I am trying to create an ML model (regression) using various techniques like SMR, Logistic Regression, and others. With all the techniques, I'm not able to get efficiency more than 35%. Here's what I'm doing:
X_data = [X_data_distance]
X_data = np.vstack(X_data).astype(np.float64)
X_data = X_data.T
y_data = X_data_orders
#print(X_data.shape)
#print(y_data.shape)
#(10000, 1)
#(10000,)
X_train, X_test, y_train, y_test = train_test_split(X_data, y_data, test_size=0.33, random_state=42)
svr_rbf = SVC(kernel= 'rbf', C= 1.0)
svr_rbf.fit(X_train, y_train)
plt.plot(X_data_distance, svr_rbf.predict(X_data), color= 'red', label= 'RBF model')
For the plot, I'm getting the following:
I have tried various parameter tuning, changing the parameter C, gamma even tried different kernels, but nothing changes the accuracy. Even tried SVR, Logistic regression instead of SVC, but nothing helps. I tried different scaling for training input data like StandardScalar() and scale().
I used this as a reference
What should I do?
As a rule of thumb, we usually follow this convention:
For little number of features, go with Logistic Regression.
For a lot of features but not a lot of data, go with SVM.
For a lot of features and a lot of data, go with Neural Network.
Because your dataset is a 10K cases, it'd be better to use Logistic Regression because SVM will take forever to finish!.
Nevertheless, because your dataset contains a lot of classes, there is a chance of classes imbalance in your implementation. Thus I tried to workaround this problem via using the StratifiedKFold instead of train_test_split which doesn't guarantee balanced classes in the splits.
Moreover, I used GridSearchCV with StratifiedKFold to perform Cross-Validation in order to tune the parameters and try all different optimizers!
So the full implementation is as follows:
import pandas as pd
from sklearn.linear_model import LogisticRegression
from sklearn.metrics import accuracy_score
from sklearn.model_selection import GridSearchCV, StratifiedKFold, StratifiedShuffleSplit
import numpy as np
def getDataset(path, x_attr, y_attr):
"""
Extract dataset from CSV file
:param path: location of csv file
:param x_attr: list of Features Names
:param y_attr: Y header name in CSV file
:return: tuple, (X, Y)
"""
df = pd.read_csv(path)
X = X = np.array(df[x_attr]).reshape(len(df), len(x_attr))
Y = np.array(df[y_attr])
return X, Y
def stratifiedSplit(X, Y):
sss = StratifiedShuffleSplit(n_splits=1, test_size=0.2, random_state=0)
train_index, test_index = next(sss.split(X, Y))
X_train, X_test = X[train_index], X[test_index]
Y_train, Y_test = Y[train_index], Y[test_index]
return X_train, X_test, Y_train, Y_test
def run(X_data, Y_data):
X_train, X_test, Y_train, Y_test = stratifiedSplit(X_data, Y_data)
param_grid = {'C': [0.01, 0.1, 1, 10, 100, 1000], 'penalty': ['l1', 'l2'],
'solver':['newton-cg', 'lbfgs', 'liblinear', 'sag', 'saga']}
model = LogisticRegression(random_state=0)
clf = GridSearchCV(model, param_grid, cv=StratifiedKFold(n_splits=10))
clf.fit(X_train, Y_train)
print(accuracy_score(Y_train, clf.best_estimator_.predict(X_train)))
print(accuracy_score(Y_test, clf.best_estimator_.predict(X_test)))
X_data, Y_data = getDataset("data - Sheet1.csv", ['distance'], 'orders')
run(X_data, Y_data)
Despite all the attempts with all different algorithms, the accuracy didn't exceed 36%!!.
Why is that?
If you want to make a person recognize/classify another person by their T-shirt color, you cannot say: hey if it's red that means he's John and if it's red it's Peter but if it's red it's Aisling!! He would say "really, what the hack is the difference"?!!.
And that's exactly what is in your dataset!
Simply, run print(len(np.unique(X_data))) and print(len(np.unique(Y_data))) and you'll find that the numbers are so weird, in a nutshell you have:
Number of Cases: 10000 !!
Number of Classes: 118 !!
Number of Unique Inputs (i.e. Features): 66 !!
All classes are sharing hell a lot of information which make it impressive to have even up to 36% accuracy!
In other words, you have no informative features which lead to a lack in the uniqueness of each class model!
What to do?
I believe you are not allowed to remove some classes, so the only two solutions you have are:
Either live with this very valid result.
Or add more informative feature(s).
Update
Having you provided same dataset but with more features (i.e. complete set of features), the situation now is different.
I recommend you do the following:
Pre-process your dataset (i.e. prepare it by imputing missing values or deleting rows containing missing values, and converting dates to some unique values (example) ...etc).
Check what features are most important to the Orders Classes, you can achieve that by using of Forests of Trees to evaluate the importance of features. Here is a complete and simple example of how to do that in Scikit-Learn.
Create a new version of the dataset but this time hold Orders as the Y response, and the above-found features as the X variables.
