Im trying to use LinearSVC model in OneVsRest in PySpark , but it seems its not supported yet.
My error msg
LinearSVC only supports binary classification. 1 classes detected in LinearSVC_43a50b0b70d60a8cbdb1__labelCol
What kind of changes do i need in order to implement it in PySpark?
Does anyone know when will OneVsRest in Pyspark will support LinearSVC?
The error message tells you that your dataset contains only one class currently, but LinearSVM is a binary classification algorithm which requires exactly two classes.
I'm not sure if the rest of your code will cause any issues, because you haven't posted anything. Just in case you or someone else needs it, have a look below.
Like alrady said, LinearSVM is a binary classification algorithm which will never support multi class classification by definition but you can always reduce a multi class classification problem to a binary classification problem. One-vs-Rest is an approach for such a reduction. It trains one classifier per class and from an engineering perspective it makes sense to seperate this to a dedicated class like spark did. The OneVsRest trains one classifier for each of your classes and a given sample is scored against this list of classifiers. The classifier with the highstest score determines the predicted label for your sample.
Have a look at the code below for the usage of OneVsRest with LinearSVC:
from pyspark.ml.feature import VectorAssembler
from pyspark.ml.feature import StringIndexer
from pyspark.ml.classification import OneVsRest, LinearSVC
from pyspark.ml.evaluation import MulticlassClassificationEvaluator
df = spark.read.csv('/tmp/iris.data', schema='sepalLength DOUBLE, sepalWidth DOUBLE, petalLength DOUBLE, petalWidth DOUBLE, class STRING')
vecAssembler = VectorAssembler(inputCols=["sepalLength", "sepalWidth", "petalLength", 'petalWidth'], outputCol="features")
df = vecAssembler.transform(df)
stringIndexer = StringIndexer(inputCol="class", outputCol="label")
si_model = stringIndexer.fit(df)
df = si_model.transform(df)
svm = LinearSVC()
ovr = OneVsRest(classifier=svm)
ovrModel = ovr.fit(df)
evaluator = MulticlassClassificationEvaluator(metricName="accuracy")
predictions = ovrModel.transform(df)
print("Accuracy: {}".format(evaluator.evaluate(predictions)))
Output:
Accuracy: 0.9533333333333334
This is a funny bug in PySpark. If you have multiple classes, they must be identified starting from zero.
I went through this bug just right now. I had a dataframe built in the same way they suggest in the LinearSVC guide.
df = sc.parallelize([
Row(label=1.0, features=Vectors.dense(1.0, 1.0, 1.0)),
Row(label=0.0, features=Vectors.dense(1.0, 2.0, 3.0))]).toDF()
Originally, it was a plain RDD, then I transformed each RDD record in a Row. I had a three-classes problem where classes were named 1, 2 and 3. I instantiated a OneVsRest object (just like #cronoik suggested) and I went through the same error as yours.
So I took the df dataframe exactly as initalised in their user guide (see above) and I decided to start playing with it by adding and removing classes. So I simply substituted in the second pattern label=0.0 with label=2.0 and the error appeared. Even with their dataframe, even with just two classes.
So I changed the naming of my classes from 1, 2, 3 to 0, 1, 2 and the error went away.
Hope this helps!
Related
I have been reading the book Hands-on Machine Learning with Scikit-Learn and Tensorflow and I found this code:
from sklearn.model_selection import StratifiedShuffleSplit
split = StratifiedShuffleSplit(n_splits=1, test_size=0.2, random_state=42)
for train_index, test_index in split.split(housing, housing["income_cat"]):
strat_train_set = housing.loc[train_index]
strat_test_set = housing.loc[test_index]
I would like to know what the argument 'n_splits' does. I searched everywhere but couldn't find a satisfactory response.
Thanks in advance!!
As the name suggests, the n_splits parameter is used to specify how many times (basically how many separate splits) you want the splits to happen.
For example, setting n_splits = 3 would make the loop generate 3 different splits (one for each iteration) so you can perform validation more effectively.
Setting n_splits = 1 would mimic what sklearn.model_selection.train_test_split would do (along with the stratify parameter mentioned). The documentation has detailed explanations of each parameter for this function.
