Given a dataset of blood results, say cholesterol level, and knowing that the instrument that produced those results is subject to a known degree of variability, how would I add that variability back into the dataset? i.e. I want to assume the result in the original dataset is the true/mean value, and then produce new results that are subject to the known variability of the instrument.
In Excel you use =NORM.INV(RAND(), mean, std_dev), where RAND() provides a random value between 0 and 1, "mean" will be the original value and I have the CV so I can calculate the SD. NORM.INV then provides the inverse of the cumulative normal distribution function.
I've done the following to create a new column with my new values, but would like to know if it is valid (i.e., will each row have a different random number between 0 and 1 as the probability? and is this formula equivalent to NORM.INV?
df8000['HDL_1'] = norm.ppf(random(), loc = df8000['HDL_0'], scale = TAE_df.loc[0,'HDL'])
Thanks in advance!
Related
Ho can I create a loop to fit models with increasing number of predictors. The first iteration should
use one predictor, then two, and so on until all predictors are included. I have to compute the RMSE
on both the training and test data for this model, and store these values in a list/array.
predictors = ['bedrooms','bathrooms','sqft_living','sqft_lot','floors',
'waterfront','view','condition','grade','sqft_above',
'sqft_basement','yr_built','yr_renovated','zipcode','lat',
'long','sqft_living15','sqft_lot15']
models = []
formula = 'price ~ bedrooms'
for p in predictors[0:19]:
formula = formula + p
print(formula)
model_linear_kc_5 = smf.ols(formula=formula, data=df_train_kc)
models.append(model_linear_kc_5.fit())
My code so far but I know this isn't right and am stuck how to do it.
I have to put print(formula) inside loop and then adjust the formula = … line until it does what I want it to.
I would really appreciate help in this regard. Thank you.
I'm building a Random Forest with Caret package on R with method = "rf". I see that every type of random forest on caret seems only tune mtry which is the number of features selected randomly for each tree. I do not understand why max_depth of each tree is not a tunable parameter (like cart) ? In my mind, it is a parameter which can limit over-fitting.
For example, my rf seems really better on train data than the test data :
model <- train(
group ~., data = train.data, method = "rf",
trControl = trainControl("repeatedcv", number = 5,repeats =10),
tuneLength=5
)
> postResample(fitted(model),train.data$group)
Accuracy Kappa
0.9574592 0.9745841
> postResample(predict(model,test.data),test.data$group)
Accuracy Kappa
0.7333333 0.5428571
As you can see my model is clearly over-fitted. However, I tried a lot of different things to handle this but nothing worked. I always have something like 0.7 accuracy on test data and 0.95 on train data. This is why I want to optimize other parameters.
I cannot share my data to reproduce this.
public BinomialModelPrediction predictBinomial(RowData data) throws PredictException {
double[] preds = this.preamble(ModelCategory.Binomial, data);
BinomialModelPrediction p = new BinomialModelPrediction();
double d = preds[0];
p.labelIndex = (int)d;
String[] domainValues = this.m.getDomainValues(this.m.getResponseIdx());
p.label = domainValues[p.labelIndex];
p.classProbabilities = new double[this.m.getNumResponseClasses()];
System.arraycopy(preds, 1, p.classProbabilities, 0, p.classProbabilities.length);
if(this.m.calibrateClassProbabilities(preds)) {
p.calibratedClassProbabilities = new double[this.m.getNumResponseClasses()];
System.arraycopy(preds, 1, p.calibratedClassProbabilities, 0, p.calibratedClassProbabilities.length);
}
return p;
}
Eg: classProbabilities =[0.82333,0,276666]
labelIndex = 1
label = true
domainValues = [false,true]
what does this labelIndex signifies and does the class probabilities
order is same as the domain value order ,If order is same then it means that here probability of false is 0.82333 and probability of true is 0.27666 but why is this labelIndex showing as 1 and label as true.
Please help me to figure out this issue.
Like Tom commented, the prediction is not "wrong". You can infer from this that the threshold H2O has chosen is less than 0.27666. You probably have imbalanced training data, otherwise H2O would have not picked a low threshold for classifying a predicted value of 0.27666 as a 1. Does your training set include fewer examples of the positive class than the negative class?
If you don't like that threshold for whatever reason, then you can manually create your own. Just make sure you know how to properly evaluate the effect of using different thresholds on the performance of your model, otherwise I'd recommend just using the default threshold.
The name, "classProbabilities" is a misnomer. These are not actual probabilities, they are predicted values, though people often use the terms interchangeably. Binary classification algorithms produce "predicted values" that look like probabilities when they're between 0 and 1, but unless a calibration process is performed, they are not going to represent the probabilities. Calibration is not necessarily a straight-forward process and there are many techniques. Here's some more info about calibration methods for imbalanced data. In H2O, you can perform calibration using Platt scaling using the calibrate_model option. But this is probably not really necessary to what you're trying to do.
The proper way to use the raw output from a binary classification model is to only look at the predicted value for the positive class (you can simply ignore the predicted value for the negative class). Then you choose a threshold which suits your needs, or you can use the default threshold in H2O, which is chosen to maximize the F1 score. Some other software will use a hardcoded threshold of 0.5, but that will be a terrible choice if you don't have an even number of positive and negative examples in your training data. If you have only a few positive examples in your training data, then the best threshold will be something much lower than 0.5.
In a fairly balanced binomial classification response problem, I am observing unusual level of error in h2o.gbm classification for determining class 0, on train set itself. It is from a competition which is over, so interest is only towards understanding what is going wrong.
