After I fit a randomForest using the train() function, I'm having problems invoking partialPlot() and plotmo(). Here's some reproducible code:
library(AER)
library(caret)
data(Mortgage)
fitControl <- trainControl(method = "repeatedcv"
,number = 5
,repeats = 10
,allowParallel = TRUE)
library(doMC)
registerDoMC(cores=10)
Final.rfModel <- train(form=networth ~ ., data=Mortgage, method = "rf", metric='RMSE', trControl = fitControl, tuneLength=10, importance = TRUE)
#### partial plots fail
partialPlot(Final.rfModel$finalModel, Mortgage, "liquid")
library(plotmo)
plotmo(Final.rfModel$finalModel)
There is some inconsistency between how some functions (including randomForest and train) handle dummy variables. Most functions in R that use the formula method will convert factor predictors to dummy variables because their models require numerical representations of the data. The exceptions to this are tree- and rule-based models (that can split on categorical predictors), naive Bayes, and a few others.
So randomForest will not create dummy variables when you use randomForest(y ~ ., data = dat) but train (and most others) will using a call like train(y ~ ., data = dat).
The error occurs because rate, married and a few other predictors are factors. The dummy variables created by train don't have the same names so partialPlot can't find them.
Using the non-formula method with train will pass the factor predictors to randomForest and everything will work.
TL;DR
Use the non-formula method with train in this case:
Final.rfModel <- train(form=networth ~ ., data=Mortgage,
method = "rf",
metric='RMSE',
trControl = fitControl,
tuneLength=10,
importance = TRUE)
Max
Related
I am unable to reproduce the only example I can find of using h2o with iml (https://www.r-bloggers.com/2018/08/iml-and-h2o-machine-learning-model-interpretability-and-feature-explanation/) as detailed here (Error when extracting variable importance with FeatureImp$new and H2O). Can anyone point to a workaround or other examples of using iml with h2o?
Reproducible example:
library(rsample) # data splitting
library(ggplot2) # allows extension of visualizations
library(dplyr) # basic data transformation
library(h2o) # machine learning modeling
library(iml) # ML interprtation
library(modeldata) #attrition data
# initialize h2o session
h2o.no_progress()
h2o.init()
# classification data
data("attrition", package = "modeldata")
df <- rsample::attrition %>%
mutate_if(is.ordered, factor, ordered = FALSE) %>%
mutate(Attrition = recode(Attrition, "Yes" = "1", "No" = "0") %>% factor(levels = c("1", "0")))
# convert to h2o object
df.h2o <- as.h2o(df)
# create train, validation, and test splits
set.seed(123)
splits <- h2o.splitFrame(df.h2o, ratios = c(.7, .15), destination_frames =
c("train","valid","test"))
names(splits) <- c("train","valid","test")
# variable names for resonse & features
y <- "Attrition"
x <- setdiff(names(df), y)
# elastic net model
glm <- h2o.glm(
x = x,
y = y,
training_frame = splits$train,
validation_frame = splits$valid,
family = "binomial",
seed = 123
)
# 1. create a data frame with just the features
features <- as.data.frame(splits$valid) %>% select(-Attrition)
# 2. Create a vector with the actual responses
response <- as.numeric(as.vector(splits$valid$Attrition))
# 3. Create custom predict function that returns the predicted values as a
# vector (probability of purchasing in our example)
pred <- function(model, newdata) {
results <- as.data.frame(h2o.predict(model, as.h2o(newdata)))
return(results[[3L]])
}
# create predictor object to pass to explainer functions
predictor.glm <- Predictor$new(
model = glm,
data = features,
y = response,
predict.fun = pred,
class = "classification"
)
imp.glm <- FeatureImp$new(predictor.glm, loss = "mse")
Error obtained:
Error in `[.data.frame`(prediction, , self$class, drop = FALSE): undefined columns
selected
traceback()
1. FeatureImp$new(predictor.glm, loss = "mse")
2. .subset2(public_bind_env, "initialize")(...)
