I've been using R-squared (coefficient of determination) and mean-absolute-percentage-error to see the difference between true output value (scalar) and the predicted output value (also scalar) that come out of a regression model.
Now, I want to see how the regressed output (vector) is close to my true output (vector) in an intuitive way. MSE is used for the regression model's training, but it is hard to tell whether your model is doing OK or not. For example, if the true output value itself is very small (close to zero) and if your predicted output is twice times bigger than the true output, the MSE will be very small even though the prediction is twice as larger than the true output.
I've been searching a while, and I found terms like "wilk's lambda test", ANOVA, MANOVA, p-value, adjusted-R-squared. But I have not figured out what is the one I can and should use.
I just decided to use MAPE by using the Euclidean distance between vectors instead of the absolute value of the difference between scalars (predicted, true value).
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I am working on a time-series prediction problem using GradientBoostingRegressor, and I think I'm seeing significant overfitting, as evidenced by a significantly better RMSE for training than for prediction. In order to examine this, I'm trying to use sklearn.model_selection.cross_validate, but I'm having problems understanding the result.
First: I was calculating RMSE by fitting to all my training data, then "predicting" the training data outputs using the fitted model and comparing those with the training outputs (the same ones I used for fitting). The RMSE that I observe is the same order of magnitude the predicted values and, more important, it's in the same ballpark as the RMSE I get when I submit my predicted results to Kaggle (although the latter is lower, reflecting overfitting).
Second, I use the same training data, but apply sklearn.model_selection.cross_validate as follows:
cross_validate( predictor, features, targets, cv = 5, scoring = "neg_mean_squared_error" )
I figure the neg_mean_squared_error should be the square of my RMSE. Accounting for that, I still find that the error reported by cross_validate is one or two orders of magnitude smaller than the RMSE I was calculating as described above.
In addition, when I modify my GradientBoostingRegressor max_depth from 3 to 2, which I would expect reduces overfitting and thus should improve the CV error, I find that the opposite is the case.
I'm keenly interested to use Cross Validation so I don't have to validate my hyperparameter choices by using up Kaggle submissions, but given what I've observed, I'm not clear that the results will be understandable or useful.
Can someone explain how I should be using Cross Validation to get meaningful results?
I think there is a conceptual problem here.
If you want to compute the error of a prediction you should not use the training data. As the name says theese type of data are used only in training, for evaluating accuracy scores you ahve to use data that the model has never seen.
About cross-validation I can tell that it's an approach to find the best training/testing set. The process is as follows: you divide your data into n groups and you do various iterating changing the testing group you pick. If you have n groups you will do n iteration and each time the training and testing set will be different. It's more understamdable in the image below.
Basically what you should do it's kile this:
Train the model using months from 0 to 30 (for example)
See the predictions made with months from 31 to 35 as input.
If the input has to be the same lenght divide feature in half (should be 17 months).
I hope I understood correctly, othewise comment.
So I am trying to explain my strictly bounded variable (percentage) with some predictors - categorical as well as numerical. I have read quite a bit about the topic, but I am still confused about some of the arguments. The purpose of my regression is explaining, not predicting.
What are the consequences of running a linear regression on a strictly bounded outcome variable?
A linear regression does not have a bounded output. It's a linear transformation of the input, so if the input is twice as large, the output will be twice as large. That way, it will always be possible to find an input that exceeds the boundaries of the output.
You can apply a sigmoid function to the output of the linear regression (this is called "logistic regression"), but this will model a binary variable and give you the probability of the variable being 1. In your case, your variable isn't binary, it can have any value between 0 and 1. For that problem, you need to apply a beta regression, which will give you a bounded output between 0 and 1.
I am trying to evaluate the performance of a regressor by means of GridSearchCV. In my implementation cv is an int, so I'm applying the K-fold validation method. Looking at cv_results_['mean_test_score'],
the best mean score on the k-fold unseen data is around 0.7, while the train scores are much higher, like 0.999. This is very normal, and I'm ok with that.
Well, following the reasoning behind this concept, when I apply the best_estimator_ on the whole data set, I expect to see at least some part of the data predicted not perfectly, right? Instead, the numerical deviations between the predicted quantities and the real values are near zero for all datapoints. And this smells of overfitting.