I have a classification problem with 10 features and I have to predict 1 or 0. When I train the SVC model, with the train test split, all the predicted values for the test portion of the data comes out to be 0. The data has the following 0-1 count:
0: 1875
1: 1463
The code to train the model is given below:
from sklearn.svm import SVC
model = SVC()
model.fit(X_train, y_train)
pred= model.predict(X_test)
from sklearn.metrics import accuracy_score
accuracy_score(y_test, pred)`
Why does it predict 0 for all the cases?
The model predicts the more frequent class, even though the dataset is nor much imbalanced. It is very likely that the class cannot be predicted from the features as they are right now.
You may try normalizing the features.
Another thing you might want to try is to have a look at how correlated the features are with each other. Having highly correlated features might also prevent the model from converging.
Also, you might have chosen the wrong features.
For a classification problem, it is always good to run a dummy classifiar as a starting point. This will give you an idea how good your model can be.
You can use this as a code:
from sklearn.dummy import DummyClassifier
dummy_classifier = DummyClassifier(strategy="most_frequent")
dummy_classifier.fit(X_train,y_train)
pred_dum= dummy_classifier.predict(X_test)
accuracy_score(y_test, pred_dum)
this will give you an accuracy, if you predict always the most frequent class. If this is for example: 100% , this would mean that you only have one class in your dataset. 80% means, that 80% of your data belongs to one class.
In a first step you can adjust your SVC:
model = SVC(C=1.0, kernel=’rbf’, random_state=42)
C : float, optional (default=1.0)Penalty parameter C of the error
term.
kernel : Specifies the kernel type to be used in the algorithm. It
must be one of ‘linear’, ‘poly’, ‘rbf’
This can give you a starting point.
On top you should run also a prediction for your training data, to see the comparison if you are over- or underfitting.
trainpred= model.predict(X_train)
accuracy_score(y_test, trainpred)
Hi I was following the Machine Learning course by Andrew Ng.
I found that in regression problems, specially logistic regression they have used integer values for the features which could be plotted in a graph. But there are so many use cases where the feature values may not be integer.
Let's consider the follow example :
I want to build a model to predict if any particular person will take a leave today or not. From my historical data I may find the following features helpful to build the training set.
Name of the person, Day of the week, Number of Leaves left for him till now (which maybe a continuous decreasing variable), etc.
So here are the following questions based on above
How do I go about designing the training set for my logistic regression model.
In my training set, I find some variables are continuously decreasing (ex no of leaves left). Would that create any problem, because I know continuously increasing or decreasing variables are used in linear regression. Is that true ?
Any help is really appreciated. Thanks !
Well, there are a lot of missing information in your question, for example, it'll be very much clearer if you have provided all the features you have, but let me dare to throw some assumptions!
ML Modeling in classification always requires dealing with numerical inputs, and you can easily infer each of the unique input as an integer, especially the classes!
Now let me try to answer your questions:
How do I go about designing the training set for my logistic regression model.
How I see it, you have two options (not necessary both are practical, it's you who should decide according to the dataset you have and the problem), either you predict the probability of all employees in the company who will be off in a certain day according to the historical data you have (i.e. previous observations), in this case, each employee will represent a class (integer from 0 to the number of employees you want to include). Or you create a model for each employee, in this case the classes will be either off (i.e. Leave) or on (i.e. Present).
Example 1
I created a dataset example of 70 cases and 4 employees which looks like this:
Here each name is associated with the day and month they took as off with respect to how many Annual Leaves left for them!