3. private$run.prediction(private$sampler$X)
4. self$predictor$predict(data.frame(dataDesign))
5. prediction[, self$class, drop = FALSE]
6. `[.data.frame`(prediction, , self$class, drop = FALSE)
7. stop("undefined columns selected")
In the iml package documentation, it says that the class argument is "The class column to be returned.". When you set class = "classification", it's looking for a column called "classification" which is not found. At least on GitHub, it looks like the iml package has gone through a fair amount of development since that blog post, so I imagine some functionality may not be backwards compatible anymore.
After reading through the package documentation, I think you might want to try something like:
predictor.glm <- Predictor$new(
model = glm,
data = features,
y = "Attrition",
predict.function = pred,
type = "prob"
)
# check ability to predict first
check <- predictor.glm$predict(features)
print(check)
Even better might be to leverage H2O's extensive functionality around machine learning interpretability.
h2o.varimp(glm) will give the user the variable importance for each feature
h2o.varimp_plot(glm, 10) will render a graphic showing the relative importance of each feature.
h2o.explain(glm, as.h2o(features)) is a wrapper for the explainability interface and will by default provide the confusion matrix (in this case) as well as variable importance, and partial dependency plots for each feature.
For certain algorithms (e.g., tree-based methods), h2o.shap_explain_row_plot() and h2o.shap_summary_plot() will provide the shap contributions.
The h2o-3 docs might be useful here to explore more
All my models are initialized with the below:
def intiailize_clf_models(self):
model = RandomForestClassifier(random_state=42)
self.clf_models.append((model))
model = ExtraTreesClassifier(random_state=42)
self.clf_models.append((model))
model = MLPClassifier(random_state=42)
self.clf_models.append((model))
model = LogisticRegression(random_state=42)
self.clf_models.append((model))
model = xgb.XGBClassifier(random_state=42)
self.clf_models.append((model))
model = lgb.LGBMClassifier(random_state=42)
self.clf_models.append((model))
Which loops through the models and performs k fold cross validation with :
def kfold_cross_validation(self):
clf_models = self.get_models()
models = []
self.results = {}
for model in clf_models:
self.current_model_name = model.__class__.__name__
cross_validate = cross_val_score(model, self.xtrain, self.ytrain, cv=4)
self.mean_cross_validation_score = cross_validate.mean()
print("Kfold cross validation for", self.current_model_name)
self.results[self.current_model_name] = self.mean_cross_validation_score
models.append(model)
Anytime i run this cross validation, i get a different result even after i have set a random state on the different models. I would like to know why i get different results in cross validation and what can be done about it
This is because you did not set the random_state for your k-fold generator. By default when you pass a int value for cv as
cross_validate = cross_val_score(model, self.xtrain, self.ytrain, cv=4)
cross_val_score will call (Stratified)KFold using a different random state with every call causing your model's parameters to differ leading to different results.
The relevant part from the source file.
cv: int, cross-validation generator or an iterable, default=None
Determines the cross-validation splitting strategy.
Possible inputs for cv are:
- None, to use the default 5-fold cross validation,
- int, to specify the number of folds in a `(Stratified)KFold`,
- :term:`CV splitter`,
- An iterable yielding (train, test) splits as arrays of indices.
For int/None inputs, if the estimator is a classifier and ``y`` is
either binary or multiclass, :class:`StratifiedKFold` is used. In all
other cases, :class:`KFold` is used.
To remedy this you can pass your own cross-validation generator with a controlled random state as stated in the documentation above. For example:
# (code untested)
from sklearn.model_selection import StratifiedKFold
skf = StratifiedKFold(n_splits=4, random_state=42)
cross_validate = cross_val_score(model, self.xtrain, self.ytrain, cv=skf)
I found the solution to my question.
