I am wondering whether Friedman's H-statistic for the importance of two features should be symmetric? If I understand the source attached correctly, then it should be symmetric. However, in my application and minimum working example it is not. Where is my mistake? In the minimum working example below, I think the result for rm:crim and crim:rm should be identical. But they aren't. The statistic I calculate is $H_{jk}^2$. In the text, the author writes about sampling. Does this explain the asymmetric results? Thanks for your help. See also the source below.
library("rpart")
library("iml")
set.seed(42)
# Fit a CART on the Boston housing data set
data("Boston", package = "MASS")
rf <- rpart(medv ~ ., data = Boston)
# Create a model object
mod <- Predictor$new(rf, data = Boston[-which(names(Boston) == "medv")])
# Measure the interaction strength
ia <- Interaction$new(mod, feature = "rm")
ia2 <- Interaction$new(mod, feature = "crim")
View(ia$results)
View(ia2$results)
https://christophm.github.io/interpretable-ml-book/interaction.html
Related
I am a plant scientist new to machine learning. I have had success writing code and following tutorials of machine learning techniques. My issue is trying to understand how to actually apply these techniques to answer real world questions. I don't really understand how to use the model outputs to answer questions.
I recently followed a tutorial creating an algorithm to detect credit card fraud. All of the models ran nicely and I understand how to build them; but, how in the world do I take this information and translate it into a definitive answer? Following the same example, lets say I wrote this code for my job how would I then take real credit card data and screen it using this algorithm? I really want to establish a link between running these models and generating a useful output from real data.
Thank you all.
In the name of being concise I will highlight some specific examples using the same data set found here:
https://drive.google.com/file/d/1CTAlmlREFRaEN3NoHHitewpqAtWS5cVQ/view
# Import
creditcard_data <- read_csv('PATH')
# Restructure
creditcard_data$Amount=scale(creditcard_data$Amount)
NewData=creditcard_data[,-c(1)]
head(NewData)
#Split
library(caTools)
set.seed(123)
data_sample = sample.split(NewData$Class,SplitRatio=0.80)
train_data = subset(NewData,data_sample==TRUE)
test_data = subset(NewData,data_sample==FALSE)
1) Decision Tree
library(rpart)
library(rpart.plot)
decisionTree_model <- rpart(Class ~ . , creditcard_data, method = 'class')
predicted_val <- predict(decisionTree_model, creditcard_data, type = 'class')
probability <- predict(decisionTree_model, creditcard_data, type = 'prob')
rpart.plot(decisionTree_model)
2) Artificial Neural Network
library(neuralnet)
ANN_model =neuralnet (Class~.,train_data,linear.output=FALSE)
plot(ANN_model)
predANN=compute(ANN_model,test_data)
resultANN=predANN$net.result
resultANN=ifelse(resultANN>0.5,1,0)
3) Gradient Boosting
library(gbm, quietly=TRUE)
# train GBM model
system.time(
model_gbm <- gbm(Class ~ .
, distribution = "bernoulli"
, data = rbind(train_data, test_data)
, n.trees = 100
, interaction.depth = 2
, n.minobsinnode = 10
, shrinkage = 0.01
, bag.fraction = 0.5
, train.fraction = nrow(train_data) / (nrow(train_data) + nrow(test_data))
)
)
# best iteration
gbm.iter = gbm.perf(model_gbm, method = "test")
model.influence = relative.influence(model_gbm, n.trees = gbm.iter, sort. = TRUE)
# plot
plot(model_gbm)
# plot
gbm_test = predict(model_gbm, newdata = test_data, n.trees = gbm.iter)
gbm_auc = roc(test_data$Class, gbm_test, plot = TRUE, col = "red")
print(gbm_auc)
You develop your model with, preferably, three data sets.
Training, Testing and Validation. (Sometimes different terminology is used.)
Here, Train and Test sets are used to develop the model.
The model you decide upon must never see any of the Validation set. This set is used to see how good your model is, in effect it would simulate real-world new data that may come to you in the future. Once you decide your model does perform to an acceptable level you can then go back to running all your data to produce the final operational model. Then any new 'live' data of interest is fed to the model and produces an output. In the case of the fraud detection it would output some probability: here you need human input to decide at what level you would flag the event as fraudulent enough to warrant further investigation.
At periodic intervals or as you data arrives or your model performance weakens (fraudsters may become more cunning!) you would repeat the whole process.