The implementation (using Scikit-Learn) would be something like this (N.B date contains only day and month):
Now we can do something like this:
import math
import pandas as pd
import numpy as np
from sklearn.linear_model import LogisticRegression
from sklearn.model_selection import GridSearchCV, RepeatedStratifiedKFold
# read dataset example
df = pd.read_csv('leaves_dataset.csv')
# assign unique integer to every employee (i.e. a class label)
mapping = {'Jack': 0, 'Oliver': 1, 'Ruby': 2, 'Emily': 3}
df.replace(mapping, inplace=True)
y = np.array(df[['Name']]).reshape(-1)
X = np.array(df[['Leaves Left', 'Day', 'Month']])
# create the model
parameters = {'penalty': ['l1', 'l2'], 'C': [0.1, 0.5, 1.0, 10, 100, 1000]}
lr = LogisticRegression(random_state=0)
cv = RepeatedStratifiedKFold(n_splits=10, n_repeats=2, random_state=0)
clf = GridSearchCV(lr, parameters, cv=cv)
clf.fit(X, y)
#print(clf.best_estimator_)
#print(clf.best_score_)
# Example: probability of all employees who have 10 days left today
# warning: date must be same format
prob = clf.best_estimator_.predict_proba([[10, 9, 11]])
print({'Jack': prob[0,0], 'Oliver': prob[0,1], 'Ruby': prob[0,2], 'Emily': prob[0,3]})
Result
{'Ruby': 0.27545, 'Oliver': 0.15032,
'Emily': 0.28201, 'Jack': 0.29219}
N.B
To make this relatively work you need a real big dataset!
Also this can be better than the second one if there are other informative features in the dataset (e.g. the health status of the employee at that day..etc).
The second option is to create a model for each employee, here the result would be more accurate and more reliable, however, it's almost a nightmare if you have too many employees!
For each employee, you collect all their leaves in the past years and concatenate them into one file, in this case you have to complete all days in the year, in other words: for every day that employee has never got it off, that day should be labeled as on (or numerically speaking 1) and for the days off they should be labeled as off (or numerically speaking 0).
Obviously, in this case, the classes will be 0 and 1 (i.e. on and off) for each employee's model!
For example, consider this dataset example for the particular employee Jack:
Example 2
Then you can do for example:
import pandas as pd
import numpy as np
from sklearn.linear_model import LogisticRegression
from sklearn.model_selection import GridSearchCV, RepeatedStratifiedKFold
# read dataset example
df = pd.read_csv('leaves_dataset2.csv')
# assign unique integer to every on and off (i.e. a class label)
mapping = {'off': 0, 'on': 1}
df.replace(mapping, inplace=True)
y = np.array(df[['Type']]).reshape(-1)
X = np.array(df[['Leaves Left', 'Day', 'Month']])
# create the model
parameters = {'penalty': ['l1', 'l2'], 'C': [0.1, 0.5, 1.0, 10, 100, 1000]}
lr = LogisticRegression(random_state=0)
cv = RepeatedStratifiedKFold(n_splits=10, n_repeats=2, random_state=0)
clf = GridSearchCV(lr, parameters, cv=cv)
clf.fit(X, y)
#print(clf.best_estimator_)
#print(clf.best_score_)
# Example: probability of the employee "Jack" who has 10 days left today
prob = clf.best_estimator_.predict_proba([[10, 9, 11]])
print({'Off': prob[0,0], 'On': prob[0,1]})
Result
{'On': 0.33348, 'Off': 0.66651}
N.B in this case you have to create a dataset for each employee + training especial model + filling all the days the never taken in the past years as off!
In my training set, I find some variables are continuously decreasing (ex no of leaves left). Would that create any problem,
because I know continuously increasing or decreasing variables are
used in linear regression. Is that true ?
Well, there is nothing preventing you from using contentious values as features (e.g. number of leaves) in Logistic Regression; actually it doesn't make any difference if it's used in Linear or Logistic Regression but I believe you got confused between the features and the response:
The thing is, discrete values should be used in the response of Logistic Regression and Continuous values should be used in the response of the Linear Regression (a.k.a dependent variable or y).
I know of a couple of classification algorithms such as decision trees, but I can't use any of them to the problem I have at hands.
I have a dataset in which each row contains information about a purchase. It's columns are:
- customer id
- store id where the purchase took place
- date and time of the event
- amount of money spent
I'm trying to make a prediction that, given the information of who, where and when, predicts how much money is going to be spent.
What are some possible ways of doing this? Are there any well-known algorithms?
