Ankur Chakravarthy
  • Member for 8 years, 4 months
  • Last seen more than 3 years ago
Measures of variable importance in random forests
11 votes

Random Forest importance metrics as implemented in the randomForest package in R have quirks in that correlated predictors get low importance values. http://bioinformatics.oxfordjournals.org/content/...

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How to perform a Binomial Test when having replicates?
Accepted answer
3 votes

Try fitting a binomial Generalized Linear Model - in R , if you have a dataframe called DF with numbers of successes (called "irregular") and failures ("regular") , and a column for treatment/group ...

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Machine learning algorithms to handle missing data
3 votes

Try imputation using nearest neighbours to get rid of missing data. Additionally, the Caret package has interfaces to a wide variety of algorithms and they all come with predict methods in R that ...

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Performance Evaluation for multiple classifiers in machine learning with Wilcoxon Rank Sum Test
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2 votes

You are right about significance and P.values. I am not sure Wilcoxon's Rank Sum test is the best way to compare accuracies, however, and hopefully someone will shed more light on this. I'd be ...

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Random Forest Regression Overfitting - Quantile Test on Test Data
2 votes

The Caret package enables you to run randomForests with a range of parameters for .mtry so you can optimise, using ROC values, the number of features that gives you optimal classification. This ...

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How should I correct for multiple comparisons?
1 votes

Depends on how much correction you want to do and how conservative you want to be. Correcting for multiple testing across all the tests you've carried out is the most conservative option, but in lots ...

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How can you visualize the relationship between 3 categorical variables?
1 votes

Similar to parallel sets, as posted by nazareno above, you can use alluvial plots which are available from the alluvial R package. http://www.r-bloggers.com/alluvial-diagrams/

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RandomForest failing to learn single-factor, 'linearly separable' model
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0 votes

So now that the situation is clearer; feeding factors in one by one is not a reliable way of quantifying what variables are driving performance; look up the "importance" function to estimate how ...

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Cluster Data based on Distribution
0 votes

Non-negative matrix factorisation seems suited to this kind of problem; alternatively, give Random Forests a go, but I generally use R and am not sure if Python implementations do pattern discovery (...

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Independent t-test: non-significant different!
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0 votes

What is the P.value of the difference between your means? This will be the number of times out of a 100 that you will expect to see that difference just by chance. So if you have a P.value of 0.13, it ...

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