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).
Related
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.
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
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.
I'm trying to get the feel for SVM regression with a toy example. I generated random numbers between 1 and 100 as the predictors, then took their log and added gaussian noise to create the target variables. Popping this data into sklearn's SVR module produces a reasonable looking model:
However, when I augment the training data by throwing in the squares of the original predictor variables, everything goes haywire:
I understand that the RBF kernel does something analogous to taking powers of the original features, so throwing in the second feature is mostly redundant. However, is it really the case the SVMs are this bad at handling feature redundancy? Or am I doing something wrong?
Here is the code I used to generate these graphs:
from sklearn.svm import SVR
import numpy as np
import matplotlib.pyplot as plt
# change to highest_power=2 to get the bad model
def create_design_matrix(x_array, highest_power=1):
return np.array([[x**k for k in range(1, highest_power + 1)] for x in x_array])
N = 1000
x_array = np.random.uniform(1, 100, N)
y_array = np.log(x_array) + np.random.normal(0,0.2,N)
model = SVR(C=1.0, epsilon=0.1)
print model
X = create_design_matrix(x_array)
#print X
#print y_array
model = model.fit(X, y_array)
test_x = np.linspace(1.0, 100.0, num=10000)
test_y = model.predict(create_design_matrix(test_x))
plt.plot(x_array, y_array, 'ro')
plt.plot(test_x, test_y)
plt.show()
I'd appreciate any help with this mystery!
It looks like your model's picking up on outliers too heavily, which is a symptom of error from variance. This makes sense, because adding polynomial features increases the variance of a model. You should try tweaking the bias-variance tradeoff via cross validation by tweaking parameters. The parameters to modify would be C, epsilon, and gamma. The gamma parameter's incredibly important when using an RBF kernel, so I'd start there.
Manually fiddling with these parameters (which is not recommended - see below) gave me the following model:
The parameters used here were C=5, epsilon=0.1, gamma=2**-15.
Choosing these parameters is really a task for a proper model selection framework. I prefer simulated annealing + cross validation. The best scikit-learn currently has is random grid search + crossval. Shameless plug for a simulated annealing module I helped with: https://github.com/skylergrammer/SimulatedAnnealing
Note: Polynomial features are actually products of all combinations of size d (with replacement), not just the squares of features. In the second degree case, since you only have a single feature, these are equivalent. Scikit-learn has a class that'll calculate these though: sklearn.preprocessing.PolynomialFeatures
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.