I'm trying to create a model that predicts whether a given team makes the playoffs in the NHL, based on a variety of available team stats. However, I'm running into a problem.
I'm using R, specifically the caret package, and I'm having fairly good success so far, with one issue: I can't limit the the number of teams that are predicted to make the playoffs.
I'm using a categorical variable as the prediction -- Y or N.
For example, using the random forest method from the caret package,
rf_fit <- train(playoff ~ ., data = train_set, method = "rf")
rf_predict <- predict(rf_fit,newdata = test_data_playoffs)
mean(rf_predict == test_data_playoffs$playoff)
gives an accuracy of approximately 90% for my test set, but that's because it's overpredicting. In the NHL, 16 teams make the playoffs, but this predicts 19 teams to make the playoffs. So I want to limit the number of "Y" predictions to 16.
Is there a way to limit the number of possible responses for a categorical variable? I'm sure there is, but google searching has given me limited success so far.
EDIT: Provide sample data which can be created with the following code:
set.seed(100) # For reproducibility
data <- data.frame(Y = sample(1:10,32,replace = T)/10, N = rep(NA,32))
data$N <- 1-data$Y
This creates a data frame similar to what you get by using the "prob" option, where you have a list of probabilities for Y and N
pred <- predict(fit,newdata = test_data_playoffs, "prob")
Related
Currently learning about cross validation through a course on DataCamp. They start the process by creating an n-fold cross validation plan. This is done with the kWayCrossValidation() function from the vtreat package. They call it as follows:
splitPlan <- kWayCrossValidation(nRows, nSplits, dframe, y)
Then, they suggest running a for loop as follows:
dframe$pred.cv <- 0
# k is the number of folds
# splitPlan is the cross validation plan
for(i in 1:k) {
# Get the ith split
split <- splitPlan[[i]]
# Build a model on the training data
# from this split
# (lm, in this case)
model <- lm(fmla, data = dframe[split$train,])
# make predictions on the
# application data from this split
dframe$pred.cv[split$app] <- predict(model, newdata = dframe[split$app,])
}
This results in a new column in the datafram with the predictions, per the last line of the above chunk of code.
My doubt is thus whether the predicted values on the data frame will be in fact averages of the 3 folds or if they will just be those of the 3rd run of the for loop?
Am I missing a detail here, or is this exactly what this code is doing, which would then defeat the purpose of the 3-fold cross validation or any-fold cross validation for that matter, as it will simply output the results of the last iteration? Shouldn't we be looking to output the average of all the folds, as laid out in the splitPlan?
Thank you.
I see there is confusion about the scope of K-fold cross-validation. The idea is not to average predictions over different folds, rather to average some measure of the prediction error, so to estimate test errors.
First of all, as you are new on SO, notice that you should always provide some data to work with. As in this case your question is not data-contingent, I just simulated some. Still, it is a good practice helping us helping you.
Check the following code, which slightly modifies what you have provided in the post:
library(vtreat)
# Simulating data.
set.seed(1986)
X = matrix(rnorm(2000, 0, 1), nrow = 1000, ncol = 2)
epsilon = matrix(rnorm(1000, 0, 0.01), nrow = 1000)
y = X[, 1] + X[, 2] + epsilon
dta = data.frame(X, y, pred.cv = NA)
# Folds.
nRows = dim(dta)[1]
nSplits = 3
splitPlan = kWayCrossValidation(nRows, nSplits)
# Fitting model on all folds but i-th.
for(i in 1:nSplits)
{
# Get the i-th split.
split = splitPlan[[i]]
# Build a model on the training data from this split.
model = lm(y ~ ., data = dta[split$train, -4])
# Make predictions on the application data from this split.
dta$pred.cv[split$app] = predict(model, newdata = dta[split$app, -4])
}
# Now compute an estimate of the test error using pred.cv.
mean((dta$y - dta$pred.cv)^2)
What the for loop does, is to fit a linear model on all folds but the i-th (i.e., on dta[split$train, -4]), and then it uses the fitted function to make predictions on the i-th fold (i.e., dta[split$app, -4]). At least, I am assuming that split$train and split$app serve such roles, as the documentation is really lacking (which usually is a bad sign). Notice I am revoming the 4-th column (dta$pred.cv) as it just pre-allocates memory in order to store all the predictions (it is not a feature!).