Setting a random seed with the below solved the problem:
seed = np.random.seed(22)
This question is very similar to preprocess within cross-validation in caret; however, in a project that i'm working on I would only like to do PCA on three predictors out of 19 in my case. Here is the example from preprocess within cross-validation in caret and I'll use this data (PimaIndiansDiabetes) for ease (this is not my project data but concept should be the same). I would then like to do the preProcess only on a subset of variables i.e. PimaIndiansDiabetes[, c(4,5,6)]. Is there a way to do this?
library(caret)
library(mlbench)
data(PimaIndiansDiabetes)
control <- trainControl(method="cv",
number=5)
p <- preProcess(PimaIndiansDiabetes[, c(4,5,6)], #only do these columns!
method = c("center", "scale", "pca"))
p
grid=expand.grid(mtry=c(1,2,3))
model <- train(diabetes~., data=PimaIndiansDiabetes, method="rf",
preProcess= p,
trControl=control,
tuneGrid=grid)
But I get this error:
Error: pre-processing methods are limited to: BoxCox, YeoJohnson, expoTrans, invHyperbolicSine, center, scale, range, knnImpute, bagImpute, medianImpute, pca, ica, spatialSign, ignore, keep, remove, zv, nzv, conditionalX, corr
The reason I'm trying to do this is so I can reduce three variables to one PCA1 and use for predicting. In the project I'm doing all three variables are correlated above 90% but would like to incorporate them as other studies have used them as well. Thanks. Trying to avoid data leakage!
As far as I know this is not possible with caret.
This might be possible using recipes. However I do not use recipes but I do use mlr3 so I will show how to do it with this package:
library(mlr3)
library(mlr3pipelines)
library(mlr3learners)
library(paradox)
library(mlr3tuning)
library(mlbench)
create a task from the data:
data("PimaIndiansDiabetes")
pima_tsk <- TaskClassif$new(id = "Pima",
backend = PimaIndiansDiabetes,
target = "diabetes")
define a pre process selector named "slct1":
pos1 <- po("select", id = "slct1")
and define the selector function within it:
pos1$param_set$values$selector <- selector_name(colnames(PimaIndiansDiabetes[, 4:6]))
now define what should happen to the selected features: scaling -> pca with 1st PC selected (param_vals = list(rank. = 1))
pos1 %>>%
po("scale", id = "scale1") %>>%
po("pca", id = "pca1", param_vals = list(rank. = 1)) -> pr1
now define an invert selector:
pos2 <- po("select", id = "slct2")
pos2$param_set$values$selector <- selector_invert(pos1$param_set$values$selector)
define the learner:
rf_lrn <- po("learner", lrn("classif.ranger")) #ranger is a faster version of rf
combine them:
gunion(list(pr1, pos2)) %>>%
po("featureunion") %>>%
rf_lrn -> graph
check if it looks ok:
graph$plot(html = TRUE)
convert graph to a learner:
glrn <- GraphLearner$new(graph)
define parameters you want tuned:
ps <- ParamSet$new(list(
ParamInt$new("classif.ranger.mtry", lower = 1, upper = 6),
ParamInt$new("classif.ranger.num.trees", lower = 100, upper = 1000)))
define resampling:
cv10 <- rsmp("cv", folds = 10)
define tuning:
instance <- TuningInstance$new(
task = pima_tsk,
learner = glrn,
resampling = cv10,
measures = msr("classif.ce"),
param_set = ps,
terminator = term("evals", n_evals = 20)
)
set.seed(1)
tuner <- TunerRandomSearch$new()
tuner$tune(instance)
instance$result
For additional details on how to tune the number of PC components to keep check this answer: R caret: How do I apply separate pca to different dataframes before training?
If you find this interesting check out the mlr3book
Also
cor(PimaIndiansDiabetes[, 4:6])
triceps insulin mass
triceps 1.0000000 0.4367826 0.3925732
insulin 0.4367826 1.0000000 0.1978591
mass 0.3925732 0.1978591 1.0000000
does not produce what you mention in the question.
I am using mlr package's resample() function to subsample a random forest model 4000 times (the code snippet below).
As you can see, to create random forest models within resample() I'm using randomForest package.
