surv.gbm in the mlr3 framework outputs linear predictors, however what I'm really interested in are predicted survival times per case, which I want to compare with the actual survival times. Is there a way to obtain actual survival times?
In the mlr3 book, there is an example of a transformation between linear predictors and a distribution.
pod = po("distrcompose", param_vals = list(form = "ph", overwrite = FALSE))
prediction = pod$predict(list(base = prediction_distr, pred = prediction_lp))$output
Is there a way to change this pipeline so that it converts "lp" to "response" ?
Any help would be appriciated.
Yes this is definitely possible it just requires another transformation. Your first step is correct to compose a distribution from a linear predictor; as you're using surv.gbm only Cox PH is possible as the underlying model so default for distrcompose works for this.
Now you need to use crankcompose in order to create a survival time prediction from the distribution, you could use the mean, median, or mode of the distribution, people usually pick mean or median but that's your choice! Just make sure to include response = TRUE, overwrite = FALSE. Example code below, includes creating predictions and scoring with RMSE (surprisingly quite good!). I think the book may need updating...
Thanks,
Raphael
library(mlr3extralearners)
library(mlr3proba)
library(mlr3pipelines)
library(mlr3)
learn = ppl("crankcompositor", ppl("distrcompositor", lrn("surv.gbm")),
response = TRUE, overwrite = FALSE, method = "mean",
graph_learner = TRUE)
set.seed(1)
task = tgen("simsurv")$generate(50)
learn$train(task)
p = learn$predict(task)
p$score(msr("surv.rmse"))
Related
I'm trying to build a regression model with R using lightGBM,
and i'm getting a bit confused with some functions and when/how to use them.
First one is what i've written in the title, what's the difference between lgb.train() and lightgbm()?
The description in the documentation(https://cran.r-project.org/web/packages/lightgbm/lightgbm.pdf) says that lgb.train is 'Logic to train with LightGBM' and lightgbm is 'Simple interface for training a LightGBM model', while both their outcome value is lgb.Booster, a trained model.
One difference I've found is that lgb.train() does not work with valids = , while lightgbm() does.
Second one is about a function lgb.cv(), regarding a cross validation in lightGBM. How do you apply the output of lgb.cv() to a model?
As I understood from the documentation i've linked above, it seems like the output of both lgb.cv and lgb.train is a model.
Is it correct to use it like the example below?
lgbcv <- lgb.cv(params,
lgbtrain,
nrounds = 1000,
nfold = 5,
early_stopping_rounds = 100,
learning_rate = 1.0)
lgbcv <- lightgbm(params,
lgbtrain,
nrounds = 1000,
early_stopping_rounds = 100,
learning_rate = 1.0)
Thank you in advance!
what's the difference between lgb.train() and lightgbm()?
These functions both train a LightGBM model, they're just slightly different interfaces. The biggest difference is in how training data are prepared. LightGBM training requires a special LightGBM-specific representation of the training data, called a Dataset. To use lgb.train(), you have to construct one of these beforehand with lgb.Dataset(). lightgbm(), on the other hand, can accept a data frame, data.table, or matrix and will create the Dataset object for you.
Choose whichever method you feel has a more friendly interface...both will produce a single trained LightGBM model (class "lgb.Booster").
that lgb.train() does not work with valids = , while lightgbm() does.
This is not correct. Both functions accept the keyword argument valids. Run ?lgb.train and ?lightgbm for documentation on those methods.
How do you apply the output of lgb.cv() to a model?
I'm not sure what you mean, but you can find an example of how to use lgb.cv() in the docs that show up when you run ?lgb.cv.
library(lightgbm)
data(agaricus.train, package = "lightgbm")
train <- agaricus.train
dtrain <- lgb.Dataset(train$data, label = train$label)
params <- list(objective = "regression", metric = "l2")
model <- lgb.cv(
params = params
, data = dtrain
, nrounds = 5L
, nfold = 3L
, min_data = 1L
, learning_rate = 1.0
)
This returns an object of class "lgb.CVBooster". That object has multiple "lgb.Booster" objects in it (the trained models that lightgbm() or lgb.train() produce).
You can extract any one of these from model$boosters. However, in practice I don't recommend using the models from lgb.cv() directly. The goal of cross-validation is to get an estimate of the generalization error for a model. So you can use lgb.cv() to figure out the expected error for a given dataset + set of parameters (by looking at model$record_evals and model$best_score).
I am going through the book "Hands-on Time series analysis with R" and I am stuck at the example using machine learning h2o package. I don't get how to use h2o.predict function. In the example it requires newdata argument, which is test data in this case. But how do you predict future values of time series if you in fact don't know these values ?
