Regression scoring decreases with Gridsearch - scikit-learn

I am new to scikit learn so please excuse my ignorance. Using GridsearchCV I am trying to optimize a DecisionTreeRegressor. The broader I make the parameter space, the worse the scoring gets.
Setting min_samples_split to range(2,10) gives me a neg_mean_squared_error of -0.04. When setting it to range(2,5) The score is -0.004.
simple_tree =GridSearchCV(tree.DecisionTreeRegressor(random_state=42), n_jobs=4, param_grid={'min_samples_split': range(2, 10)}, scoring='neg_mean_squared_error', cv=10, refit='neg_mean_squared_error')
simple_tree.fit(x_tr,y_tr).score(x_tr,y_tr)
I expect an equal or more positive score for a more extensive grid search compared to the less extensive one.

You're right, you should have a metric that gets closer to 0 when having more parameters.. if you were really comparing the same model each time. This is not the case in the code you provided, because you have not set the random_state parameter in your Decision Tree.
Do DecisionTreeRegressor(random_state = 42) (or any integer) and you should have more sensible results.

Using simple_tree.best_score_ gives the mean best score over all CV folds.

Related

Using cross-validation to calculate feature importance "Some Questions"

I am currently working on a project. I already selected my features and want to check their importance. I have some questions if anyone can help me please.
1- Does it make sense if I use RandomForestClassifier with cross-validation to calculate the feature importance?
2- I tried it to calculate the feature Importance using the cross_validate function
https://scikit-learn.org/stable/modules/generated/sklearn.model_selection.cross_validate.html . The function provides the test_score and train_score results. The results I got with a 10 Fold cross-validation were as follows:
test_score [0.99950158, 0.9997231 , 0.9997231 , 0.99994462, 0.99977848, 0.99983386, 0.99977848, 0.9997231 , 0.99977847, 1.]
train_score [0.99998769, 0.99998154, 0.99997539, 0.99997539, 0.99998154,0.99997539, 0.99998154, 0.99997539, 0.99998154, 0.99997539],
Can anyone explain these results? And what does it indicate?
3- The cross_validate function has a parameter called scoring, which has different scoring values such as accuracy, balanced_accuracy and f1. What does the scoring parameter do? And what do these values mean? And how should I decide which one to choose? I already read the scikit-learn documentation but wasn't clear to me.
Thank you.
Your question 1 is slightly out of scope here. For each run (fold) of cross-validation, you will get an array of importance for your features. Then how would you combine those into single importance per feature? There may be outputs which can show that a specific feature is important based on higher scores on different folds. But that may vary.
Now, cross_validate will return the default score of the estimator used inside it, unless the scoring param is set. So if you leave the scoring, it will use RandomForestClassifier's score() method which returns accuracy.
(In scikit, all classifiers will return accuracy in score() and all regressors will return r-squared value)
So for your question 2: the returned scores are accuracies per cv fold.
If you do not want to use accuracy and want some other score, you may set the scoring param incross_validate.

GridSearchCV: based on mean_test_score results, predict should perform much worse, but it does not

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.

Improving linear regression model by taking absolute value of predicted output?

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.

