Develop Reference

Why is this accuracy of this Random forest sentiment classification so low?

I want to use RandomForestClassifier for sentiment classification. The x contains data in string text, so I used LabelEncoder to convert strings. Y contains data in numbers. And my code is this:
import pandas as pd
import numpy as np
from sklearn.model_selection import *
from sklearn.ensemble import *
from sklearn import *
from sklearn.preprocessing.label import LabelEncoder
data = pd.read_csv('data.csv')
x = data['Reviews']
y = data['Ratings']
le = LabelEncoder()
x_encoded = le.fit_transform(x)
x_train, x_test, y_train, y_test = train_test_split(x_encoded,y, test_size = 0.2)
x_train = x_train.reshape(-1,1)
x_test = x_test.reshape(-1,1)
clf = RandomForestClassifier(n_estimators=100)
clf.fit(x_train, y_train)
y_pred = clf.predict(x_test)
Then I printed out the accuracy like below:
print("Accuracy:", metrics.accuracy_score(y_test, y_pred))
And here's the output:
Accuracy: 0.5975
I have read that Random forests has high accuracy, because of the number of decision trees participating in the process. But I think that the accuracy is much lower than it should be. I have looked for some similar questions on Stack Overflow, but I couldn't find a solution for my problem.
Is there any problem in my code using Random Forest library? Or is there any exceptions of cases when using Random forest?

It is not a problem regarding Random Forests or the library, it is rather a problem how you transform your text input into a feature or feature vector.
What LabelEncoding does is; given some labels like ["a", "b", "c"] it transforms those labels into numeric values between 0 and n-1 with n-being the number of distinct input labels. However, I assume Reviews contain texts and not pure labels so to say. This means, all your reviews (if not 100% identical) are transformed into different labels. Eventually, this leads to your classifier doing random stuff. give that input. This means you need something different to transform your textual input into a numeric input that Random Forests can work on.
As a simple start, you can try something like TfIDF or also some simple count vectorizer. Those are available from sklearn https://scikit-learn.org/stable/modules/feature_extraction.html section 6.2.3. Text feature extraction. There are more sophisticated ways of transforming texts into numeric vectors but that should be a good start for you to understand what has to happen conceptually.
A last important note is that you fit those vectorizers only on the training set and not on the full dataset. Otherwise, you might leak information from training to evaluation/testing. A good way of doing this would be to build a sklearn pipeline that consists of a feature transformation step and the classifier.

Backward stepwise selection to choose an optimal subset of the predictors with the AUC as a criterion

I am looking to perform a backward feature selection process on a logistic regression with the AUC as a criterion. For building the logistic regression I used the scikit library, but unfortunately this library does not seem to have any methods for backward feature selection. My dependent variable is a binary banking crisis variable and I have 13 predictors. Does anybody have any suggestions on how to handle this?
The code below states the method to compute the AUC. The problem is that I do not know how to decide which feature I can prune because it is less important than the other.
def cv_loop(X, y, model, N):
mean_auc = 0.
for i in range(N):
X_train, X_cv, y_train, y_cv = train_test_split(
X, y, test_size=.20,
random_state = i*SEED)
model.fit(X_train, y_train)
preds = model.predict_proba(X_cv)[:,1]
fpr, tpr, _ = metrics.roc_curve(y_cv, preds)
auc = metrics.auc(fpr, tpr)
print("AUC (fold %d/%d): %f" % (i + 1, N, auc))
mean_auc += auc
return mean_auc/N
If you need more background information let me know!
Many thanks in advance,
Joris

scikit-learn has Recursive Feature Elimination (RFE) in its feature_selection module, which almost does what you described.
Given an external estimator that assigns weights to features (e.g.,
the coefficients of a linear model), the goal of recursive feature
elimination (RFE) is to select features by recursively considering
smaller and smaller sets of features. First, the estimator is trained
on the initial set of features and the importance of each feature is
obtained either through a coef_ attribute or through a
feature_importances_ attribute. Then, the least important features are
pruned from current set of features. That procedure is recursively
repeated on the pruned set until the desired number of features to
select is eventually reached.
This doesn't explicitly work on AUC however. It does the prunining by looking at coefficients of the logistic regression.

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

Calculate evaluation metrics using cross_val_predict sklearn

In the sklearn.model_selection.cross_val_predict page it is stated:
Generate cross-validated estimates for each input data point. It is
not appropriate to pass these predictions into an evaluation metric.
Can someone explain what does it mean? If this gives estimate of Y (y prediction) for every Y (true Y), why can't I calculate metrics such as RMSE or coefficient of determination using these results?

It seems to be based on how samples are grouped and predicted. From the user guide linked in the cross_val_predict docs:
Warning Note on inappropriate usage of cross_val_predict
The result of
cross_val_predict may be different from those obtained using
cross_val_score as the elements are grouped in different ways. The
function cross_val_score takes an average over cross-validation folds,
whereas cross_val_predict simply returns the labels (or probabilities)
from several distinct models undistinguished. Thus, cross_val_predict
is not an appropriate measure of generalisation error.
The cross_val_score seems to say that it averages across all of the folds, while the cross_val_predict groups individual folds and distinct models but not all and therefore it won't necessarily generalize as well. For example, using the sample code from the sklearn page:
from sklearn import datasets, linear_model
from sklearn.model_selection import cross_val_predict, cross_val_score
from sklearn.metrics import mean_squared_error, make_scorer
diabetes = datasets.load_diabetes()
X = diabetes.data[:200]
y = diabetes.target[:200]
lasso = linear_model.Lasso()
y_pred = cross_val_predict(lasso, X, y, cv=3)
print("Cross Val Prediction score:{}".format(mean_squared_error(y,y_pred)))
print("Cross Val Score:{}".format(np.mean(cross_val_score(lasso, X, y, cv=3, scoring = make_scorer(mean_squared_error)))))
Cross Val Prediction score:3993.771257795029
Cross Val Score:3997.1789145156217

Just to add a little more clarity, it is easier to understand the difference if you consider a non-linear scoring function such as Maximum-Absolute-Error instead of something like a mean-absolute error.
cross_val_score() would compute the maximum-absolute-error on each off the 3-folds (assuming 3 fold cross-validator) and report the aggregate (say mean?) over 3 such scores. That is, something like mean of (a, b, c) where a , b, c are the max-abs-errors for the 3 folds respectively. I guess it is safe to conclude the returned value as the max-absolute-error of your estimator in the average or general case.
with cross_val_predict() you would get 3-sets of predictions corresponding to 3-folds and taking the maximum-absolute-error over the aggregate (concatenation) of these 3-sets of predictions is certainly not the same as above. Even if the predicted values are identical in both the scenarios, what you end up with here is max of (a, b,c ). Also, max(a,b,c) would be an unreasonable and overly pessimistic characterization of the max-absolute-error score of your model.

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.