At each iteration, we are not filling the whole dta$pred.cv, but only a subset of that (corresponding to the rows of the i-th fold, stored each time in split$app). Thus, at the end that column just stores predictions from the K iteration.
The real rationale for cross-validation jumps in here. Let me introduce the concepts of training, validation, and test set. In data analysis, the ideal is to have such a huge data set so that we can divide it in three subsamples. The first one could then be used to train the algorithms (fitting models), the second to validate the models (tuning the models), the third to choose the best model in terms on some perfomance measure (usually mean-squared-error for regression, or MSE).
However, we often do not have all these data points (especially if you are an economist). Thus, we seek an estimator for the test MSE, so that the need for splitting data disappears. This is what K-fold cross-validation does: at once, each fold is treated as the test set, and the union of all the others as the training set. Then, we make predictions as in your code (in the loop), and save them. What you miss is the last line in the code I provided: the average of the MSE across folds. That provides us with as estimate of the test MSE, where we choose the model yielding the lowest value.
That being said, I never heard before of the vtreat package. If you are into data analysis, I suggest to have a look at the tidiyverse and the caret packages. As far as I know (and I see here on SO), they are widely used and super-well documented. May be worth learning them.
I have a stratified cox-model and want predicted survival-curves for certain profiles, based on that model.
Now, because I'm working with a large dataset with a lot of strata, I want predictions for very specific strata only, to save time and memory.
The help-page of survfit.coxph states: ... If newdata does contain strata variables, then the result will contain one curve per row of newdata, based on the indicated stratum of the original model.
When I run the code below, where newdata does contain the stratum-variable, I still get predictions for both strata, which contradicts the help-page
df <- data.frame(X1 = runif(200),
X2 = sample(c("A", "B"), 200, replace = TRUE),
Ev = sample(c(0,1), 200, replace = TRUE),
Time = rexp(200))
testfit <- coxph( Surv(Time, Ev) ~ X1 + strata(X2), df)
out <- survfit(testfit, newdata = data.frame(X1 = 0.6, X2 = "A"))
Is there anything I fail to see or understand here?
I'm not sure if this is a bug or a feature in survival:::survfit.coxph. It looks like the intended behaviour in the code is that only requested strata are returned. In the function:
strata(X2) is evaluated in an environment containing newdata and the result, A is returned.
The full curve is then created.
There is then some logic to split the curve into strata, but only if result$surv is a matrix.
In your example it is not a matrix. I can't find any documentation on the expected usage of this if it's not a bug. Perhaps it would be worth dropping the author/maintainer a note.
maintainer("survival")
# [1] "Terry M Therneau <xxxxxxxx.xxxxx#xxxx.xxx>"
Some comments that may be helpfull:
My example was not big enough (and I seem not to have read the related github post very well, but that was after I posted my question here): if newdata has at least two lines (and of course the strata-variable), predictions are returned only for the requested strata
There is an inefficiency inside survfit.coxph, where the baseline-hazard is calculated for every stratum in the original dataset, not only for the requested strata (see my contribution to the same github post). However, that doesn't seem to be a big issue (a test on a dataset with roughly half a million observation, 50% events and 1000 strata), takes less than a minute
The problem is memory allocation somewhere during calculations (in the above example, things collapse once I want predictions for 100 observations - 1 stratum each - while the final output of predictions for 80 is only a few MB)
My work-around:
Select all observations you want predictions for
use lp <- predict(..., type='lp') to get the linear predictor for all these observations
use survfit only on the first observation: survfit(fit, newdata = expand_grid(newdf, strat = strata_list))
Store the resulting survival estimates in a data.frame (or not, that's up to you)
To calculate predicted survival for other observations, use the PH-assumption (see formula below). This invokes the overhead of survfit.coxph only once and if you focus on survival on only a few times (e.g. 5- and 10-year survival), you can reduce the computer time even more
I'm fitting a k-nearest neighbor model using R's caret package.
