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Getting probability values for random forest and Gradient Boosting in python

I have been learning about classification techniques and studied about random forest, gradient boosting etc.Based on some help from codes available online,i tried to write code in python3 for random forest and GBM. My objective is to get the probability values from the model and not just look at accuracy as i intend to use the probability values to create KS later on.
I used the readily available titanic data set to start practicing.
Following are some of the steps i did :
/**load train data**/
train_df=pd.read_csv('***/classification/titanic/train.csv')
/**load test data**/
test_df =pd.read_csv('***/Desktop/classification/titanic/test.csv')
/**drop some variables in train data**/
train_df = train_df.drop(['Ticket', 'Cabin'], axis=1)
/**drop some variables in test data**/
test_df = test_df.drop(['Ticket', 'Cabin'], axis=1)
/** i calculated the title variable (again based on multiple threads in kaggle**/
train_df=pd.get_dummies(train_df,columns=['Pclass','Sex','Title'],drop_first=True)
test_df=pd.get_dummies(test_df,columns=['Pclass','Sex','Title'],drop_first=True)
/**i checked for missing and IV values next (not including that code here***/
predictors=[x for x in train.columns if x not in ['Survived','PassengerID']]
predictors
# create classifier object (GBM)
from sklearn.ensemble import GradientBoostingClassifier
clf = GradientBoostingClassifier(random_state=10)
# fit the classifier with x and y data
clf.fit(train[predictors],train.Survived)
prob=pd.DataFrame({'prob':clf.predict_proba(train[predictors])[:,1]})
prob['prob'].value_counts()
# create classifier object (RF)
from sklearn.ensemble import RandomForestClassifier
clf = RandomForestClassifier(random_state=10)
# fit the classifier with x and y data
clf.fit(train[predictors],train.Survived)
prob=pd.DataFrame({'prob':clf.predict_proba(train[predictors])[:,1]})
prob['prob'].value_counts()
Now when i check the probability values from the two different models, i noticed that for the Random forest output, a significant chunk had a 0 probability score whereas that was not the case for the GBM model.
I understand that the techniques are different, but how can the results be so far off ? Am i missing out on something ?
With a large chunk of the population getting tagged with '0' as probability score, my KS table goes for a toss.

Welcome to SO! Since you don't seem to be having an issue with code execution in specific, or totally incorrect outputs, this looks like it is more appropriate for CrossValidated, where you can find answers to questions of statistical concerns.
In fact, I'd suggest that answers to this question might give you some good insights into why you are seeing very different values from the predict_proba method. In short: while both GradientBoostingClassifier and RandomForestClassifier both use tree methods, what they do is very different, so direct comparison of the model parameters is not necessarily appropriate.

Limitations of Regression in Machine Learning?

I've been learning some of the core concepts of ML lately and writing code using the Sklearn library. After some basic practice, I tried my hand at the AirBnb NYC dataset from kaggle (which has around 40000 samples) - https://www.kaggle.com/dgomonov/new-york-city-airbnb-open-data#New_York_City_.png
I tried to make a model that could predict the price of a room/apt given the various features of the dataset. I realised that this was a regression problem and using this sklearn cheat-sheet, I started trying the various regression models.
I used the sklearn.linear_model.Ridge as my baseline and after doing some basic data cleaning, I got an abysmal R^2 score of 0.12 on my test set. Then I thought, maybe the linear model is too simplistic so I tried the 'kernel trick' method adapted for regression (sklearn.kernel_ridge.Kernel_Ridge) but they would take too much time to fit (>1hr)! To counter that, I used the sklearn.kernel_approximation.Nystroem function to approximate the kernel map, applied the transformation to the features prior to training and then used a simple linear regression model. However, even that took a lot of time to transform and fit if I increased the n_components parameter which I had to to get any meaningful increase in the accuracy.
So I am thinking now, what happens when you want to do regression on a huge dataset? The kernel trick is extremely computationally expensive while the linear regression models are too simplistic as real data is seldom linear. So are neural nets the only answer or is there some clever solution that I am missing?
P.S. I am just starting on Overflow so please let me know what I can do to make my question better!