Also, I'm currently learning RapidMiner, and I'm experimenting with some of its features. Everything that I've tried there doesn't allow me to have a real number (amount spent) as a label. Maybe I'm doing something wrong?
You could use a Decision Tree Regressor for this. Using a toolkit like scikit-learn, you could use the DecisionTreeRegressor algo where your features would be store id, date and time, and customer id, and your target would be the amount spent.
You could turn this into a supervised learning problem. This is untested code, but it could probably get you started
# Load libraries
import numpy as np
import pylab as pl
from sklearn import datasets
from sklearn.tree import DecisionTreeRegressor
from sklearn import cross_validation
from sklearn import metrics
from sklearn import grid_search
def fit_predict_model(data_import):
"""Find and tune the optimal model. Make a prediction on housing data."""
# Get the features and labels from your data
X, y = data_import.data, data_import.target
# Setup a Decision Tree Regressor
regressor = DecisionTreeRegressor()
parameters = {'max_depth':(4,5,6,7), 'random_state': [1]}
scoring_function = metrics.make_scorer(metrics.mean_absolute_error, greater_is_better=False)
## fit your data to it ##
reg = grid_search.GridSearchCV(estimator = regressor, param_grid = parameters, scoring=scoring_function, cv=10, refit=True)
fitted_data = reg.fit(X, y)
print "Best Parameters: "
print fitted_data.best_params_
# Use the model to predict the output of a particular sample
x = [## input a test sample in this list ##]
y = reg.predict(x)
print "Prediction: " + str(y)
fit_predict_model(##your data in here)
I took this from a project I was working on almost directly to predict housing prices so there are probably some unnecessary libraries and without doing validation you have no clue how accurate this case would be, but this should get you started.
Check out this link:
http://scikit-learn.org/stable/modules/generated/sklearn.tree.DecisionTreeRegressor.html
Yes, as comments have pointed out it's regression that you need. Linear regression does sound like a good starting point as you don't have a huge number of variables.
In RapidMiner type regression into the Operators menu and you'll see several options under Modelling-> Functions. Linear Regression, Polynomical, Vector, etc. (There's more, but as a beginner let's start here).
Right click any of these operators and press Show Operator Info and you'll see numerical labels are allowed.
Next scroll through the help documentation of the operator and you'll see a link to a tutorial process. It's really simple to use, but it's good to get you started with an example.
Let me know if you need any help.
I am still very new to machine learning and trying to figure things out myself. I am using SciKit learn and have a data set of tweets with around 20,000 features (n_features=20,000). So far I achieved a precision, recall and f1 score of around 79%. I would like to use RFECV for feature selection and improve the performance of my model. I have read the SciKit learn documentation but am still a bit confused on how to use RFECV.
This is the code I have so far:
from sklearn.feature_extraction.text import CountVectorizer
from sklearn.feature_extraction.text import TfidfTransformer
from sklearn.naive_bayes import MultinomialNB
from sklearn.cross_validation import StratifiedShuffleSplit
from sklearn.cross_validation import cross_val_score
from sklearn.feature_selection import RFECV
from sklearn import metrics
# cross validation
sss = StratifiedShuffleSplit(y, 5, test_size=0.2, random_state=42)
for train_index, test_index in sss:
docs_train, docs_test = X[train_index], X[test_index]
y_train, y_test = y[train_index], y[test_index]
# feature extraction
count_vect = CountVectorizer(stop_words='English', min_df=3, max_df=0.90, ngram_range=(1,3))
X_CV = count_vect.fit_transform(docs_train)
tfidf_transformer = TfidfTransformer()
X_tfidf = tfidf_transformer.fit_transform(X_CV)
# Create the RFECV object
nb = MultinomialNB(alpha=0.5)
# The "accuracy" scoring is proportional to the number of correct classifications
rfecv = RFECV(estimator=nb, step=1, cv=2, scoring='accuracy')
rfecv.fit(X_tfidf, y_train)
X_rfecv=rfecv.transform(X_tfidf)
print("Optimal number of features : %d" % rfecv.n_features_)
# train classifier
clf = MultinomialNB(alpha=0.5).fit(X_rfecv, y_train)
# test clf on test data
X_test_CV = count_vect.transform(docs_test)
X_test_tfidf = tfidf_transformer.transform(X_test_CV)
X_test_rfecv = rfecv.transform(X_test_tfidf)
y_predicted = clf.predict(X_test_rfecv)
#print the mean accuracy on the given test data and labels
print ("Classifier score is: %s " % rfecv.score(X_test_rfecv,y_test))
Three questions:
1) Is this the correct way to use cross validation and RFECV? I am especially interested to know if I am running any risk of overfitting.