If I just ignore newdata argument I get : predictions with a missing newdata argument is not implemented yet.
library(h2o)
h2o.init(max_mem_size = "16G")
train_h <- as.h2o(train_df)
test_h <- as.h2o(test_df)
forecast_h <- as.h2o(forecast_df)
x <- c("month", "lag12", "trend", "trend_sqr")
y <- "y"
rf_md <- h2o.randomForest(training_frame = train_h,
nfolds = 5,
x = x,
y = y,
ntrees = 500,
stopping_rounds = 10,
stopping_metric = "RMSE",
score_each_iteration = TRUE,
stopping_tolerance = 0.0001,
seed = 1234)
h2o.varimp_plot(rf_md)
rf_md#model$model_summary
library(plotly)
tree_score <- rf_md#model$scoring_history$training_rmse
plot_ly(x = seq_along(tree_score), y = tree_score,
type = "scatter", mode = "line") %>%
layout(title = "Random Forest Model - Trained Score History",
yaxis = list(title = "RMSE"),
xaxis = list(title = "Num. of Trees"))
test_h$pred_rf <- h2o.predict(rf_md, test_h)
test_1 <- as.data.frame(test_h)
mape_rf <- mean(abs(test_1$y - test_1$pred_rf) / test_1$y)
mape_rf
H2O-3 does not support traditional time series algorithms (e.g., ARIMA). Instead, the recommendation is to treat the time series use case as a supervised learning problem and perform time-series specific pre-processing.
For example, if your goal was to predict the sales for a store tomorrow, you could treat this as a regression problem where your target would be the Sales. If you try to train a supervised learning model on the raw data, however, chances are your performance would be pretty poor. So the trick is to add historical attributes like lags as a pre-processing step.
If we trained a model on an unaltered dataset, the Mean Absolute Error is around 35%.
If we start adding historical features like the sales from the previous day for that store, we can reduce the Mean Absolute Error to about 15%.
While H2O-3 does not support lagging, you can leverage Sparkling Water to perform this pre-processing. You can use Spark to generate the lags per group and then use H2O-3 to train the regression model. Here is an example of this process: https://github.com/h2oai/h2o-tutorials/tree/master/best-practices/forecasting
The training data, train_df has to have all the columns listed in both x (c("month", "lag12", "trend", "trend_sqr")) and y ("y"), whereas the data you give to h2o.predict() just has to have the columns in x; the y-column is what will be returned as the prediction.
As you have features (in x) that are things like lag, trend, etc. the fact that it is a time series does not matter. (But you do have to be very careful when preparing those features to make sure you do not use any information in them that was not known at that point in time - but I would imagine the book has already been emphasizing that.)
Normally with a time series, for a given row in the training data, your x data is the data known at time t, and the value in the y column is the value of interest at time t+1. So when doing a prediction, you give the x values as the values at the moment, and the prediction returned is what will happen next.
The question is more or less as the title indicates. I would like to use the caret::train function with beta-binomial models made with glmmTMB package (although I am not opposed to other functions capable of fitting beta-binomial models) to calculate median absolute error (MdAE) estimates through jack-knife (leave-one-out) cross-validation. The glmmTMBControl function is already capable of estimating the optimal dispersion parameter but I was hoping to retain this information somehow as well... or having caret do the calculation possibly?
The dataset I am working with looks like this:
df <- data.frame(Effect = rep(seq(from = 0.05, to = 1, by = 0.05), each = 5), Time = rep(seq(1:20), each = 5))
Ideally I would be able to pass the glmmTMB function to trainControl like so:
BB.glmm1 <- train(Time ~ Effect,
data = df, method = "glmmTMB",
method = "", metric = "MAD")
The output would be as per the examples contained in train, although possibly with estimates for the dispersion parameter.
Although I am in no way opposed to work arounds - Thank you in advance!
I am unsure how to perform the required operation with caret without creating a custom method but I trust it is fairly easy to implement it with a for (lapply) loop.
In the example I will use the sleepstudy data set since your example data throws a bunch of warnings.
library(glmmTMB)
to perform LOOCV - for every row, create a model without that row and predict on that row:
data(sleepstudy,package="lme4")
LOOCV <- lapply(1:nrow(sleepstudy), function(x){
m1 <- glmmTMB(Reaction ~ Days + (Days|Subject),
data = sleepstudy[-x,])
return(predict(m1, sleepstudy[x,], type = "response"))
})
get the median of the residuals (I think this is MdAE? if not post a comment on how its calculated):
median(abs(unlist(LOOCV) - sleepstudy$Reaction))
In my problem dataset response variable is extremely skewed to the left. I have tried to fit the model with h2o.randomForest() and h2o.gbm() as below. I can give tune min_split_improvement and min_rows to avoid overfitting in these two cases. But with these models, I see very high errors on the tail observations. I have tried using weights_column to oversample the tail observations and undersample other observations, but it does not help.