library(caret)
set.seed(0)
y = rnorm(20, 100, 15)
predictors = matrix(rnorm(80, 10, 5), ncol=4)
data = data.frame(cbind(y, predictors))
colnames(data)=c('Price', 'Distance', 'Cost', 'Tax', 'Transport')
I left one observation as the test data and fit the model using the training data.
id = sample(nrow(data)-1)
train = data[id, ]
test = data[-id,]
knn.model = train(Price~., method='knn', train)
predict(knn.model, test)
When I display knn.model, it tells me it uses k=9. I would love to know which 9 observations are actually the "nearest" to the test observation. Besides manually calculating the distances, is there an easier way to display the nearest neighbors?
Thanks!
When you are using knn you are creating clusters with points that are near based on independent variables. Normally, this is done using train(Price~., method='knn', train), such that the model chooses the best prediction based on some criteria (taking into account also the dependent variable as well). Given the fact I have not checked whether the R object stores the predicted price for each of the trained values, I just used the model trained to predicte the expected price given the model (where the expected price is located in the space).
At the end, the dependent variable is just a representation of all the other variables in a common space, where the price associated is assumed to be similar since you cluster based on proximity.
As a summary of steps, you need to calculate the following:
Get the distance for each of the training data points. This is done through predicting over them.
Calculate the distance between the trained data and your observation of interest (in absolut value, since you do not care about the sign but just about the absolut distances).
Take the indexes of the N smaller ones(e.g.N= 9). you can get the observations and related to this lower distances.
TestPred<-predict(knn.model, newdata = test)
TrainPred<-predict(knn.model, train)
Nearest9neighbors<-order(abs(TestPred-TrainPred))[1:9]
train[Nearest9neighbors,]
Price Distance Cost Tax Transport
15 95.51177 13.633754 9.725613 13.320678 12.981295
7 86.07149 15.428847 2.181090 2.874508 14.984934
19 106.53525 16.191521 -1.119501 5.439658 11.145098
2 95.10650 11.886978 12.803730 9.944773 16.270416
4 119.08644 14.020948 5.839784 9.420873 8.902422
9 99.91349 3.577003 14.160236 11.242063 16.280094
18 86.62118 7.852434 9.136882 9.411232 17.279942
11 111.45390 8.821467 11.330687 10.095782 16.496562
17 103.78335 14.960802 13.091216 10.718857 8.589131
I'm working with Support Vector Machines from the e1071 package in R. This is my first project using SVM.
I have a dataset containing order histories of ~1k customers over 1 year and I want to predict costumer purchases. For every customer I have the information if a certain item (out of ~50) was bought or not in a certain week (for 52 weeks aka 1 yr).
My goal is to predict next month's purchases for every single customer.
I believe that a purchase let's say 1 month ago is more meaningful for my prediction than a purchase 10 months ago.
My question is now how I can give more recent data a higher impact? There is a 'weight' option in the svm-function but I'm not sure how to use it.
Anyone who can give me a hint? Would be much appreciated!
That's my code
# Fit model using Support Vecctor Machines
# install.packages("e1071")
library(e1071)
response <- train[,5]; # purchases
formula <- response ~ .;
tuned.svm <- tune.svm(train, response, probability=TRUE,
gamma=10^(-6:-3), cost=10^(1:2));
gamma.k <- tuned.svm$best.parameter[[1]];
cost.k <- tuned.svm$best.parameter[[2]];
svm.model <- svm(formula, data = train,
type='eps-regression', probability=TRUE,
gamma=gamma.k, cost=cost.k);
svm.pred <- predict(svm.model, test, probability=TRUE);
Side notes: I'm fitting a model for every single customer. Also, since I'm interested in the probability, that customer i buys item j in week k, I put
probability=TRUE
click here to see a sccreenshot of my data
Weights option in the R SVM Model is more towards assigning weights to solve the problem of imbalance classes. its class.Weights parameter and is used to assign weightage to different classes 1/0 in a biased dataset.
