grid of mtry values while training random forests with ranger - random-forest

I am working with a subset of the 'Ames Housing' dataset and have originally 17 features. Using the 'recipes' package, I have engineered the original set of features and created dummy variables for nominal predictors with the following code. That has resulted in 35 features in the 'baked_train' dataset below.
blueprint <- recipe(Sale_Price ~ ., data = _train) %>%
step_nzv(Street, Utilities, Pool_Area, Screen_Porch, Misc_Val) %>%
step_impute_knn(Gr_Liv_Area) %>%
step_integer(Overall_Qual) %>%
step_normalize(all_numeric_predictors()) %>%
step_other(Neighborhood, threshold = 0.01, other = "other") %>%
step_dummy(all_nominal_predictors(), one_hot = FALSE)
prepare <- prep(blueprint, data = ames_train)
baked_train <- bake(prepare, new_data = ames_train)
baked_test <- bake(prepare, new_data = ames_test)
Now, I am trying to train random forests with the 'ranger' package using the following code.
cv_specs <- trainControl(method = "repeatedcv", number = 5, repeats = 5)
param_grid_rf <- expand.grid(mtry = seq(1, 35, 1),
splitrule = "variance",
min.node.size = 2)
rf_cv <- train(blueprint,
data = ames_train,
method = "ranger",
trControl = cv_specs,
tuneGrid = param_grid_rf,
metric = "RMSE")
I have set the grid of 'mtry' values based on the number of features in the 'baked_train' data. It is my understanding that 'caret' will apply the blueprint within each resample of 'ames_train' creating a baked version at each CV step.
The text Hands-On Machine Learning with R by Boehmke & Greenwell says on section 3.8.3,
Consequently, the goal is to develop our blueprint, then within each resample iteration we want to apply prep() and bake() to our resample training and validation data. Luckily, the caret package simplifies this process. We only need to specify the blueprint and caret will automatically prepare and bake within each resample.
However, when I run the code above I get an error,
mtry can not be larger than number of variables in data. Ranger will EXIT now.
I get the same error when I specify 'tuneLength = 20' instead of the 'tuneGrid'. Although the code works fine when the grid of 'mtry' values is specified to be from 1 to 17 (the number of features in the original training data 'ames_train').
When I specify a grid of 'mtry' values from 1 to 17, info about the final model after CV is shown below. Notice that it mentions Number of independent variables: 35 which corresponds to the 'baked_train' data, although specifying a grid from 1 to 35 throws an error.
Type: Regression
Number of trees: 500
Sample size: 618
Number of independent variables: 35
Mtry: 15
Target node size: 2
Variable importance mode: impurity
Splitrule: variance
OOB prediction error (MSE): 995351989
R squared (OOB): 0.8412147
What am I missing here? Specifically, why do I have to specify the number of features in 'ames_train' instead of 'baked_train' when essentially 'caret' is supposed to create a baked version before fitting and evaluating the model for each resample?
Thanks.

Related

Overcoming compatibility issues with using iml from h2o models

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

LSTM sequence prediction overfits on one specific value only

hello guys i am new in machine learning. I am implementing federated learning on with LSTM to predict the next label in a sequence. my sequence looks like this [2,3,5,1,4,2,5,7]. for example, the intention is predict the 7 in this sequence. So I tried a simple federated learning with keras. I used this approach for another model(Not LSTM) and it worked for me, but here it always overfits on 2. it always predict 2 for any input. I made the input data so balance, means there are almost equal number for each label in last index (here is 7).I tested this data on simple deep learning and greatly works. so it seems to me this data mybe is not suitable for LSTM or any other issue. Please help me. This is my Code for my federated learning. Please let me know if more information is needed, I really need it. Thanks
def get_lstm(units):
"""LSTM(Long Short-Term Memory)
Build LSTM Model.
# Arguments
units: List(int), number of input, output and hidden units.
# Returns
model: Model, nn model.
"""
model = Sequential()
inp = layers.Input((units[0],1))
x = layers.LSTM(units[1], return_sequences=True)(inp)
x = layers.LSTM(units[2])(x)
x = layers.Dropout(0.2)(x)
out = layers.Dense(units[3], activation='softmax')(x)
model = Model(inp, out)
optimizer = keras.optimizers.Adam(lr=0.01)
seqLen=8 -1;
global_model = Mymodel.get_lstm([seqLen, 64, 64, 15]) # 14 categories we have , array start from 0 but never can predict zero class
global_model.compile(loss="sparse_categorical_crossentropy", optimizer=optimizer, metrics=tf.keras.metrics.SparseTopKCategoricalAccuracy(k=1))
def main(argv):
for comm_round in range(comms_round):
print("round_%d" %( comm_round))
scaled_local_weight_list = list()
global_weights = global_model.get_weights()
np.random.shuffle(train)
temp_data = train[:]
# data divided among ten users and shuffled
for user in range(10):
user_data = temp_data[user * userDataSize: (user+1)*userDataSize]
X_train = user_data[:, 0:seqLen]
X_train = np.asarray(X_train).astype(np.float32)
Y_train = user_data[:, seqLen]
Y_train = np.asarray(Y_train).astype(np.float32)
local_model = Mymodel.get_lstm([seqLen, 64, 64, 15])
X_train = np.reshape(X_train, (X_train.shape[0], X_train.shape[1], 1))
local_model.compile(loss="sparse_categorical_crossentropy", optimizer=optimizer, metrics=tf.keras.metrics.SparseTopKCategoricalAccuracy(k=1))
local_model.set_weights(global_weights)
local_model.fit(X_train, Y_train)
scaling_factor = 1 / 10 # 10 is number of users
scaled_weights = scale_model_weights(local_model.get_weights(), scaling_factor)
scaled_local_weight_list.append(scaled_weights)
K.clear_session()
average_weights = sum_scaled_weights(scaled_local_weight_list)
global_model.set_weights(average_weights)
predictions=global_model.predict(X_test)
for i in range(len(X_test)):
print('%d,%d' % ((np.argmax(predictions[i])), Y_test[i]),file=f2 )
I could find some reasons for my problem, so I thought I can share it with you:
1- the proportion of different items in sequences are not balanced. I mean for example I have 1000 of "2" and 100 of other numbers, so after a few rounds the model fitted on 2 because there are much more data for specific numbers.
2- I changed my sequences as there are not any two items in a sequence while both have same value. so I could remove some repetitive data from the sequences and make them more balance. maybe it is not the whole presentation of activities but in my case it makes sense.

