Develop Reference

Getting feature importances out of an Adaboosted linear regression

I have the following code:
modelClf = AdaBoostRegressor(base_estimator=LinearRegression(), learning_rate=2, n_estimators=427, random_state=42)
modelClf.fit(X_train, y_train)
While trying to interpret and improve the results, I wanted to see the feature importances, however I get an error saying that linear regressions don't really do that kind of thing.
Alright, sounds reasonable, so I tried using .coef_ since it should work for linear regressions, but it, in place, turned out incompatible with the adaboost regressor.
Is there any way to find the feature importances or is it impossible when adaboost it used on a linear regression?

Issue12137 suggests to add support for this using the coefs_, although a choice needs to be made how to normalize negative coefficients. There's also the question of when coefficients are really good representatives of importance (you should at least scale your data first). And then there's the question of when adaptive boosting helps a linear model in the first place.
One way to do this quickly is to modify the LinearRegression class:
class MyLinReg(LinearRegression):
#property
def feature_importances_(self):
return self.coef_ # assuming one output
modelClf = AdaBoostRegressor(base_estimator=MyLinReg(), ...)

Checked with below code, there is an attribute for feature importance:
import pandas as pd
import random
from sklearn.ensemble import AdaBoostRegressor
df = pd.DataFrame({'x1':random.choices(range(0, 100), k=10), 'x2':random.choices(range(0, 100), k=10)})
df['y'] = df['x2'] * .5
X = df[['x1','x2']].values
y = df['y'].values
regr = AdaBoostRegressor(random_state=0, n_estimators=100)
regr.fit(X, y)
regr.feature_importances_
Output: You can see feature 2 is more important as Y is nothing but half of it (as the data is created in such way).

Large Margin Classifier

I am building a classifier to maximize the margin between positively and negatively labelled points.
I am using sklearn.LinearSVC to do this. I have to find both the weights (a vector, theta) and intercept ( a scalar theta_0). I also need to calculate the maximum margin. So, I wrote the below code.
import numpy as np
import sklearn
from sklearn.svm import LinearSVC
# training data
X_train = np.array([[0,0],[2,0],[3,0],[0,2],[2,2],[5,1],[5,2],[2,4],[4,4],[5,5]])
y_train = [-1,-1,-1,-1,-1,1,1,1,1,1]
classifier = LinearSVC(random_state = 0, C=1.0, fit_intercept= True)
classifier.fit(X_train, y_train)
theta = classifier.coef_
theta_0.intercept_
norm = np.linalg.norm(theta)
margin = 2/norm
As per my understanding, LinearSVC is the right package for this; though I see some tutorials in which people use SVC and then kernel = 'linear'.
I am not sure whether I should set the fit_intercept parameter to True. I am getting a different value for theta and theta_0 when I default it to False.
Can somebody guide me on the understanding of this parameter and also whether the margin calculation is correct? Lastly, whether LinearSVC is the right model. Thanks.

This statement is wrong:
theta_0.intercept_
I assume that it should be:
theta_0 = classifier.intercept_

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

How to improve speed of the program for large data in python

I am trying to calculate the prediction probability. I have wrote a program which is calculating but speed is very slow and taking so much time for large dataset.
The aim is to calculate each prediction probability in the SVM model by using LinearSVC and OneVsRestClassifier but getting the error
AttributeError: 'LinearSVC' object has no attribute 'predict_proba'
Due to the above error, I have tried below
Code
from sklearn import svm
model_1 = svm.SVC(kernel='linear', probability=True)
from sklearn.preprocessing import LabelEncoder
X_1 = df["Property Address"]
lb = LabelEncoder()
X_2 = lb.fit_transform(X_1)
y_1 = df["Location_Name"]
y_2 = lb.fit_transform(y_1)
test_1 = test["Property Address"]
lb = LabelEncoder()
test_1 = lb.fit_transform(test_1)
X_2= X_2.reshape(-1, 1)
y_2= y_2.reshape(-1, 1)
test_1 = test_1.reshape(-1, 1)
model_1.fit(X_2, y_2)
results = model_1.predict_proba(test_1)[0]
# gets a dictionary of {'class_name': probability}
prob_per_class_dictionary = dict(zip(model.classes_, results))
Is there any other way for the same task? please suggest

You could use sklearns CalibratedClassifierCV if you need to use to the predict_proba method.
Or you could use Logistic Regression.
If your issue is related to speed, try consider using the LinearSVC in sklearn.svm instead of SVC(kernel='linear'). It is faster.

As suggested in another answer, LinearSVC is faster than SVC(kernel='linear').
Regarding probability, SVC doesn't have predict_proba(). Instead, you have to set its probability hyperparameter to True. Link
Tip: SVM is preferred for small datasets, so prefer to use other algorithms to handle large datasets.

Scikit-learn cross validation scoring for regression

How can one use cross_val_score for regression? The default scoring seems to be accuracy, which is not very meaningful for regression. Supposedly I would like to use mean squared error, is it possible to specify that in cross_val_score?
Tried the following two but doesn't work:
scores = cross_validation.cross_val_score(svr, diabetes.data, diabetes.target, cv=5, scoring='mean_squared_error')
and
scores = cross_validation.cross_val_score(svr, diabetes.data, diabetes.target, cv=5, scoring=metrics.mean_squared_error)
The first one generates a list of negative numbers while mean squared error should always be non-negative. The second one complains that:
mean_squared_error() takes exactly 2 arguments (3 given)

I dont have the reputation to comment but I want to provide this link for you and/or a passersby where the negative output of the MSE in scikit learn is discussed - https://github.com/scikit-learn/scikit-learn/issues/2439
In addition (to make this a real answer) your first option is correct in that not only is MSE the metric you want to use to compare models but R^2 cannot be calculated depending (I think) on the type of cross-val you are using.
If you choose MSE as a scorer, it outputs a list of errors which you can then take the mean of, like so:
# Doing linear regression with leave one out cross val
from sklearn import cross_validation, linear_model
import numpy as np
# Including this to remind you that it is necessary to use numpy arrays rather
# than lists otherwise you will get an error
X_digits = np.array(x)
Y_digits = np.array(y)
loo = cross_validation.LeaveOneOut(len(Y_digits))
regr = linear_model.LinearRegression()
scores = cross_validation.cross_val_score(regr, X_digits, Y_digits, scoring='mean_squared_error', cv=loo,)
# This will print the mean of the list of errors that were output and
# provide your metric for evaluation
print scores.mean()

The first one is correct. It outputs the negative of the MSE, as it always tries to maximize the score. Please help us by suggesting an improvement to the documentation.