I want to get random forest model's importance results (mean decrease in accuracy over all classes) for each of the subsample iterations. What I can get right now as the importance measure is the mean decrease in Gini index.
What I can see from the source code of mlr, getFeatureImportanceLearner.classif.randomForest() function (line 69) in makeRLearner.classif.randomForest uses randomForest::importance() function (line 83) to get importance value from the resulting object of randomForest class. But as you can see from the source code (line 73) it uses 2L as the default value. I want it to use 1L (line 75) as the value (mean decrease in accuracy).
How can I pass the value of 2L to resample() function, ("extract = getFeatureImportance" line in the code below) so that getFeatureImportanceLearner.classif.randomForest() function gets that value and sets ctrl$type = 2L (line 73)?
rf_task <- makeClassifTask(id = 'task',
data = data[, -1], target = 'target_var',
positive = 'positive_var')
rf_learner <- makeLearner('classif.randomForest', id = 'random forest',
par.vals = list(ntree = 1000, importance = TRUE),
predict.type = 'prob')
base_subsample_instance <- makeResampleInstance(rf_boot_desc, rf_task)
rf_subsample_result <- resample(rf_learner, rf_task,
base_subsample_instance,
extract = getFeatureImportance,
measures = list(acc, auc, tpr, tnr,
ppv, npv, f1, brier))
My solution: Downloaded source code of the mlr package. Changed the source file line 73 to 1L (https://github.com/mlr-org/mlr/blob/v2.15.0/R/RLearner_classif_randomForest.R). Installed the package from command line and used it. Not an optimal solution but a solution.
You provide a lot of specifics that do not actually relate to your question, at least how I understood it.
So I wrote a simple MWE that includes the answer.
The idea is that you have to write a short wrapper for getFeatureImportance so that you can pass your own arguments. Fans of purrr can do that with purrr::partial(getFeatureImportance, type = 2) but here I wrote myExtractor manually.
library(mlr)
rf_learner <- makeLearner('classif.randomForest', id = 'random forest',
par.vals = list(ntree = 100, importance = TRUE),
predict.type = 'prob')
measures = list(acc, auc, tpr, tnr,
ppv, npv, f1, brier)
myExtractor = function(.model, ...) {
getFeatureImportance(.model, type = 2, ...)
}
res = resample(rf_learner, sonar.task, cv10,
measures = measures, extract = myExtractor)
# first feature importance result:
res$extract[[1]]
# all values in a matrix:
sapply(res$extract, function(x) x$res)
If you want to do a bootstraped learenr maybe you should also have a look at makeBaggingWrapper instead of solving this problem through resample.
In every book and example always they show only binary classification (two classes) and new vector can belong to any one class.
Here the problem is I have 4 classes(c1, c2, c3, c4). I've training data for 4 classes.
For new vector the output should be like
C1 80% (the winner)
c2 10%
c3 6%
c4 4%
How to do this? I'm planning to use libsvm (because it most popular). I don't know much about it. If any of you guys used it previously please tell me specific commands I'm supposed to use.
LibSVM uses the one-against-one approach for multi-class learning problems. From the FAQ:
Q: What method does libsvm use for multi-class SVM ? Why don't you use the "1-against-the rest" method ?
It is one-against-one. We chose it after doing the following comparison: C.-W. Hsu and C.-J. Lin. A comparison of methods for multi-class support vector machines, IEEE Transactions on Neural Networks, 13(2002), 415-425.
"1-against-the rest" is a good method whose performance is comparable to "1-against-1." We do the latter simply because its training time is shorter.
Commonly used methods are One vs. Rest and One vs. One.
In the first method you get n classifiers and the resulting class will have the highest score.
In the second method the resulting class is obtained by majority votes of all classifiers.
AFAIR, libsvm supports both strategies of multiclass classification.
