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 want to select top K features using SelectKBest and run GaussianNB:
selection = SelectKBest(mutual_info_classif, k=300)
data_transformed = selection.fit_transform(data, labels)
new_data_transformed = selection.transform(new_data)
classifier = GaussianNB()
classifier.fit(data_transformed, labels)
y_predicted = classifier.predict(new_data)
acc = accuracy_score(new_data_labels, y_predicted)
However, I do not get consistent results for accuracy on the same data.
The accuracy has been:
0.61063743402354853
0.60678034916768164
0.61733658140479086
0.61652456354039786
0.64778725131952908
0.58384084449857898
For the SAME data. I don't do splits etc. I just use two static sets of data and new_data.
Why do the results vary? How do I make sure I get the same accuracy for the same data?
This is because their is some randomness in the data or variables. This depends on the Random number generator used internally by the estimators or functions, in your case it is mutual_info_classif which you pass into SelectKBest.
Have a look at the usage of random_state here and in this answer
As a workaround you can insert the following line on top of your code.
np.random.seed(some_integer)
This will set the numpy's seed to the some_integer and as far as I know, scikit estimators uses numpy's random number generator. See this for more details
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
I am taking dataquest.io and I observed something strange (but could not get any answer back there). I am wondering why I can't use a code snippet that worked before in a situation that use the same kind/type of data, and should not cause an exception.
The lesson first teach to fit a regressor on a same training set and to predict on the same values, the calculating MSE.
Then it shows that it would overfit and propose a randomization process to avoid that. Problem being, apart from the random splitting, the dataframes generated are very similar, but if I try to calculate my MSE on the final results, it fails poorly, and I have to change the code for an alternative.
Here are both codes:
First code
# Import the linear regression class
from sklearn.linear_model import LinearRegression
# Initialize the linear regression class.
regressor = LinearRegression()
# We're using 'value' as a predictor, and making predictions for 'next_day'.
# The predictors need to be in a dataframe.
# We pass in a list when we select predictor columns from "sp500" to
# force pandas not to generate a series.
# (?) I could not figure out why it is not necessary for "to_predict"
predictors = sp500[["value"]]
to_predict = sp500["next_day"]
# Train the linear regression model on our dataset.
regressor.fit(predictors, to_predict)
# Generate a list of predictions with our trained linear regression model
next_day_predictions = regressor.predict(predictors)
print(next_day_predictions)
MSE_frame=(next_day_predictions-to_predict)**2
#(?) can math.pow(frame_difference, 2) be used on a dataframe?
mse=MSE_frame.sum()/len(MSE_frame.index)
______________________________________________________________________________
Second code
import numpy as np
import random
# Set a random seed to make the shuffle deterministic.
np.random.seed(1)
random.seed(1)
#(?) are there any difference between both of these statements? Are they
# both necessary or just one out of two?
# Randomly shuffle the rows in our dataframe
sp500 = sp500.loc[np.random.permutation(sp500.index)]
# Select 70% of the dataset to be training data
highest_train_row = int(sp500.shape[0] * .7)
train = sp500.loc[:highest_train_row,:]
# Select 30% of the dataset to be test data.
test = sp500.loc[highest_train_row:,:]
regressor = LinearRegression()
regressor.fit(train[["value"]], train["next_day"])
predictions = regressor.predict(test[["value"]])
mse = sum((predictions - test["next_day"]) ** 2) / len(predictions)
regressor = LinearRegression()
predictors = train[["value"]]
to_predict = train["next_day"]
# Train the linear regression model on our dataset.
regressor.fit(predictors, to_predict)
# Generate a list of predictions with our trained linear regression model
next_day_predictions = regressor.predict(test[["value"]])
print(next_day_predictions)
sqr=(next_day_predictions-test["next_day"])**2
Mistake was here, I was passing a with test[["next_day"]] while it was not done in the first code. Stupid me
mse=sum(sqr)/len(sqr.index)
#or
mse=sqr.sum()/len(sqr.index)
# This is the line which failed while it was identical to what was
#done before.
** it is worth noting both mse expressions don't yield the same results, They are identical for first ten decimals, but comparison with == doesn't give True.
So, the problem was there:
sqr=(next_day_predictions-test["next_day"])**2
I originally wrote
sqr=(next_day_predictions-test[["next_day"]])**2
thus passing a list into calculation, which was not done in the first code.
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.