p1=0.15 for both functions. For any fixed value of lambda (between 0.01 and 0.99), I solve f1(p1=0.15, lambda=lambda, z)=0 for the corresponding cl and cu values. Then I plug the calculated cl and cu back into the function f2. Eventually, I want to find the lambda value

Estimados miembros de la lista, Estoy trabajando con la función spin del paquete knitr y no puedo entender por que a pesar de usar #+ para iniciar cada chunk no reconoce el inicio de un nuevo chunk al menos que incluya texto entre cada inicio de chunk. En el ejemplo abajo #+ data(cars) #+ cars No los separa como 2 chunks diferentes Ejemplo de mi archivo en r (prueba.R)

Muchas gracias Carlos. Tengo la versión de knitr 1.14 Corrí el ejemplo que me dices y resulta en lo mismo, no se crean chunks diferentes. Parece que para separar chunks se debe incluir #' Por ejemplo, estas 2 líneas de códigos del ejemplo: #+ test-chisq5 sum(x^2) # chi-square distribution with df 5 #+ test-chisq4 sum((x - mean(x))^2) # df is 4 now Dan como resultado en el Rmd (todo queda

Hi R help, Hi R help, Which is the easiest (most elegant) way to force "aov" to treat numerical variables as categorical ? Sincerely, Andrea Bernasconi DG PROBLEM EXAMPLE I consider the latin squares example described at page 157 of the book: Statistics for Experimenters: Design, Innovation, and Discovery by George E. P. Box, J. Stuart Hunter, William G. Hunter. This example use

I am just getting the hang of this, so forgive me if this is a stupid question... - I have two models/tables: Owners and Cars - Owners may have many cars. Cars have one owner. - Cars have a property named Color. Question: how do I get a list of all the Owners who have RED cars? I want something like this to work: Owner.find_by_car.color("RED") Thanks for any help! -- Posted via

Hello all and thank you for reading this message. I am new to RoR, and ruby too, but i am quite excited about it. I am developing a RoR project and with active record associations in the following manner # Tables (all tables have only one record) CREATE TABLE `webcars_development`.`cars` ( `id` int(11) NOT NULL auto_increment, `model_id` int(11) NOT NULL default ''0'',

Muchas gracias Carlos. Tengo las mismas versiones pero no logro separar los chunks cuando uso #+, pero si lo logro cuando uso #". Mi script es spin("prueba.R", knit = FALSE) y si me hace un Rmd. #+ a = c(1,2,3) #+ mean(a) Ejecuto, > spin("prueba.R", knit = FALSE) y obtengo un prueba.Rmd ```{r } a = c(1,2,3) ```{r } mean(a) ``` Pero si escribo: #' a =

Dear all, In my dataset I have 12 columns and 5824 rows. The second column contains information about the height of a claim: it might be zero or positive. I would like to do an analysis on the positive part of this matrix, but I do need the other colums with this. So if there are like 1000 rows where the second column is positive, I would like to extract this 1000x13 matrix into another matrix.

I like the functionality provided by outer=TRUE, but when it comes time to place separate xlabs or ylabs, I always end up 'eyeballing' it on a case-by-case basis. For example, ##begin example require(lattice) cars.lo <- loess(dist ~ speed, cars) print(xyplot(cars.lo$residuals+cars.lo$fitted~cars.lo$x, strip=FALSE, outer=TRUE, layout=c(1,2),

Hi First off, I''m using Ruby 1.9.1p378 and Rails 2.3.8. I was creating a minimal application to test handling of Norwegian special characters when I bumped into this strange problem... I have a simple Car model with fields maker:string and model:string. For the controller I planned to just have an index action do all the work: class CarsController < ApplicationController def index

Hello, I have two dataframes DF1 and DF2 that should be identical but are not (DF1 has some rows that aren't in DF2, and vice versa). I would like to produce a new dataframe DF3 containing rows in DF1 that aren't in DF2 (and similarly DF4 would contain rows in DF2 that aren't in DF1). I have a solution for this problem (see self contained example below) but it's awkward and

if i have a table called cars with columns called year and type, and i create a model and do this @mycars = Cars.find(:all) in my view i have something like this for cars in @mycars do something like print out cars.type i know that @mycars in a hash where each key is a column name paired with its corresponding value right? without the for loop, how can i access the 2nd record directly?

can someone help me with this i have a table cars ( id int model_id int make_id int color_id int description text constraint fk_cars_carmake foreign key(make_id) references carmakes(id), constraint fk_cars_carmodel foreign key(model_id) references carmodels(id), constraint fk_cars_color foreign key(color_id) references colors(id) ) carmakes ( id int carmake varchar ) carmodels ( id

Hello, I know it''s off-topic. But I''m sure you are using SQL and can help me ;) I''ve a table CARS and a table KEYS and a LOCKS table. CARS id|name 1|audi 2|ford 3|mazda 4|porsche ... KEYS id|car_id|lock_id 1|1|1 2|2|1 3|2|2 4|3|1 5|3|2 6|4|1 7|4|2 8|4|3 ... LOCKS id|name 1|main 2|spare 3|engine ... A car can have many keys. Keys are for different locks. How can I

Dear All, I've been?searching for?appropriate codes to compute the rate of change and the curvature?of ?nonparametric regression model whish was denoted by a smooth function?but?unfortunately?don't manage to?do?it. I presume that such characteristics from a smooth curve can be determined by the first and second derivative operators. The following are the example of fitting a

Hi, I want to perform t-test in R for each individual car. Here is what i have. Is there a way i can test each car? The original table (test.csv) Car Tester2 Tester2 Controller1 Controller2 Audi 0.56 0.9 0.5 0.9 Toyota 0.2 0.9 0.9 0.2 Honda 0.5 0.9 0.1 0.5 BMW 0.2 0.3 0.5 0.7 > cars<-read.table("c://test.csv",header=T,sep=",", + skip=2,comment.char="")

Hi, I'm trying to create a plot using pair. Currently I'm doing the following pairs(~mpg + hp + wt , data=cars, labels = c("Miles per Gallon","Horsepower","Weight"),pch = c(24,25)[unclass(cars$tr + 1)], bg = c("red", "green3")[unclass(cars$tr + 1)],panel=panel.smooth) for the attached dataset. However,instead of using panel.smooth I

Dear List, This is just a programming problem which i cannot seem to figure out. I am trying to get a set of power from a test (say, kolmogorov smirnov) out of a distribution (say, G-K distribution) as follows. I am trying to reduce to pain of writing the whole set of data points (p# below) using "for" loop. However, I seem to have some problem in it as the output "M" does not