2) The accuracy of my model before and after I implemented RFECV with the above code are almost the same (around 78-79%), which puzzles me. I would expect performance to improve by using RFECV. Anything I might have missed here or could do differently to improve the performance of my model?
3) What other feature selection methods could you recommend me to try? I have tried RFE and SelectKBest so far, but they both haven't given me any improvement in terms of model accuracy.
To answer your questions:
There is a cross-validation built in the RFECV feature selection (hence the name), so you don't really need to have additional cross-validation for this single step. However since I understand you are running several tests, it's good to have an overall cross-validation to ensure you're not overfitting to a specific train-test split. I'd like to mention 2 points here:
I doubt the code behaves exactly like you think it does ;).
# cross validation
sss = StratifiedShuffleSplit(y, 5, test_size=0.2, random_state=42)
for train_index, test_index in sss:
docs_train, docs_test = X[train_index], X[test_index]
y_train, y_test = y[train_index], y[test_index]
# feature extraction
count_vect = CountVectorizer(stop_words='English', min_df=3, max_df=0.90, ngram_range=(1,3))
X_CV = count_vect.fit_transform(docs_train)
Here we first go through the loop, that has 5 iterations (n_iter parameter in StratifiedShuffleSplit). Then we go out of the loop and we just run all your code with the last values of train_index, test_index. So this is equivalent to a single train-test split where you probably meant to have 5. You should move your code back into the loop if you want it to run like a 'proper' cross validation.
You are worried about overfitting: indeed when 'looking for the best method' the risk exists that we're going to pick the method that works best... only on the small sample we're testing the method on.
Here the best practice is to have a first train-test split, then to perform cross-validation only using the train set. The test set can be used 'sparingly' when you think you found something, to make sure the scores you get are consistent and you're not overfitting.
It may look like you're throwing away 30% of your data (your test set), but it's absolutely worth it.
It can be puzzling to see feature selection does not have that big an impact. To introspect a bit more you could look into the evolution of the score with the number of selected features (see the example from the docs).
That being said, I don't think this is the right use case for RFE. Basically with your code you are eliminating features one by one, which probably takes a long time to run and does not make so much sense when you have 20000 features.
Other feature selection methods: here you mention SelectKBest but you don't tell us which method you use to score your features! SelectKBest will pick the K best features according to a score function. I'm guessing you were using the default which is ok, but it's better to have an idea of what the default does ;).
I would try SelectPercentile with chi2 as a score function. SelectPercentile is probably a bit more convenient than SelectKBest because if your dataset grows a percentage probably makes more sense than a hardcoded number of features.
Another example from the docs that does just that (and more).
Additional remarks:
You could use a TfidfVectorizer instead of a CountVectorizer followed by a TfidfTransformer. This is strictly equivalent.
You could use a pipeline object to pack the different steps of your classifier into a single object you can run cross validation on (I encourage you to read the docs, it's pretty useful).
from sklearn.feature_selection import chi2_sparse
from sklearn.feature_selection import SelectPercentile
from sklearn.pipeline import Pipeline
from sklearn.feature_extraction.text import TfidfVectorizer
pipeline = Pipeline(steps=[
("vectorizer", TfidfVectorizer(stop_words='English', min_df=3, max_df=0.90, ngram_range=(1,3))),
("selector", SelectPercentile(score_func=chi2, percentile=70)),
('NB', MultinomialNB(alpha=0.5))
])
Then you'd be able to run cross validation on the pipeline object to find the best combination of alpha and percentile, which is much harder to do with separate estimators.
Hope this helps, happy learning ;).