This is a great question but as it often happens there is no simple answer to complex problems. Regression is not a simple as it is often presented. It involves a number of assumptions and is not limited to linear least squares models. It takes couple university courses to fully understand it. Below I'll write a quick (and far from complete) memo about regressions:
Nothing will replace proper analysis. This might involve expert interviews to understand limits of your dataset.
Your model (any model, not limited to regressions) is only as good as your features. If home price depends on local tax rate or school rating, even a perfect model would not perform well without these features.
Some features cannot be included in the model by design, so never expect a perfect score in real world. For example, it is practically impossible to account for access to grocery stores, eateries, clubs etc. Many of these features are also moving targets, as they tend to change over time. Even 0.12 R2 might be great if human experts perform worse.
Models have their assumptions. Linear regression expects that dependent variable (price) is linearly related to independent ones (e.g. property size). By exploring residuals you can observe some non-linearities and cover them with non-linear features. However, some patterns are hard to spot, while still addressable by other models, like non-parametric regressions and neural networks.
So, why people still use (linear) regression?
it is the simplest and fastest model. There are a lot of implications for real-time systems and statistical analysis, so it does matter
often it is used as a baseline model. Before trying a fancy neural network architecture, it would be helpful to know how much we improve comparing to a naive method.
sometimes regressions are used to test certain assumptions, e.g. linearity of effects and relations between variables
To summarize, regression is definitely not the ultimate tool in most cases, but this is usually the cheapest solution to try first
UPD, to illustrate the point about non-linearity.
After building a regression you calculate residuals, i.e. regression error predicted_value - true_value. Then, for each feature you make a scatter plot, where horizontal axis is feature value and vertical axis is the error value. Ideally, residuals have normal distribution and do not depend on the feature value. Basically, errors are more often small than large, and similar across the plot.
This is how it should look:
This is still normal - it only reflects the difference in density of your samples, but errors have the same distribution:
This is an example of nonlinearity (a periodic pattern, add sin(x+b) as a feature):
Another example of non-linearity (adding squared feature should help):
The above two examples can be described as different residuals mean depending on feature value. Other problems include but not limited to:
different variance depending on feature value
non-normal distribution of residuals (error is either +1 or -1, clusters, etc)
Some of the pictures above are taken from here:
http://www.contrib.andrew.cmu.edu/~achoulde/94842/homework/regression_diagnostics.html
This is an great read on regression diagnostics for beginners.

I'll take a stab at this one. Look at my notes/comments embedded in the code. Keep in mind, this is just a few ideas that I tested. There are all kinds of other things you can try (get more data, test different models, etc.)
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import seaborn as sns
#%matplotlib inline
import sklearn
from sklearn.linear_model import RidgeCV, LassoCV, Ridge, Lasso
from sklearn.datasets import load_boston
#boston = load_boston()
# Predicting Continuous Target Variables with Regression Analysis
df = pd.read_csv('C:\\your_path_here\\AB_NYC_2019.csv')
df
# get only 2 fields and convert non-numerics to numerics
df_new = df[['neighbourhood']]
df_new = pd.get_dummies(df_new)
# print(df_new.columns.values)
# df_new.shape
# df.shape
# let's use a feature selection technique so we can see which features (independent variables) have the highest statistical influence on the target (dependent variable).
from sklearn.ensemble import RandomForestClassifier
features = df_new.columns.values
clf = RandomForestClassifier()
clf.fit(df_new[features], df['price'])
# from the calculated importances, order them from most to least important
# and make a barplot so we can visualize what is/isn't important
importances = clf.feature_importances_
sorted_idx = np.argsort(importances)
# what kind of object is this
# type(sorted_idx)
padding = np.arange(len(features)) + 0.5
plt.barh(padding, importances[sorted_idx], align='center')
plt.yticks(padding, features[sorted_idx])
plt.xlabel("Relative Importance")
plt.title("Variable Importance")
plt.show()
X = df_new[features]
y = df['price']
reg = LassoCV()
reg.fit(X, y)
print("Best alpha using built-in LassoCV: %f" % reg.alpha_)
print("Best score using built-in LassoCV: %f" %reg.score(X,y))
coef = pd.Series(reg.coef_, index = X.columns)
print("Lasso picked " + str(sum(coef != 0)) + " variables and eliminated the other " + str(sum(coef == 0)) + " variables")
Result:
Best alpha using built-in LassoCV: 0.040582
Best score using built-in LassoCV: 0.103947
Lasso picked 78 variables and eliminated the other 146 variables
Next step...
imp_coef = coef.sort_values()
import matplotlib
matplotlib.rcParams['figure.figsize'] = (8.0, 10.0)
imp_coef.plot(kind = "barh")
plt.title("Feature importance using Lasso Model")
# get the top 25; plotting fewer features so we can actually read the chart
type(imp_coef)
imp_coef = imp_coef.tail(25)
matplotlib.rcParams['figure.figsize'] = (8.0, 10.0)
imp_coef.plot(kind = "barh")
plt.title("Feature importance using Lasso Model")
X = df_new
y = df['price']
from sklearn.model_selection import train_test_split
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size = 0.2, random_state = 10)
# Training the Model
# We will now train our model using the LinearRegression function from the sklearn library.
from sklearn.linear_model import LinearRegression
lm = LinearRegression()
lm.fit(X_train, y_train)
# Prediction
# We will now make prediction on the test data using the LinearRegression function and plot a scatterplot between the test data and the predicted value.
prediction = lm.predict(X_test)
plt.scatter(y_test, prediction)
from sklearn import metrics
from sklearn.metrics import r2_score
print('MAE', metrics.mean_absolute_error(y_test, prediction))
print('MSE', metrics.mean_squared_error(y_test, prediction))
print('RMSE', np.sqrt(metrics.mean_squared_error(y_test, prediction)))
print('R squared error', r2_score(y_test, prediction))
Result:
MAE 1004799260.0756996
MSE 9.87308783180938e+21
RMSE 99363412943.64531
R squared error -2.603867717517002e+17
This is horrible! Well, we know this doesn't work. Let's try something else. We still need to rowk with numeric data so let's try lng and lat coordinates.
X = df[['longitude','latitude']]
y = df['price']
from sklearn.model_selection import train_test_split
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size = 0.2, random_state = 10)
# Training the Model
# We will now train our model using the LinearRegression function from the sklearn library.
from sklearn.linear_model import LinearRegression
lm = LinearRegression()
lm.fit(X_train, y_train)
# Prediction
# We will now make prediction on the test data using the LinearRegression function and plot a scatterplot between the test data and the predicted value.
prediction = lm.predict(X_test)
plt.scatter(y_test, prediction)
df1 = pd.DataFrame({'Actual': y_test, 'Predicted':prediction})
df2 = df1.head(10)
df2
df2.plot(kind = 'bar')
from sklearn import metrics
from sklearn.metrics import r2_score
print('MAE', metrics.mean_absolute_error(y_test, prediction))
print('MSE', metrics.mean_squared_error(y_test, prediction))
print('RMSE', np.sqrt(metrics.mean_squared_error(y_test, prediction)))
print('R squared error', r2_score(y_test, prediction))
# better but not awesome
Result:
MAE 85.35438165291622
MSE 36552.6244271195
RMSE 191.18740655994972
R squared error 0.03598346983552425
Let's look at OLS:
import statsmodels.api as sm
model = sm.OLS(y, X).fit()
# run the model and interpret the predictions
predictions = model.predict(X)
# Print out the statistics
model.summary()
I would hypothesize the following:
One hot encoding is doing exactly what it is supposed to do, but it is not helping you get the results you want. Using lng/lat, is performing slightly better, but this too, is not helping you achieve the results you want. As you know, you must work with numeric data for a regression problem, but none of the features is helping you to predict price, at least not very well. Of course, I could have made a mistake somewhere. If I did make a mistake, please let me know!
Check out the links below for a good example of using various features to predict housing prices. Notice: all variables are numeric, and the results are pretty decent (just around 70%, give or take, but still much better than what we're seeing with the Air BNB data set).
https://bigdata-madesimple.com/how-to-run-linear-regression-in-python-scikit-learn/
https://towardsdatascience.com/linear-regression-on-boston-housing-dataset-f409b7e4a155

