search for: pearsoned

Hello, I want to use pearson's correlation as distance between observations and then use any centroid based linkage distance (ex. Ward's distance) When linkage distances are formed as the Lance-Williams recursive formulation, they just require the initial distance between observations. See here: http://en.wikipedia.org/wiki/Ward%27s_method It is said that you have to use euclidean

R version 1.3.0 OS: SunOS 5.7, but I think the same problem occurs with Windows An incorrect formula seems to be used to calculate the pearson residuals for a generalized linear model with a binomial response. Here is a simple program which gives (a) the pearson residuals calculated directly, (b) the pearson residuals from glm, and (c) the deviance residuals from glm. The first and last

Hi all, I have a data set (df, n=10 for the sake of simplicity here) where I have two continuous variables (age and weight) and I also have a grouping variable (group, with two levels). I want to run correlations for each group separately (kind of similar to "split file" in SPSS). I've been experimenting with different functions, and I was able to do this correctly using ddply

Dear all, I would like to conduct a Pearson's correlation using cor.test separately for each individual in my data set. Each individual has nine observations of the measurement variable 'Delta13C'. I can figure our how to do it manually one 'Individual_ID' at a time, but I cannot figure out how to automate the process (a loop I am guessing). In addition, I would like to

Hi all, This is a very silly question or something escapes me: Let obj a simple gam poisson model. Let >obj<-gam(....,family=poisson) >obj1<-update(obj, family=quasi(link="log", var="mu")) >From summary.glm(obj1) the dispersion parameter is estimated 1.165; In fact it is: > (predict(obj1, se.fit=T)$se.fit[1:5]/predict(obj, se.fit=T)$se.fit[1:5])^2 4

In running the following example of a loglinear model for the Titanic data, I was surprised to see NaN reported for the Pearson chisq > loglin(Titanic, margin=list(1:3, 4)) 2 iterations: deviation 2.273737e-13 $lrt [1] 671.9622 $pearson [1] NaN $df [1] 15 $margin $margin[[1]] [1] "Class" "Sex" "Age" $margin[[2]] [1] "Survived" Tracing it back,

Hi, Is there any R library/package that calculates tetrachoric correlations from given marginals and Pearson correlations among ordinal variables? Inputs to polychor function in polycor package are either contingency tables or ordinal data themselves. I am looking for something that takes marginal distributions and Pearson correlation as inputs. For example, Y1=(1,2,3) with P(Y1=1)=0.3,

Is there any reason that rstandard.glm doesn't have a "pearson" option? And if not, can it be added? Background: I'm currently teaching an undergrad/grad-service course from Agresti's "Introduction to Categorical Data Analysis (2nd edn)" and deviance residuals are not used in the text. For now I'll just provide the students with a simple function to use, but I

Can anyone show me how to add a log Pearson type III plot using the evdistq() command to an extreme value plot using the lmom package? Attached sample code below... Thanks in advance, Dave library(lmom) # annual maximum daily streamflows Mackenzie River mackenzieRiver = c(26600, 30300, 34000, 32000, 29200, 28300, 28600, 26400, 28300, 28800, 29000, 22100, 32900, 31800, 21600, 32100, 27000,

Full_Name: Bin Nan Version: 1.3.0 OS: Win32 Submission from: (NULL) (141.211.15.110) The function resid(fit, "pearson") seems not giving the same Pearson's residuals for logistic regression as what Splus does. I found the problem when I fitted glm with family = binomial using Crowder's seed data.

Thank you for your reply. I am aware of the good reasons not to use the deviance estimate in binomial, Poisson, and gamma families. However, for the inverse Gaussian, the choice seems to me less clear cut. So I just wanted to compare two different options. I have used the dispersion parameter to compute the standardized deviance residuals: summary(model.gamma)$deviance.resid

Hi all, I'm not quite sure if this is a feature or a bug or if I just fail to understand the documentation: If I use cor() with pairwise.complete.obs and method=pearson, the result is a scalar: ->cor(c(1,2,3),c(3,4,6),use="pairwise.complete.obs",method="pearson") [1] 0.9819805 The documentation says that " '"pairwise.complete.obs"' only

Hi again List, Well this time I’m writing for a friend (really J). He needs to create a distance matrix based on an abundance matrix using the 1-Pearson’s R index. Well I told him to look at the proxy package, but there is only Pearson Index. He needs it to perform a clustering. Well, as soon as he told me there proxy only had the Pearson index I thought: “He could just do something like

Full_Name: Carmen Fernandez Version: 1.3.1 OS: Submission from: (NULL) (138.251.202.115) I think there is a problem when computing Pearson residuals, in that they seem to be computed at the raw residuals divided by the square root of the corresponding diagonal element of the weight matrix W evaluated at the last step of the iterative model fitting procedure (IWLS), instead of dividing by the

Hello, I used the function cor to calculate the pearson correlation coefficient between variables. However, the resulting values do not correspond to the outcome of my excel-calculations, for which I used the formula Cor(x,y)=Cov(x,y)/(SD(x)*SD(y)) So my question is: How does the function "cor" compute the pearson correlation coefficient? Thank you in advance, Ake Nauta

I have used the correlation analysis (pearson) in the agricolae package to analyse my data and got unexpectedly low p-values (therefore making many more highly significant correlations in my data than I had expected). I am wondering if the p-values given should be subtracted from 1 to give the real p-value, because for each variable compared against itself has a p-value of 1 and I thought it

Dear all, I am new to programming in R. I deal with microarray data,which is a data frame object type. I need to carry out a few statistical procedures on this, one of them being the pearson corelation. I need to do this between each row which is a gene. So the desired result is a square matrix with the pearson corelation value between each row. So the first column would be (1,1)=0,(1,2),(1,3)

Dear colleagues, In checking a function I am adding to an R package, I get the following warning pair: ... Bad \usage lines found in documentation object 'nominal': "\\method{print}{nominal}"(x, max.print = 10, posthoc = "std.pearson.residuals.sign", assoc = ifelse("univariate" list(c("N", "alpha.X2",

Is Pearson's Chi-Square test for contingency tables asymptotically unbiased for large tables (large degrees of freedom) regardless of the expected values in each cell? The rule of thumb is that Pearson's Chi-square should not be used when large numbers of cells have expected values < 5. However, I compared the results on 4x4 contingency tables for R's chisq.test using chi-square

Hi all: There was a concern raised by reviewers of a manuscript of mine over the proper execution of a Pearson's correlation. In brief, this was undertaken in order to determine the relationship between the extent of wheel running (y axis) and ethanol intake (x axis) across three, separate 10 day periods in 7 animals. In the paper, the correlational plots for each 10 day-period had 70 data