To answer your question: to give more weightage in a SVM Model for recent data, a simple trick in absence of an ibuild weight functionality at observation level is to repeat the recent columns (i.e. create duplicate rows for recent data) hence indirectly assigning them higher weight
Try this package: https://CRAN.R-project.org/package=WeightSVM
It uses a modified version of 'libsvm' and is able to deal with instance weighting. You can assign higher weights to recent data.
For example. You have simulated data (x,y)
x <- seq(0.1, 5, by = 0.05)
y <- log(x) + rnorm(x, sd = 0.2)
This is an unweighted SVM:
model1 <- wsvm(x, y, weight = rep(1,99))
Blue dots is the unweighted SVM and do not fit the first instance well. We want to put more weights on the first several instances.
So we can use a weighted SVM:
model2 <- wsvm(x, y, weight = seq(99,1,length.out = 99))
Green dots is the weighted SVM and fit the first instance better.
I'm trying to apply glm on a given dataset,but the summary(model1) is not giving me the correct output , it's not giving coefficient values for Estimate Std. Error z value Pr(>|z|) etc, it's just giving me NA as an output for individual attribute element.
TEXT <- c('Learned a new concept today : metamorphic testing. t.co/0is1IUs3aW','BMC Bioinformatics BioMed Central: Detecting novel ncRNAs by experimental #RNomics is not an easy task... http:/t.co/ui3Unxpx #bing #MyEN','BMC Bioinformatics BioMed Central: small #RNA with a regulatory function as a scientific ... Detecting novel… http:/t.co/wWHOEkR0vc #bing','True or false? link(#Addition, #Classification) http:/t.co/zMJuTFt8iq #Oxytocin','Biologists do have a sense of humor, especially computational bio people http:/t.co/wFZqaaFy')
NAME <- c('QSoft Consulting','Fabrice Leclerc','Sungsam Gong','Frederic','Zach Stednick')
SCREEN_NAME <-c ('QSoftConsulting','rnomics','sunggong','rnomics','jdwasmuth')
FOLLOWERS_COUNT <- c(734,1900,234,266,788)
RETWEET <- c(1,3,5,0,2)
FRIENDS_COUNT <-c(34,532,77,213,422)
STATUSES_COUNT <- c(234,643,899,222,226)
FAVOURITES_COUNT <- c(144,2677,445,930,254)
df <- data.frame(TEXT,NAME,SCREEN_NAME,RETWEET,FRIENDS_COUNT,STATUSES_COUNT,FAVOURITES_COUNT)
mydata<-df
mydata$FAVOURITES_COUNT <- ifelse( mydata$FAVOURITES_COUNT >= 445, 1, 0) #converting fav_count to binary values
Splitting data
library(caret)
split=0.60
trainIndex <- createDataPartition(mydata$FAVOURITES_COUNT, p=split, list=FALSE)
data_train <- mydata[ trainIndex,]
data_test <- mydata[-trainIndex,]
glm model
library(e1071)
model1 <- glm(FAVOURITES_COUNT~.,family = binomial, data = data_train)
summary(model1)
I want to get the p value for further analysis so far i think my code is right, how can i get the correct output?
A binomial distribution will only work if the dependent variable has two outcomes. You should consider a Poisson distribution when the dependent variable is a count. See here for more details: http://www.statmethods.net/advstats/glm.html
Your code for fitting the GLM is programmatically correct. However, there are a few issues:
As mentioned in the comments, for every variable that is categorical, you should use as.factor() to make it into a factor. GLM doesn't know what a "string" variable is.
As MorganBall indicated, if your data truly is count data, you may consider fitting it using a Poisson GLM, instead of converting to binary and using Logistic regression.
You indicate that you have 13 parameters and 1000 observations. While this may seem like enough data, note that some of these parameters may have very few (close to 0?) observations in them. This is a problem.
In addition, did you make sure that your data does not perfectly separate the response? Because if there are some combinations of parameters that do separate the response perfectly, the maximum likelihood estimate won't converge and theoretically goes to infinity. Practically speaking, you'll get very large standard errors for your estimates.