h2o.model <- h2o.gbm(x = predictors, y = response, training_frame = train,valid = valid, seed = 1,
ntrees =150, max_depth = 10, min_rows = 2, model_id = "GBM_DD", balance_classes = T, nbins = 20, stopping_metric = "MSE",
stopping_rounds = 10, min_split_improvement = 0.0005)
h2o.model <- h2o.randomForest(x = predictors, y = response, training_frame = train,valid = valid, seed = 1,ntrees =150, max_depth = 10, min_rows = 2, model_id = "DRF_DD", balance_classes = T, nbins = 20, stopping_metric = "MSE",
stopping_rounds = 10, min_split_improvement = 0.0005)
I have tried the h2o.automl() function of h2o package for the problem for better performance. However, I see significant overfitting. I don't know of any parameters in h2o.automl() to control overfitting.
Does anyone know of a way to avoid overfitting with h2o.automl()?
EDIT
The distribution of the log transformed response is given below. After the suggestion from Erin
EDIT2:
Distribution of original response.
H2O AutoML uses H2O algos (e.g. RF, GBM) underneath, so if you're not able to get good models there, you will suffer from the same issues using AutoML. I am not sure that I would call this overfitting -- it's more that your models are not doing well at predicting outliers.
My recommendation is to log your response variable -- that's a useful thing to do when you have a skewed response. In the future, H2O AutoML will try to detect a skewed response automatically and take the log, but that's not a feature of the the current version (H2O 3.16.*).
Here's a bit more detail if you are not familiar with this process. First, create a new column, e.g. log_response, as follows and use that as the response when training (in RF, GBM or AutoML):
train[,"log_response"] <- h2o.log(train[,response])
Caveats: If you have zeros in your response, you should use h2o.log1p() instead. Make sure not to include the original response in your predictors. In your case, you don't need to change anything because you are already explicitly specifying the predictors using a predictors vector.
Keep in mind that when you log the response that your predictions and model metrics will be on the log scale. So if you need to convert your predictions back to the normal scale, like this:
model <- h2o.randomForest(x = predictors, y = "log_response",
training_frame = train, valid = valid)
log_pred <- h2o.predict(model, test)
pred <- h2o.exp(log_pred)
This gives you the predictions, but if you also want to see the metrics, you will have to compute those using the h2o.make_metrics() function using the new preds rather than extracting the metrics from the model.
perf <- h2o.make_metrics(predicted = pred, actual = test[,response])
h2o.mse(perf)
You can try this using RF like I showed above, or a GBM, or with AutoML (which should give better performance than a single RF or GBM).
Hopefully that helps improve the performance of your models!
When your target variable is skewed, mse is not a good metric to use. I would try changing the loss function because gbm tries to fit the model to the gradient of the loss function and you want to make sure that you are using the correct distribution. if you have a spike on zero and right skewed positive target, probably Tweedie would be a better option.
The xgboost package allows to build a random forest (in fact, it chooses a random subset of columns to choose a variable for a split for the whole tree, not for a nod, as it is in a classical version of the algorithm, but it can be tolerated). But it seems that for regression only one tree from the forest (maybe, the last one built) is used.
To ensure that, consider just a standard toy example.
library(xgboost)
library(randomForest)
data(agaricus.train, package = 'xgboost')
dtrain = xgb.DMatrix(agaricus.train$data,
label = agaricus.train$label)
bst = xgb.train(data = dtrain,
nround = 1,
subsample = 0.8,
colsample_bytree = 0.5,
num_parallel_tree = 100,
verbose = 2,
max_depth = 12)
answer1 = predict(bst, dtrain);
(answer1 - agaricus.train$label) %*% (answer1 - agaricus.train$label)
forest = randomForest(x = as.matrix(agaricus.train$data), y = agaricus.train$label, ntree = 50)
answer2 = predict(forest, as.matrix(agaricus.train$data))
(answer2 - agaricus.train$label) %*% (answer2 - agaricus.train$label)
Yes, of course, the default version of the xgboost random forest uses not a Gini score function but just the MSE; it can be changed easily. Also it is not correct to do such a validation and so on, so on. It does not affect a main problem. Regardless of which sets of parameters are being tried results are suprisingly bad compared with the randomForest implementation. This holds for another data sets as well.
Could anybody provide a hint on such strange behaviour? When it comes to the classification task the algorithm does work as expected.
#
Well, all trees are grown and all are used to make a prediction. You may check that using the parameter 'ntreelimit' for the 'predict' function.
The main problem remains: is the specific form of the Random Forest algorithm that is produced by the xgbbost package valid?