Should GridSearchCV score results be equal to score of cross_validate using same input?

I am playing around with scikit-learn a bit and wanted to reproduce the cross-validation scores for one specific hyper-parameter combination of a carried out grid search.
For the grid search, I used the GridSearchCV class and to reproduce the result for one specific hyper-parameter combination I used the cross_validate function with the exact same split and classifier settings.
My problem is that I do not get the expected score results, which to my understanding should be exactly the same as the same computations are carried out to obtain the scores in both methods.
I made sure to exclude any randomness sources from my script by fixing the used splits on the training data.
In the following code snippet, an example of the stated problem is given.
import numpy as np
from sklearn.model_selection import cross_validate, StratifiedKFold, GridSearchCV
from sklearn.svm import NuSVC
np.random.seed(2018)
# generate random training features
X = np.random.random((100, 10))
# class labels
y = np.random.randint(2, size=100)
clf = NuSVC(nu=0.4, gamma='auto')
# Compute score for one parameter combination
grid = GridSearchCV(clf,
cv=StratifiedKFold(n_splits=10, random_state=2018),
param_grid={'nu': [0.4]},
scoring=['f1_macro'],
refit=False)
grid.fit(X, y)
print(grid.cv_results_['mean_test_f1_macro'][0])
# Recompute score for exact same input
result = cross_validate(clf,
X,
y,
cv=StratifiedKFold(n_splits=10, random_state=2018),
scoring=['f1_macro'])
print(result['test_f1_macro'].mean())
Executing the given snippet results in the output:
0.38414468864468865
0.3848840048840049
I would have expected these scores to be exactly the same, as they are computed on the same split, using the same training data with the same classifier.