I don't understand that, because if I remove a small part of the data and apply GridSearchCV to the remaining part, I find almost identical results as above, but the best regressor applied to the totally unseen data predicts with much higher errors, like 10%, 30% or 50%. Which is what I expected, at least for some points, fitting GridSearchCV on the whole set, based on the results of k-fold test sets.
Now, I understand that this forces the predictor to see all datapoints, but the best estimator is the result of k fits, each of them never saw 1/k fraction of data. Being the mean_test_score the average between these k scores, I expect to see a bunch of predictions (depending on cv value) which show errors distributed around a mean error that justifies a 0.7 score.
The refit=True parameter of GridSearchCV makes the estimator with the found best set of hyperparameters be refit on the full data. So if your training error is almost zero in the CV folds, you would expect it to be near zero in the best_estimator_ as well.
I have a particular classification problem that I was able to improve using Python's abs() function. I am still somewhat new when it comes to machine learning, and I wanted to know if what I am doing is actually "allowed," so to speak, for improving a regression problem. The following line describes my method:
lr = linear_model.LinearRegression()
predicted = abs(cross_val_predict(lr, features, labels_postop_IS, cv=10))
I attempted this solution because linear regression can sometimes produce negative predictions values, even though my particular case, these predictions should never be negative, as they are a physical quantity.
Using the abs() function, my predictions produce a better fit for the data.
Is this allowed?
Why would it not be "allowed". I mean if you want to make certain statistical statements (like a 95% CI e.g.) you need to be careful. However, most ML practitioners do not care too much about underlying statistical assumptions and just want a blackbox model that can be evaluated based on accuracy or some other performance metric. So basically everything is allowed in ML, you just have to be careful not to overfit. Maybe a more sensible solution to your problem would be to use a function that truncates at 0 like f(x) = x if x > 0 else 0. This way larger negative values don't suddenly become large positive ones.
On a side note, you should probably try some other models as well with more parameters like a SVR with a non-linear kernel. The thing is obviously that a LR fits a line, and if this line is not parallel to your x-axis (thinking in the single variable case) it will inevitably lead to negative values at some point on the line. That's one reason for why it is often advised not to use LRs for predictions outside the "fitted" data.
A straight line y=a+bx will predict negative y for some x unless a>0 and b=0. Using logarithmic scale seems natural solution to fix this.
In the case of linear regression, there is no restriction on your outputs.
If your data is non-negative (as in your case the values are physical quantities and cannot be negative), you could model using a generalized linear model (GLM) with a log link function. This is known as Poisson regression and is helpful for modeling discrete non-negative counts such as the problem you described. The Poisson distribution is parameterized by a single value λ, which describes both the expected value and the variance of the distribution.
I cannot say your approach is wrong but a better way is to go towards the above method.
This results in an approach that you are attempting to fit a linear model to the log of your observations.
I am using Spark ML to optimise a Naive Bayes multi-class classifier.
I have about 300 categories and I am classifying text documents.
The training set is balanced enough and there is about 300 training examples for each category.
All looks good and the classifier is working with acceptable precision on unseen documents. But what I am noticing that when classifying a new document, very often, the classifier assigns a high probability to one of the categories (the prediction probability is almost equal to 1), while the other categories receive very low probabilities (close to zero).
What are the possible reasons for this?
I would like to add that in SPARK ML there is something called "raw prediction" and when I look at it, I can see negative numbers but they have more or less comparable magnitude, so even the category with the high probability has comparable raw prediction score, but I am finding difficulties in interpreting this scores.
Lets start with a very informal description of Naive Bayes classifier. If C is a set of all classes and d is a document and xi are the features, Naive Bayes returns:
Since P(d) is the same for all classes we can simplify this to
where
Since we assume that features are conditionally independent (that is why it is naive) we can further simplify this (with Laplace correction to avoid zeros) to:
Problem with this expression is that in any non-trivial case it is numerically equal to zero. To avoid we use following property:
and replace initial condition with:
These are the values you get as the raw probabilities. Since each element is negative (logarithm of the value in (0, 1]) a whole expression has negative value as well. As you discovered by yourself these values are further normalized so the maximum value is equal to 1 and divided by the sum of the normalized values
It is important to note that while values you get are not strictly P(c|d) they preserve all important properties. The order and ratios are exactly (ignoring possible numerical issues) the same. If none other class gets prediction close to one it means that, given the evidence, it is a very strong prediction. So it is actually something you want to see.