In the past few days I have developed multiple PLS models in R for spectral data (wavebands as explanatory variables) and various vegetation parameters (as individual response variables). In total, the dataset comprises of 56. The first 28 (training set) have been used for model calibration, now all I want to do is to predict the response values for the remaining 28 observations in the tesset. For some reason, however, R keeps on the returning the fitted values of the calibration set for a given number of components rather than predictions for the independent test set. Here is what the model looks like in short.
# first simulate some data
set.seed(123)
bands=101
data <- data.frame(matrix(runif(56*bands),ncol=bands))
colnames(data) <- paste0(1:bands)
data$height <- rpois(56,10)
data$fbm <- rpois(56,10)
data$nitrogen <- rpois(56,10)
data$carbon <- rpois(56,10)
data$chl <- rpois(56,10)
data$ID <- 1:56
data <- as.data.frame(data)
caldata <- data[1:28,] # define model training set
valdata <- data[29:56,] # define model testing set
# define explanatory variables (x)
spectra <- caldata[,1:101]
# build PLS model using training data only
library(pls)
refl.pls <- plsr(height ~ spectra, data = caldata, ncomp = 10, validation =
"LOO", jackknife = TRUE)
It was then identified that a model comprising of 3 components yielded the best performance without over-fitting. Hence, the following command was used to predict the values of the 28 observations in the testing set using the above calibrated PLS model with 3 components:
predict(refl.pls, ncomp = 3, newdata = valdata)
Sensible as the output may seem, I soon discovered that all this piece of code generates are the fitted values of the PLS model for the calibration/training data, rather than predictions. I discovered this because the below code, in which newdata = is omitted, yields identical results.
predict(refl.pls, ncomp = 3)
Surely something must be going wrong, although I cannot seem to find out what specifically is. Is there someone out there who can, and is willing to help me move in the right direction?
I think the problem is with the nature of the input data. Looking at ?plsr and str(yarn) that goes with the example, plsr requires a very specific data frame that I find tricky to work with. The input data frame should have a matrix as one of its elements (in your case, the spectral data). I think the following works correctly (note I changed the size of the training set so that it wasn't half the original data, for troubleshooting):
library("pls")
set.seed(123)
bands=101
spectra = matrix(runif(56*bands),ncol=bands)
DF <- data.frame(spectra = I(spectra),
height = rpois(56,10),
fbm = rpois(56,10),
nitrogen = rpois(56,10),
carbon = rpois(56,10),
chl = rpois(56,10),
ID = 1:56)
class(DF$spectra) <- "matrix" # just to be certain, it was "AsIs"
str(DF)
DF$train <- rep(FALSE, 56)
DF$train[1:20] <- TRUE
refl.pls <- plsr(height ~ spectra, data = DF, ncomp = 10, validation =
"LOO", jackknife = TRUE, subset = train)
res <- predict(refl.pls, ncomp = 3, newdata = DF[!DF$train,])
Note that I got the spectral data into the data frame as a matrix by protecting it with I which equates to AsIs. There might be a more standard way to do this, but it works. As I said, to me a matrix inside of a data frame is not completely intuitive or easy to grok.
As to why your version didn't work quite right, I think the best explanation is that everything needs to be in the one data frame you pass to plsr for the data sources to be completely unambiguous.
Here is the updated code. My issue is with the output of "results". I'll post below as the format for readability.
library("neuralnet")
library("ggplot2")
setwd("C:/Users/Aaron/Documents/UMUC/R/Data For Assignments")
trainset <- read.csv("SOTS.csv")
head(trainset)
## val data classification
str(trainset)
## building the neural network
risknet <- neuralnet(Overall.Risk.Value ~ Finance + Personnel + Information.Dissemenation.C, trainset, hidden = 10, lifesign = "minimal", linear.output = FALSE, threshold = 0.1)
##plot nn
plot(risknet, rep="best")
##import scoring set
score_set <- read.csv("SOSS.csv")
##select subsets-training and scoring match
score_test <- subset(score_set, select = c("Finance", "Personnel", "Information.Dissemenation.C"))
##display values of score_test
head(score_test)
##neural network compute function score_test and the neural net "risknet"
risknet.results <- compute(risknet, score_test)
##Actual value of Overall.Risk.Value variable wanting to predict. net.result = a matrix containing the overall result of the neural network
results <- data.frame(Actual = score_set$Overall.Risk.Value, Prediction = risknet.results$net.result)
results[1:14, ]
The output of results is not as expected. For instance, the actual data is a number between 5 and 8, whereas "Prediction" displays outputs of .9995...for each result.
