similar to: error about MCA

Dear Users, I helped to install R Commander and FactoMineR to one of my collegaues. He wanted to do an MCA. Selecting three variables and using the default settings results in only one graph, the variables representation, where he gets three points for the three variables (which is totally fine). Running the code (output of the point-and-click method) EuTop100.MCA<-EuTop100[,

¿Alguien ha utilizado el paquete dynGraph? ¡No logro nada! Gracias anticipadas Jorge *************************************************************************************************** 1 de Agosto de 1961, creación del Ministerio del Transporte. [[alternative HTML version deleted]]

Hi! Apologies if this is not the correct place to ask. I am attempting a MFA analysis of a dataset based on wine chemical and sensory analysis, based on the STHDA tutorial [1]. (I am using this dataset here too, as an example dataset to work on without posting my actual data. I've tried this with both my data and the example data, with the exact same results.) The only issue I am having is

Dear all, i'm trying to label specific individuals (supplementary ones) after a PCA with the FactoMiner package. There is not much details (possibilities?) in the R-help of the plot.pca function. There is indeed a "label" parameter but i could only manage to label the supplementary individuals with there "row.names" (i.e. label="indiv.sup") and not with the

Cordial saludo adjunto la base de datos y el script: eu60<- read.csv(file.choose(), header=T, sep=";", dec=".", row.names=1) eu60.pca <- PCA(eu60, quali.sup=19) eu60.data <- cbind.data.frame(eu60[,19], eu60.pca$ind$coord) eu60.ellipse <- coord.ellipse(eu60.data, bary=TRUE) plot.PCA(ellipse=eu60.ellipse, cex=0.8) El 4 de diciembre de 2013 17:35, Camilo Calle

Cordial saludo, soy principiante en R y estoy haciendo un análisis multivariado de datos, y necesito hacer las elipses de confianza donde la variable 19 es cualitativa suplementaria y el resto son variables activas; tengo el siguiente código: eu60 <- read.csv(file.choose(), header=T, sep=";", dec=",", row.names=1) ### capturar datos eu60.pca <- PCA(eu60, quali.sup=19) ##

Hola, ¿qué tal? Tu problema es que tienes agrupaciones de países (p.e., ASIA_MERIDIOL) con un solo respresentante. Las matrices con las que construyes las elipses son degeneradas (más bien, no existen: sospecho que tienen que ver con la covarianza... ¿de un único caso?). Elimina las agrupaciones triviales. Un saludo, Carlos J. Gil Bellosta http://www.datanalytics.com El día 4 de diciembre de

If there are many character variables,and I want to get the mosaic plot of every pair of each variable,how to do then? If the variables are numeric, I can use pairs to get paired scatter plot. But as to the character variables, how to get the "paired mosaic plot"? Many thanks. -- QQ: 1733768559 At 2015-02-07 17:04:26,"Jim Lemon" <drjimlemon at gmail.com>

hi all: Here's a dataframe(dat) and a vector(z): dat: x1 x2 x3 0.2 1.2 2.5 0.5 2 5 0.8 3 6.2 > z [1] 10 100 100 I wanna do the following: 10*x1,100*x2,1000*x3 My solution is using the loop for z and dat(since the length of z is the same as ncol of dat),which is tedious. I wanna an efficient solution to do it . Any help? Many thanks! My best

Hi,all: I met a problem of nls. My data: x y 60 0.8 80 6.5 100 20.5 120 45.9 I want to fit exp curve of data. My code: > nls(y ~ exp(a + b*x)+d,start=list(a=0,b=0,d=1)) Error in nlsModel(formula, mf, start, wts) : singular gradient matrix at initial parameter estimates I can't find out the reason for the error. Any suggesions are welcome. Many thanks. [[alternative HTML

Hi all: My data is in the attachment. I want to analysis the mean difference of y between 2 sex. My code: result_lm<-lm(y~factor(sex) + x1 + x2) summary(result_lm) The result of "factor(sex)m" 136.83, is the mean difference of y between 2 sex,and the corresponding p value is 0.07618. My question is: how to get the mean y of sex(m) and sex(f) respectively via lm function? Many

Hello everybody, Without any loop and any package, I would like to return N products of M rows in a matrix A : Today, I managed to do it with a loop : B <- matrix(NA, ncol = ncol(A), nrow = 0) for (i in 1 : N) B <- rbind(B, apply(A[sample(1 : nrow(A), M, replace = T), ], 2, prod)) Do you have a solution ? Thank you in advance ! [[alternative HTML version deleted]]

> On 1 Jun 2017, at 22:42, Roy Mendelssohn - NOAA Federal <roy.mendelssohn at noaa.gov> wrote: > > Thanks to all for responses/. There was a question of exactly what was wanted. It is the generalization of the obvious example I gave, > >>>> junk1 <- junk[, rev(seq_len(10), ] > > > so that > > junk[1,1,1 ] = junk1[1,10,1] > junk[1,2,1] =

###Dear R users ###I have been using SensoMineR package from CRAN for most of my work in sensory data analysis and from my usage experience, I encountered some areas for improvement and considered ###modifying the function in SensoMineR package for my personal use. I felt that it could be useful to share this to the community for enabling adoption by other users where they might require a

My error. Clearly I did not do enough testing. z <- array(1:24,dim=2:4) > all.equal(f(z,1),f2(z,1)) [1] TRUE > all.equal(f(z,2),f2(z,2)) [1] TRUE > all.equal(f(z,3),f2(z,3)) [1] "Attributes: < Component ?dim?: Mean relative difference: 0.4444444 >" [2] "Mean relative difference: 0.6109091" # Your earlier example > z <- array(1:120, dim=2:5) >

Hi all: If there are two numeric variable:x,y, and I can get paired scatter plot by function "pairs".But if x and y are character, and I want to get paired mosaic plot,which function should be used then? Many thanks! My best. -- QQ: 1733768559 [[alternative HTML version deleted]]

Hi all: I have a question about multi-comparison. The data is in the attachment. My purpose: Compare the predicted means of the 3 methods(a,b,c) pairwisely. I have 3 ideas: #idea1 result_aov<-aov(y~ method + x1 + x2) TukeyHSD(result_aov) diff lwr upr p adj b-a 0.845 0.5861098 1.1038902 0.0000001 c-a 0.790 0.5311098 1.0488902 0.0000002 c-b -0.055 -0.3138902

On the off chance that anyone is still interested, here is the corrected function using aperm(): z <- array(1:120,dim=2:5) f2 <- function(a, wh) { idx <- seq_len(length(dim(a))) dims <- setdiff(idx, wh) idx <- append(idx[-1], idx[1], wh-1) aperm(apply(a, dims, rev), idx) } all.equal(f(z, 1), f2(z, 1)) # [1] TRUE all.equal(f(z, 2), f2(z, 2)) # [1] TRUE

Thanks again. I am going to try the different versions. But I probably won't be able to get to it till next week. This is probably at the point where anything further should be sent to me privately. -Roy > On Jun 1, 2017, at 1:56 PM, David L Carlson <dcarlson at tamu.edu> wrote: > > On the off chance that anyone is still interested, here is the corrected function using

Here is an alternative approach using apply(). Note that with apply() you are reversing rows or columns not indices of rows or columns so apply(junk, 2, rev) reverses the values in each column not the column indices. We actually need to use rev() on everything but the index we are interested in reversing: f2 <- function(a, wh) { dims <- seq_len(length(dim(a))) dims <-