Cross-validation, parameter tunning and other crap have nothing to do with that -- every one may add necessary corrections to the code and see what happens.
You may specify the 'objective' option like this:
mse = function(predict, dtrain)
{
real = getinfo(dtrain, 'label')
return(list(grad = 2 * (predict - real),
hess = rep(2, length(real))))
}
This provides that you use the MSE when choosing a variable for the split. Even after that, results are suprisingly bad compared to those of randomForest.
Maybe, the problem is of academical nature and concerns the way how a random subset of features to make a split is chosen. The classical implementation chooses a subset of features (the size is specified with 'mtry' for the randomForest package) for EVERY split separately and the xgboost implementation chooses one subset for a tree (specified with 'colsample_bytree').
So this fine difference appears to be of great importance, at least for some types of datasets. It is interesting, indeed.
xgboost(random forest style) does use more than one tree to predict. But there are many other differences to explore.
I myself am new to xgboost, but curious. So I wrote the code below to visualize the trees. You can run the code yourself to verify or explore other differences.
Your data set of choice is a classification problem as labels are either 0 or 1. I like to switch to a simple regression problem to visualize what xgboost does.
true model: $y = x_1 * x_2$ + noise
If you train a single tree or multiple tree, with the code examples below you observe that the learned model structure does contain more trees. You cannot argue alone from the prediction accuracy how many trees are trained.
Maybe the predictions are different because the implementations are different. None of the ~5 RF implementations I know of are exactly alike, and this xgboost(rf style) is as closest a distant "cousin".
I observe the colsample_bytree is not equal to mtry, as the former uses the same subset of variable/columns for the entire tree. My regression problem is one big interaction only, which cannot be learned if trees only uses either x1 or x2. Thus in this case colsample_bytree must be set to 1 to use both variables in all trees. Regular RF could model this problem with mtry=1, as each node would use either X1 or X2
I see your randomForest predictions are not out-of-bag cross-validated. If drawing any conclusions on predictions you must cross-validate, especially for fully grown trees.
NB You need to fix the function vec.plot as does not support xgboost out of the box, because xgboost out of some other box do not take data.frame as an valid input. The instruction in the code should be clear
library(xgboost)
library(rgl)
library(forestFloor)
Data = data.frame(replicate(2,rnorm(5000)))
Data$y = Data$X1*Data$X2 + rnorm(5000)*.5
gradientByTarget =fcol(Data,3)
plot3d(Data,col=gradientByTarget) #true data structure
fix(vec.plot) #change these two line in the function, as xgboost do not support data.frame
#16# yhat.vec = predict(model, as.matrix(Xtest.vec))
#21# yhat.obs = predict(model, as.matrix(Xtest.obs))
#1 single deep tree
xgb.model = xgboost(data = as.matrix(Data[,1:2]),label=Data$y,
nrounds=1,params = list(max.depth=250))
vec.plot(xgb.model,as.matrix(Data[,1:2]),1:2,col=gradientByTarget,grid=200)
plot(Data$y,predict(xgb.model,as.matrix(Data[,1:2])),col=gradientByTarget)
#clearly just one tree
#100 trees (gbm boosting)
xgb.model = xgboost(data = as.matrix(Data[,1:2]),label=Data$y,
nrounds=100,params = list(max.depth=16,eta=.5,subsample=.6))
vec.plot(xgb.model,as.matrix(Data[,1:2]),1:2,col=gradientByTarget)
plot(Data$y,predict(xgb.model,as.matrix(Data[,1:2])),col=gradientByTarget) ##predictions are not OOB cross-validated!
#20 shallow trees (bagging)
xgb.model = xgboost(data = as.matrix(Data[,1:2]),label=Data$y,
nrounds=1,params = list(max.depth=250,
num_parallel_tree=20,colsample_bytree = .5, subsample = .5))
vec.plot(xgb.model,as.matrix(Data[,1:2]),1:2,col=gradientByTarget) #bagged mix of trees
plot(Data$y,predict(xgb.model,as.matrix(Data[,1:2]))) #terrible fit!!
#problem, colsample_bytree is NOT mtry as columns are only sampled once
# (this could be raised as an issue on their github page, that this does not mimic RF)
#20 deep tree (bagging), no column limitation
xgb.model = xgboost(data = as.matrix(Data[,1:2]),label=Data$y,
nrounds=1,params = list(max.depth=500,
num_parallel_tree=200,colsample_bytree = 1, subsample = .5))
vec.plot(xgb.model,as.matrix(Data[,1:2]),1:2,col=gradientByTarget) #boosted mix of trees
plot(Data$y,predict(xgb.model,as.matrix(Data[,1:2])))
#voila model can fit data