Thanks again for the help.
This is how you train and predict:
Use training data to learn model parameters (the variable risknet in your case)
Use parameters to predict scores on test data
Here is an example very much similar to yours that explains how this is done.
The default activation function in neuralnet is "logistic". When linear.output is set as FALSE, it ensures that the output is mapped by the activation function to the interval [0,1].(R_Journal (neuralnet)- Frauke Günther)
I just updated linear.output=TRUE in your code and final result looks much better.
Thanks for the help!
I am learning how to use various random forest packages and coded up the following from example code:
library(party)
library(randomForest)
set.seed(415)
#I'll try to reproduce this with a public data set; in the mean time here's the existing code
data = read.csv(data_location, sep = ',')
test = data[1:65] #basically data w/o the "answers"
m = sample(1:(nrow(factor)),nrow(factor)/2,replace=FALSE)
o = sample(1:(nrow(data)),nrow(data)/2,replace=FALSE)
train2 = data[m,]
train3 = data[o,]
#random forest implementation
fit.rf <- randomForest(train2[,66] ~., data=train2, importance=TRUE, ntree=10000)
Prediction.rf <- predict(fit.rf, test) #to see if the predictions are accurate -- but it errors out unless I give it all data[1:66]
#cforest implementation
fit.cf <- cforest(train3[,66]~., data=train3, controls=cforest_unbiased(ntree=10000, mtry=10))
Prediction.cf <- predict(fit.cf, test, OOB=TRUE) #to see if the predictions are accurate -- but it errors out unless I give it all data[1:66]
Data[,66] is the is the target factor I'm trying to predict, but it seems that by using "~ ." to solve for it is causing the formula to use the factor in the prediction model itself.
How do I solve for the dimension I want on high-ish dimensionality data, without having to spell out exactly which dimensions to use in the formula (so I don't end up with some sort of cforest(data[,66] ~ data[,1] + data[,2] + data[,3}... etc.?
EDIT:
On a high level, I believe one basically
loads full data
breaks it down to several subsets to prevent overfitting
trains via subset data
generates a fitting formula so one can predict values of target (in my case data[,66]) given data[1:65].
so my PROBLEM is now if I give it a new set of test data, let’s say test = data{1:65], it now says “Error in eval(expr, envir, enclos) :” where it is expecting data[,66]. I want to basically predict data[,66] given the rest of the data!
I think that if the response is in train3 then it will be used as a feature.
I believe this is more like what you want:
crtl <- cforest_unbiased(ntree=1000, mtry=3)
mod <- cforest(iris[,5] ~ ., data = iris[,-5], controls=crtl)
I was trying to implement the use of train function in R using nnet as method on monthly consumption data. But the output (the predicted values) are all showing to be equal to some mean value.
I have data for 24 time points (each representing a month's data) and I have used first 20 for training and the rest 4 for testing the model. Here is my code:
a<-read.csv("...",header=TRUE)
tem<-a[,5]
hum<-a[,4]
con<- a[,3]
require(quantmod)
require(nnet)
require(caret)
y<-con
plot(con,type="l")
dat <- data.frame( y, x1=tem, x2=hum)
names(dat) <- c('y','x1','x2')
#Fit model
model <- train(y ~ x1+x2,
dat[1:20,],
method='nnet',
linout=TRUE,
trace = FALSE)
ps <- predict(model2, dat[21:24,])
plot(1:24,y,type="l",col = 2)
lines(1:24,c(y[1:20],ps), col=3,type="o")
legend(5, 70, c("y", "pred"), cex=1.5, fill=2:3)
Any suggestion on how can I approach this problem alternatively? Is there any way to use Neural Network more efficiently? Or is there any other better method for this?
The problem is likely to be not enough data. 24 data points is quite low, for any machine learning problem. If the curve/shape/surface of the data is eg a simple sin wave, then 24 would be enough.
But for any more complex function, the more data the better. Can you accurately model eg sin^2 x * cos^0.3 x / sinh x with only 6 data points? No, because the available data does not capture enough detail.
If you can acquire daily data, use that instead.