Getting a specific random forest variable importance measure from mlr package's resample function

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.

How to add cluster label columns back into original dataframe- python, for supervised learning

I have a column in my data frame which contains Url information. It has 1200+ unique values. I wanted to use text mining to generate features from these values. I have used tfidfvectorizer to generate vectors and then used kmeans to identify clusters. I now want to assign these cluster labels back into my original dataframe, so that I can bin the URL information into these clusters.
Below code to generate vectors and cluster labels
from scipy.spatial.distance import cdist
vectorizer = TfidfVectorizer(min_df = 1,lowercase = False, ngram_range = (1,1), use_idf = True, stop_words='english')
X = vectorizer.fit_transform(sample\['lead_lead_source_modified'\])
X = X.toarray()
distortions=\[\]
K = range(1,10)
for k in K:
kmeanModel = KMeans(n_clusters=k).fit(X)
kmeanModel.fit(X)
distortions.append(sum(np.min(cdist(X, kmeanModel.cluster_centers_, 'euclidean'), axis=1)) / X.shape\[0\])
#append cluster labels
km = KMeans(n_clusters=4, random_state=0)
km.fit_transform(X)
cluster_labels = km.labels_
cluster_labels = pd.DataFrame(cluster_labels, columns=\['ClusterLabel_lead_lead_source'\])
cluster_labels
Through the elbow method, I decided on 4 clusters. I now have cluster labels, but I am not sure how to add them bank to dataframe on its respective index. Concatenating along axis=1 is creating Nans due to indexing issues. Below is the sample output after concatenation.
lead_lead_source_modified ClusterLabel_lead_lead_source
0 NaN 3.0
1 NaN 0.0
2 NaN 0.0
3 ['direct', 'salesline', 'website', ''] 0.0
I want to know if this approach is the right way to do, if so then how to solve this issue. If not, is there a better way to do.
Adding index value during dataframe conversion solved the issue.
But it still want to know if this is the right approach

What is the meaning of the GridSearchCV best_score_ attribute? (the value is different from the mean of the cross validation array)

I'm confused with the results, probably I'm not getting the concept of cross validation and GridSearch right. I had followed the logic behind this post:
https://randomforests.wordpress.com/2014/02/02/basics-of-k-fold-cross-validation-and-gridsearchcv-in-scikit-learn/
argd = CommandLineParser(argv)
folder,fname=argd['dir'],argd['fname']
df = pd.read_csv('../../'+folder+'/Results/'+fname, sep=";")
explanatory_variable_columns = set(df.columns.values)
response_variable_column = df['A']
explanatory_variable_columns.remove('A')
y = np.array([1 if e else 0 for e in response_variable_column])
X =df[list(explanatory_variable_columns)].as_matrix()
kf_total = KFold(len(X), n_folds=5, indices=True, shuffle=True, random_state=4)
dt=DecisionTreeClassifier(criterion='entropy')
min_samples_split_range=[x for x in range(1,20)]
dtgs=GridSearchCV(estimator=dt, param_grid=dict(min_samples_split=min_samples_split_range), n_jobs=1)
scores=[dtgs.fit(X[train],y[train]).score(X[test],y[test]) for train, test in kf_total]
# SAME AS DOING: cross_validation.cross_val_score(dtgs, X, y, cv=kf_total, n_jobs = 1)
print scores
print np.mean(scores)
print dtgs.best_score_
RESULTS OBTAINED:
# score [0.81818181818181823, 0.78181818181818186, 0.7592592592592593, 0.7592592592592593, 0.72222222222222221]
# mean score 0.768
# .best_score_ 0.683486238532
ADDITIONAL NOTE:
I ran it using another combination of the explanatory variables (using only some of them) and I got the inverse problem. Now the .best_score_ is higher than all the values in the cross validation array.
# score [0.74545454545454548, 0.70909090909090911, 0.79629629629629628, 0.7407407407407407, 0.64814814814814814]
# mean score 0.728
# .best_score_ 0.802752293578
The code is confusing several things.
dtgs.fit(X[train_],y[train_]) does internal 3-fold cross-validation for every parameter combination from param_grid, producing a grid of 20 results, which you can open by calling dtgs.grid_scores_.
[dtgs.fit(X[train_],y[train_]).score(X[test],y[test]) for train_, test in kf_total] Therefore this line fits grid search five times and then takes its score using 5-Fold cross validation. The result is the array of scores of 5-Fold validation.
And when you call dtgs.best_score_ you get the best score in the grid of the results of 3-fold validation of hyperparameters for the last fit (of 5).

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