How to get started with Tensorflow

I am pretty new to Tensorflow, and I am currently learning it through given website https://www.tensorflow.org/get_started/get_started
It is said in the manual that:
We've created a model, but we don't know how good it is yet. To evaluate the model on training data, we need a y placeholder to provide the desired values, and we need to write a loss function.
A loss function measures how far apart the current model is from the provided data. We'll use a standard loss model for linear regression, which sums the squares of the deltas between the current model and the provided data. linear_model - y creates a vector where each element is the corresponding example's error delta. We call tf.square to square that error. Then, we sum all the squared errors to create a single scalar that abstracts the error of all examples using tf.reduce_sum:"
q1."we don't know how good it is yet.", I didn't understand this
quote as the simple model created is a simple slope equation and on
what it should train for?, as the model is a simple slope. Is it
require an perfect slope or what? why am I training that model and
for what?
q2.what is a loss function? Is loss function is used to determine the
accuracy of the model? Why is it required?
q3. I didn't understand " 'sums the squares of the deltas' between
the current model and the provided data."
q4.I didn't understood this part of code,"squared_deltas =
tf.square(linear_model - y)
this is the code:
y = tf.placeholder(tf.float32)
squared_deltas = tf.square(linear_model - y)
loss = tf.reduce_sum(squared_deltas)
print(sess.run(loss, {x:[1,2,3,4], y:[0,-1,-2,-3]}))
this may be simple questions, but I am a beginner to Tensorflow and having a hard time understanding it.
1) So you're kind of right about "Why should we train for a simple problem" but this is just an introduction piece. With any machine learning task you need to evaluate your model to see how good it is. In this case you are just trying to train to find the coefficients for the line of best fit.
2) A loss function in any machine learning context represents your error with your model. This usually means a function of your "distance" of your calculated value to the ground truth value. Think of it as an internal evaluation score. You want to minimise your loss so the gradients and parameter changes are based on your loss.
3/4) Your question here is more to do with least square regression. It's a statistical method to create lines of best fit between points. The deltas represent the differences between your calculated values and the truth values. The aim is to minimise the area of the squares and hence minise the error and have a better line of best fit.
What you are doing in this Tensorflow example is creating a machine learning model that will learn the coefficients for the line of best fit automatically using a least squares based system.
Pretty much all of your question have to-do with the loss function.
The loss function is a function that determines how far apart your output are from the expected (correct) output.
It has two usages:
Help the algorithm determine if the tweaking of the weight is helping going in the good or bad direction
Determinate the accuracy (~the number of time your system guesses the correct answer)
The loss function is the sum of the deltas witch is: the addition of the diff (delta) between the expected output and the actual output.
I think It's squared to magnifies the error the algorithm makes.

Sklearn overfitting

I have a data set containing 1000 points each with 2 inputs and 1 output. It has been split into 80% for training and 20% for testing purpose. I am training it using sklearn support vector regressor. I have got 100% accuracy with training set but results obtained with test set are not good. I think it may be because of overfitting. Please can you suggest me something to solve the problem.
You may be right: if your model scores very high on the training data, but it does poorly on the test data, it is usually a symptom of overfitting. You need to retrain your model under a different situation. I assume you are using train_test_split provided in sklearn, or a similar mechanism which guarantees that your split is fair and random. So, you will need to tweak the hyperparameters of SVR and create several models and see which one does best on your test data.
If you look at the SVR documentation, you will see that it can be initiated using several input parameters, each of which could be set to a number of different values. For the simplicity, let's assume you are only dealing with two parameters that you want to tweak: 'kernel' and 'C', while keeping the third parameter 'degree' set to 4. You are considering 'rbf' and 'linear' for kernel, and 0.1, 1, 10 for C. A simple solution is this:
for kernel in ('rbf', 'linear'):
for c in (0.1, 1, 10):
svr = SVR(kernel=kernel, C=c, degree=4)
svr.fit(train_features, train_target)
score = svr.score(test_features, test_target)
print kernel, c, score
This way, you can generate 6 models and see which parameters lead to the best score, which will be the best model to choose, given these parameters.
A simpler way is to let sklearn to do most of this work for you, using GridSearchCV (or RandomizedSearchCV):
parameters = {'kernel':('linear', 'rbf'), 'C':(0.1, 1, 10)}
clf = GridSearchCV(SVC(degree=4), parameters)
clf.fit(train_features, train_target)
print clf.best_score_
print clf.best_params_
model = clf.best_estimator_ # This is your model
I am working on a little tool to simplify using sklearn for small projects, and make it a matter of configuring a yaml file, and letting the tool do all the work for you. It is available on my github account. You might want to take a look and see if it helps.
Finally, your data may not be linear. In that case you may want to try using something like PolynomialFeatures to generate new nonlinear features based on the existing ones and see if it improves your model quality.
Try fitting your data using training data split Sklearn K-Fold cross-validation, this provides you a fair split of data and better model , though at a cost of performance , which should really matter for small dataset and where the priority is accuracy.
A few hints:
Since you have only two inputs, it would be great if you plot your data. Try either a scatter with alpha = 0.3 or a heatmap.
Try GridSearchCV, as mentioned by #shahins.
Especially, try different values for the C parameter. As mentioned in the docs, if you have a lot of noisy observations you should decrease it. It corresponds to regularize more the estimation.
If it's taking too long, you can also try RandomizedSearchCV
As a side note from #shahins answer (I am not allowed to add comments), both implementations are not equivalent. GridSearchCV is better since it performs cross-validation in the training set for tuning the hyperparameters. Do not use the test set for tuning hyperparameters!
Don't forget to scale your data

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