You can always reduce a multi-class classification problem to a binary problem by choosing random partititions of the set of classes, recursively. This is not necessarily any less effective or efficient than learning all at once, since the sub-learning problems require less examples since the partitioning problem is smaller. (It may require at most a constant order time more, e.g. twice as long). It may also lead to more accurate learning.
I'm not necessarily recommending this, but it is one answer to your question, and is a general technique that can be applied to any binary learning algorithm.
Use the SVM Multiclass library. Find it at the SVM page by Thorsten Joachims
It does not have a specific switch (command) for multi-class prediction. it automatically handles multi-class prediction if your training dataset contains more than two classes.
Nothing special compared with binary prediction. see the following example for 3-class prediction based on SVM.
install.packages("e1071")
library("e1071")
data(iris)
attach(iris)
## classification mode
# default with factor response:
model <- svm(Species ~ ., data = iris)
# alternatively the traditional interface:
x <- subset(iris, select = -Species)
y <- Species
model <- svm(x, y)
print(model)
summary(model)
# test with train data
pred <- predict(model, x)
# (same as:)
pred <- fitted(model)
# Check accuracy:
table(pred, y)
# compute decision values and probabilities:
pred <- predict(model, x, decision.values = TRUE)
attr(pred, "decision.values")[1:4,]
# visualize (classes by color, SV by crosses):
plot(cmdscale(dist(iris[,-5])),
col = as.integer(iris[,5]),
pch = c("o","+")[1:150 %in% model$index + 1])
data=load('E:\dataset\scene_categories\all_dataset.mat');
meas = data.all_dataset;
species = data.dataset_label;
[g gn] = grp2idx(species); %# nominal class to numeric
%# split training/testing sets
[trainIdx testIdx] = crossvalind('HoldOut', species, 1/10);
%# 1-vs-1 pairwise models
num_labels = length(gn);
clear gn;
num_classifiers = num_labels*(num_labels-1)/2;
pairwise = zeros(num_classifiers ,2);
row_end = 0;
for i=1:num_labels - 1
row_start = row_end + 1;
row_end = row_start + num_labels - i -1;
pairwise(row_start : row_end, 1) = i;
count = 0;
for j = i+1 : num_labels
pairwise( row_start + count , 2) = j;
count = count + 1;
end
end
clear row_start row_end count i j num_labels num_classifiers;
svmModel = cell(size(pairwise,1),1); %# store binary-classifers
predTest = zeros(sum(testIdx),numel(svmModel)); %# store binary predictions
%# classify using one-against-one approach, SVM with 3rd degree poly kernel
for k=1:numel(svmModel)
%# get only training instances belonging to this pair
idx = trainIdx & any( bsxfun(#eq, g, pairwise(k,:)) , 2 );
%# train
svmModel{k} = svmtrain(meas(idx,:), g(idx), ...
'Autoscale',true, 'Showplot',false, 'Method','QP', ...
'BoxConstraint',2e-1, 'Kernel_Function','rbf', 'RBF_Sigma',1);
%# test
predTest(:,k) = svmclassify(svmModel{k}, meas(testIdx,:));
end
pred = mode(predTest,2); %# voting: clasify as the class receiving most votes
%# performance
cmat = confusionmat(g(testIdx),pred);
acc = 100*sum(diag(cmat))./sum(cmat(:));
fprintf('SVM (1-against-1):\naccuracy = %.2f%%\n', acc);
fprintf('Confusion Matrix:\n'), disp(cmat)
For multi class classification using SVM;
It is NOT (one vs one) and NOT (one vs REST).
Instead learn a two-class classifier where the feature vector is (x, y) where x is data and y is the correct label associated with the data.
The training gap is the Difference between the value for the correct class and the value of the nearest other class.
At Inference choose the "y" that has the maximum
value of (x,y).
y = arg_max(y') W.(x,y') [W is the weight vector and (x,y) is the feature Vector]
Please Visit link:
https://nlp.stanford.edu/IR-book/html/htmledition/multiclass-svms-1.html#:~:text=It%20is%20also%20a%20simple,the%20label%20of%20structural%20SVMs%20.