Confusion Matrix (vertical: actual; across: predicted) for F1-optimal threshold:
0 1 Error Rate
0 147857 234035 0.612830 =234035/381892
1 44782 271661 0.141517 =44782/316443
Totals 192639 505696 0.399260 =278817/698335
Any expert suggestions to treat the data and reduce the error is welcome.
Following approaches are tried and error is not found decreasing.
Approach 1: Selecting top 5 important variables via h2o.varimp(gbm)
Approach 2: Converting the negative normalized variable as zero and possitive as 1.
#Data Definition
# Variable Definition
#Independent Variables
# ID Unique ID for each observation
# Timestamp Unique value representing one day
# Stock_ID Unique ID representing one stock
# Volume Normalized values of volume traded of given stock ID on that timestamp
# Three_Day_Moving_Average Normalized values of three days moving average of Closing price for given stock ID (Including Current day)
# Five_Day_Moving_Average Normalized values of five days moving average of Closing price for given stock ID (Including Current day)
# Ten_Day_Moving_Average Normalized values of ten days moving average of Closing price for given stock ID (Including Current day)
# Twenty_Day_Moving_Average Normalized values of twenty days moving average of Closing price for given stock ID (Including Current day)
# True_Range Normalized values of true range for given stock ID
# Average_True_Range Normalized values of average true range for given stock ID
# Positive_Directional_Movement Normalized values of positive directional movement for given stock ID
# Negative_Directional_Movement Normalized values of negative directional movement for given stock ID
#Dependent Response Variable
# Outcome Binary outcome variable representing whether price for one particular stock at the tomorrow’s market close is higher(1) or lower(0) compared to the price at today’s market close
temp <- tempfile()
download.file('https://github.com/meethariprasad/trikaal/raw/master/Competetions/AnalyticsVidhya/Stock_Closure/test_6lvBXoI.zip',temp)
test <- read.csv(unz(temp, "test.csv"))
unlink(temp)
temp <- tempfile()
download.file('https://github.com/meethariprasad/trikaal/raw/master/Competetions/AnalyticsVidhya/Stock_Closure/train_xup5Mf8.zip',temp)
#Please wait for 60 Mb file to load.
train <- read.csv(unz(temp, "train.csv"))
unlink(temp)
summary(train)
#We don't want the ID
train<-train[,2:ncol(train)]
# Preserving Test ID if needed
ID<-test$ID
#Remove ID from test
test<-test[,2:ncol(test)]
#Create Empty Response SalePrice
test$Outcome<-NA
#Original
combi.imp<-rbind(train,test)
rm(train,test)
summary(combi.imp)
#Creating Factor Variable
combi.imp$Outcome<-as.factor(combi.imp$Outcome)
combi.imp$Stock_ID<-as.factor(combi.imp$Stock_ID)
combi.imp$timestamp<-as.factor(combi.imp$timestamp)
summary(combi.imp)
#Brute Force NA treatment by taking only complete cases without NA.
train.complete<-combi.imp[1:702739,]
train.complete<-train.complete[complete.cases(train.complete),]
test.complete<-combi.imp[702740:804685,]
library(h2o)
y<-c("Outcome")
features=names(train.complete)[!names(train.complete) %in% c("Outcome")]
h2o.shutdown(prompt=F)
#Adjust memory size based on your system.
h2o.init(nthreads = -1,max_mem_size = "5g")
train.hex<-as.h2o(train.complete)
test.hex<-as.h2o(test.complete[,features])
#Models
gbmF_model_1 = h2o.gbm( x=features,
y = y,
training_frame =train.hex,
seed=1234
)
h2o.performance(gbmF_model_1)
You've only trained a single GBM with the default parameters, so it doesn't look like you've put enough effort into tuning your model. I'd recommend a random grid search on GBM using the h2o.grid() function. Here is an H2O R code example you can follow.
I have a question,
In Matlab, I have a vector of 20 years of daily data (X) and a vector of the relevant dates (DATES). In order to find the mean value of the daily data per year, I use the following script:
A = fints(DATES,X); %convert to financial time series
B = toannual(A,'CalcMethod', 'SimpAvg'); %calculate average value per year
C = fts2mat(B); %Convert fts object to vector
C is a vector of 20 values. showing the average value of the daily data for each of the 20 years. So far, so good.. Now I am trying to do the same thing but instead of calculating mean values annually, i need to calculate std annually but it seems there is not such an option with function "toannual".
Any ideas on how to do this?
THANK YOU IN ADVANCE
I'm assuming that X is the financial information and it is an even distribution across each year. You'll have to modify this if that isn't the case. Just to clarify, by even distribution, I mean that if there are 20 years and X has 200 values, each year has 10 values to it.
You should be able to do something like this:
num_years = length(C);
span_size = length(X)/num_years;
for n = 0:num_years-1
std_dev(n+1,1) = std(X(1+(n*span_size):(n+1)*span_size));
end
The idea is that you simply pass the date for the given year (the day to day values) into matlab's standard deviation function. That will return the std-dev for that year. std_dev should be a column vector that correlates 1:1 with your C vector of yearly averages.
unique_Dates = unique(DATES) %This should return a vector of 20 elements since you have 20 years.
std_dev = zeros(size(unique_Dates)); %Just pre allocating the standard deviation vector.
for n = 1:length(unique_Dates)
std_dev(n) = std(X(DATES==unique_Dates(n)));
end
Now this is assuming that your DATES matrix is passable to the unique function and that it will return the expected list of dates. If you have the dates in a numeric form I know this will work, I'm just concerned about the dates being in a string form.
In the event they are in a string form you can look at using regexp to parse the information and replace matching dates with a numeric identifier and use the above code. Or you can take the basic theory behind this and adapt it to what works best for you!