It is because the mean_test_f1_macro is not a simple average of all combination of folds, it is a weight average, with weights being the size of the test fold. To know more about the actual implementation of refer this answer.
Now, to replicate the GridSearchCV result, try this!
print('grid search cv result',grid.cv_results_['mean_test_f1_macro'][0])
# grid search cv result 0.38414468864468865
print('simple mean: ', result['test_f1_macro'].mean())
# simple mean: 0.3848840048840049
weights= [len(test) for (_, test) in StratifiedKFold(n_splits=10, random_state=2018).split(X,y)]
print('weighted mean: {}'.format(np.average(result['test_f1_macro'], axis=0, weights=weights)))
# weighted mean: 0.38414468864468865

Getting probabilities of best model for RandomizedSearchCV

I'm using RandomizedSearchCV to get the best parameters with a 10-fold cross-validation and 100 iterations. This works well. But now I would like to also get the probabilities of each predicted test data point (like predict_proba) from the best performing model.
How can this be done?
I see two options. First, perhaps it is possible to get these probabilities directly from the RandomizedSearchCV or second, getting the best parameters from RandomizedSearchCV and then doing again a 10-fold cross-validation (with the same seed so that I get the same splits) with this best parameters.
Edit: Is the following code correct to get the probabilities of the best performing model? X is the training data and y are the labels and model is my RandomizedSearchCV containing a Pipeline with imputing missing values, standardization and SVM.
cv_outer = StratifiedKFold(n_splits=10, shuffle=True, random_state=0)
y_prob = np.empty([y.size, nrClasses]) * np.nan
best_model = model.fit(X, y).best_estimator_
for train, test in cv_outer.split(X, y):
probas_ = best_model.fit(X[train], y[train]).predict_proba(X[test])
y_prob[test] = probas_

If I understood it right, you would like to get the individual scores of every sample in your test split for the case with the highest CV score. If that is the case, you have to use one of those CV generators which give you control over split indices, such as those here: http://scikit-learn.org/stable/tutorial/statistical_inference/model_selection.html#cross-validation-generators
If you want to calculate scores of a new test sample with the best performing model, the predict_proba() function of RandomizedSearchCV would suffice, given that your underlying model supports it.
Example:
import numpy
skf = StratifiedKFold(n_splits=10, random_state=0, shuffle=True)
scores = cross_val_score(svc, X, y, cv=skf, n_jobs=-1)
max_score_split = numpy.argmax(scores)
Now that you know that your best model happens at max_score_split, you can get that split yourself and fit your model with it.
train_indices, test_indices = k_fold.split(X)[max_score_split]
X_train = X[train_indices]
y_train = y[train_indices]
X_test = X[test_indices]
y_test = y[test_indices]
model.fit(X_train, y_train) # this is your model object that should have been created before
And finally get your predictions by:
model.predict_proba(X_test)
I haven't tested the code myself but should work with minor modifications.

You need to look in cv_results_ this will give you the scores, and mean scores for all of your folds, along with a mean, fitting time etc...
If you want to predict_proba() for each of the iterations, the way to do this would be to loop through the params given in cv_results_, re-fit the model for each of then, then predict the probabilities, as the individual models are not cached anywhere, as far as I know.
best_params_ will give you the best fit parameters, for if you want to train a model just using the best parameters next time.
See cv_results_ in the information page http://scikit-learn.org/stable/modules/generated/sklearn.model_selection.RandomizedSearchCV.html

Train scikit SVM, customize score assessment

I plan on using scikit svm for class prediction.
I have a two-class dataset consisting of about 100 experiments. Each experiment encapsulates my data-points (vectors) + classification.
Training of an SVM according to http://scikit-learn.org/stable/modules/svm.html should straight forward.
I will have to put all vectors in an array and generate another array with the corresponding class labels, train SVM. However, in order to run leave-one-out error estimation, I need to leave out a specific subset of vectors - one experiment.
How do I achieve that with the available score function?
Cheers,
EL

You could manually train on everything but the one observation, using numpy indexing to drop it out. Then you can use any of sklearn's helpers to evaluate the classification. For example:
import numpy as np
from sklearn import svm
clf = svm.SVC(...)
idx = np.arange(len(observations))
preds = np.zeros(len(observations))
for i in idx:
is_train = idx != i
clf.fit(observations[is_train, :], labels[is_train])
preds[i] = clf.predict(observations[i, :])
Alternatively, scikit-learn has a helper to do leave-one-out, and another helper to get cross-validation scores:
from sklearn import svm, cross_validation
clf = svm.SVC(...)
loo = cross_validation.LeaveOneOut(len(observations))
was_right = cross_validation.cross_val_score(clf, observations, labels, cv=loo)
total_acc = np.mean(was_right)
See the user's guide for more. cross_val_score actually returns a score for each fold (which is a little strange IMO), but since we have one fold per observation, this will just be 0 if it was wrong and 1 if it was right.
Of course, leave-one-out is very slow and has terrible statistical properties to boot, so you should probably use KFold instead.