search for: choularton

Hi I am trying to do a large glm and running into this message. Error: cannot allocate vector of size 3725426 Kb In addition: Warning message: Reached total allocation of 494Mb: see help(memory.size) Am I simply out of memory (I only have .5 gig)? Is there something I can do? Stephen [[alternative HTML version deleted]]

Hi I am just starting to look at R and trading in futures, stock, etc Can anyone point me to useful background material? Stephen Choularton 02 9999 2226 0413 545 182 -- 11:39 PM [[alternative HTML version deleted]]

Is there some decision tree method available with R, like ripper, that ends up producing a list of the rules and can be used for prediction? Stephen Choularton 02 9999 2226 0413 545 182 -- 5:40 PM [[alternative HTML version deleted]]

I want to make a glm and then use predict. I have a fairly small sample (4000 cases) and I want to train on 90% and test on 10% but I want to do it in slices so I test on every 10th case and train on the others. Is there some simple way to get these elements? Stephen -- 21/10/2005 [[alternative HTML version deleted]]

Hi Say I have some data, two columns in a table being a binary outcome plus a predictor and I want to plot a graph that shows the percentage positives of the binary outcome within bands of the predictor, e.g. Outcome predictor 0 1 1 2 1 2 0 3 0 3 0

Hi I am trying to do this: chisq.test(c(11, 13, 12, 18, 21, 43, 15, 12, 9, 10, 5, 28, 22, 11, 15, 11, 18, 28, 16, 8, 15, 19, 44, 18, 11, 23, 15, 23, 2, 5, 4, 14, 3, 22, 9, 0, 6, 19, 15, 32, 3, 16, 14, 10, 24, 16, 24, 31, 29, 28, 16, 26, 11, 11, 4, 17, 16, 13, 20, 26, 16, 19, 34, 19, 17, 14, 22, 25, 17, 12, 23, 14, 19, 30, 18, 10, 23, 21, 17, 16, 10, 14, 6, 17, 17, 10, 21, 25, 20, 4, 11, 4,

I am working with a largish dataset of 25k lines and I am now tying to use predict. pred = predict(cuDataGlmModel, length + meanPitch + minimumPitch + maximumPitch + meanF1 + meanF2 + meanF3 + meanF4 + meanF5 + ratioF1ToF2 + rationF3ToF1 + jitter + shimmer + percentUnvoicedFrames + numberOfVoiceBreaks + percentOfVoiceBreaks + meanIntensity + minimumIntensity + maximumIntensity +

Hi I am looking for a couple of pointers using glm (family = binary). 1. I want to add all the products of my predictive features as additional features (and I have 23 of them). Is there some easy way to add them? 2. I want to drop each feature in turn and get the most significant, then drop two and get the next most significant, etc. Is there some function that allows me to do this?

Hi I am using R for logistic regression and finding it very useful. However, I wondered if anyone could point me to any course or notes on this subject using R. All help most welcome. Stephen -- Internal Virus Database is out-of-date. Checked by AVG Anti-Virus. [[alternative HTML version deleted]]

Hi Is there some function R that multiplies each coefficient by the standard deviation of the corresponding variable and produces a ranking? Stephen -- No virus found in this outgoing message. Checked by AVG Anti-Virus. [[alternative HTML version deleted]]

Hi I am trying to use this function. Can anyone show me how I would input the following example? Chi-Squared = (40-30)^2 + (20-30)^2 + (30-30)^2 30 30 30 = 3.333 + 3.333 + 0 = 6.666 (p value = 0.036) I want to be able to use different denominators so can you show me how I can do it to accommodate these rather than assuming they are all the

Hi I am running chisq as below and getting a warning. Can anyone tell me the significance or the warning? > chisq.test(c(10 ,4 ,2 ,6 ,5 ,3 ,4 ,4 ,6 ,3 ,2 ,2 ,2 ,4 ,7 ,10 ,0 ,6 ,19 ,3 ,2 ,7 ,2 ,2 ,2 ,1 ,32 ,2 ,3 ,10 ,1 ,3 ,9 ,4 ,10 ,2 ,2 ,4 ,5 ,7 ,6 ,3 ,7 ,4 ,3 ,3 ,7 ,1 ,4 ,2 ,2 ,3 ,3 ,5 ,5 ,4 ), p =c(0.01704142 ,0.017988166 ,0.018224852 ,0.017751479 ,0.017988166 ,0.018224852 ,0.017278107

Hi I am trying to use randomForest for classification. I am using this code: > set.seed(71) > rf.model <- randomForest(similarity ~ ., data=set1[1:100,], importance=TRUE, proximity=TRUE) Warning message: The response has five or fewer unique values. Are you sure you want to do regression? in: randomForest.default(m, y, ...) > rf.model Call: randomForest(x = similarity ~ .,

...ope this isn't off topics, but I have always found when I stepAIC() some glm I get an improvement in accuracy and kappa, but I have just done a case where I got a marginal deterioration. Is this possible, or should I be going through my figures carefully to see if I have messed up? Stephen Choularton 02 9999 2226 0413 545 182 -------------- next part -------------- Checked by AVG Free Edition.

....386 = 8346 / 6022 Misclassification error rate: 0.487 = 2933 / 6023 > plot(tree.model); text(tree.model) Error in plot.tree(tree.model) : cannot plot singlenode tree > I'm not getting any sort of tree formed. I wondered if anyone could point me in the right direction. Thanks. Stephen Choularton [[alternative HTML version deleted]]

Hi I can get all my features by doing this: > logistic.model = glm(similarity ~ ., family=binomial, data = cData[3001:3800,]) I can get the product of all my features by this: logistic.model = glm(similarity ~ . ^ 2, family=binomial, data = cData[3001:3800,]) I don't seem to be able to get polys by doing this: logistic.model = glm(similarity ~ poly(.,2), family=binomial, data

Hi I am working with this data: my data summary is: > summary(spi) open high low close volume Min. :4315 Min. :4365 Min. :4301 Min. :4352 Min. : 0 1st Qu.:4480 1st Qu.:4497 1st Qu.:4458 1st Qu.:4475 1st Qu.:11135 Median :4609 Median :4631 Median :4594 Median :4614 Median :14439 Mean :4620

I am trying to do some verification across a large dataset, cuData, that has 23 columns. Column 23 (similarity) is the outcome 0 or 1 and the other columns are the features. I do this: verificationglm.model <- glm(formula = similarity ~ ., family=binomial, data=cuData[1:1000,]) and produce the model: > summary(verificationglm.model) Call: glm(formula = similarity ~ ., family =

Hi I am trying to fit an svm to predict speech recognition errors. I am using best.svm like this: svm.model = best.svm(data[1:3000,1:23],data[1:3000,24],tunecontrol = tune.control()) I got this: > print(svm.model) Call: best.svm(x = data[1:3000, 1:23], tunecontrol = tune.control(), data[1:3000, 24]) Parameters: SVM-Type: eps-regression SVM-Kernel: radial cost: 1

I want to use the LPC to estimate the vocal tract of various speakers making short utterances. Can anyone point me in the direction of a script that extracts the correct features Stephen -- Internal Virus Database is out-of-date. Checked by AVG Anti-Virus. [[alternative HTML version deleted]]