Follow the same GrdiSearchCV and StratifiedKFold procedure that I showed you in the implementation above.
Hint
As per mentioned by Vivek Kumar in the comment below, stratify parameter has been added in Scikit-learn update to the train_test_split function.
It works by passing the array-like ground truth, so you don't need my workaround in the function stratifiedSplit(X, Y) above.
Related
I have a very imbalanced dataset. I used sklearn.train_test_split function to extract the train dataset. Now I want to oversample the train dataset, so I used to count number of type1(my data set has 2 categories and types(type1 and tupe2) but approximately all of my train data are type1. So I cant oversample.
Previously I used to split train test datasets with my written code. In that code 0.8 of all type1 data and 0.8 of all type2 data were in the train dataset.
How I can use this method with train_test_split function or other spliting methods in sklearn?
*I should just use sklearn or my own written methods.
You're looking for stratification. Why?
There's a parameter stratify in method train_test_split to which you can give the labels list e.g. :
from sklearn.model_selection import train_test_split
X_train, X_test, y_train, y_test = train_test_split(X, y,
stratify=y,
test_size=0.2)
There's also StratifiedShuffleSplit.
It seems like we both had similar issues here. Unfortunately, imbalanced-learn isn't always what you need and scikit does not offer the functionality you want. You will want to implement your own code.
This is what I came up for my application. Note that I have not had extensive time to debug it but I believe it works from the testing I have done. Hope it helps:
def equal_sampler(classes, data, target, test_frac):
# Find the least frequent class and its fraction of the total
_, count = np.unique(target, return_counts=True)
fraction_of_total = min(count) / len(target)
# split further into train and test
train_frac = (1-test_frac)*fraction_of_total
test_frac = test_frac*fraction_of_total
# initialize index arrays and find length of train and test
train=[]
train_len = int(train_frac * data.shape[0])
test=[]
test_len = int(test_frac* data.shape[0])
# add values to train, drop them from the index and proceed to add to test
for i in classes:
indeces = list(target[target ==i].index.copy())
train_temp = np.random.choice(indeces, train_len, replace=False)
for val in train_temp:
train.append(val)
indeces.remove(val)
test_temp = np.random.choice(indeces, test_len, replace=False)
for val in test_temp:
test.append(val)
# X_train, y_train, X_test, y_test
return data.loc[train], target[train], data.loc[test], target[test]
For the input, classes expects a list of possible values, data expects the dataframe columns used for prediction, target expects the target column.
Take care that the algorithm may not be extremely efficient, due to the triple for-loop(list.remove takes linear time). Despite that, it should be reasonably fast.
You may also look into stratified shuffle split as follows:
# We use a utility to generate artificial classification data.
from sklearn.datasets import make_classification
from sklearn.model_selection import StratifiedShuffleSplit
from sklearn.svm import SVC
from sklearn.pipeline import make_pipeline
X, y = make_classification(n_samples=100, n_informative=10, n_classes=2)
sss = StratifiedShuffleSplit(n_splits=5, test_size=0.5, random_state=0)
for train_index, test_index in sss.split(X, y):
print("TRAIN:", train_index, "TEST:", test_index)
X_train, X_test = X[train_index], X[test_index]
y_train, y_test = y[train_index], y[test_index]
clf = make_pipeline(StandardScaler(), SVC(gamma='auto'))
clf.fit(X_train, y_train)
y_pred = clf.predict(X_test)
I am trying to understand how the best coefficients are calculated in a logistic regression cross-validation, where the "refit" parameter is True.
If I understand the docs correctly, the best coefficients are the result of first determining the best regularization parameter "C", i.e., the value of C that has the highest average score over all folds. Then, the best coefficients are simply the coefficients that were calculated on the fold that has the highest score for the best C. I assume that if the maximum score is achieved by several folds, the coefficients of these folds would be averaged to give the best coefficients (I didn't see anything on how this case is handled in the docs).
To test my understanding, I determined the best coefficients in two different ways:
directly from the coef_ attribute of the fitted model, and
from the coefs_paths attribute, which contains the path of the coefficients obtained during cross-validating across each fold and then across each C.
The results I get from 1. and 2. are similar but not identical, so I was hoping someone could point out what I am doing wrong here.
Thanks!
An example to demonstrate the issue:
from sklearn.datasets import load_breast_cancer
import numpy as np
import pandas as pd
from sklearn.linear_model import LogisticRegressionCV
from sklearn.preprocessing import StandardScaler
from sklearn.model_selection import train_test_split
# Set parameters
n_folds = 10
C_values = [0.001, 0.01, 0.05, 0.1, 1., 100.]
# Load and preprocess data
cancer = load_breast_cancer()
X, y = cancer.data, cancer.target
X_train, X_test, y_train, y_test = train_test_split(X, y, random_state=0)
X_train_scaled = StandardScaler().fit_transform(X_train)
# Fit model
clf = LogisticRegressionCV(Cs=C_values, cv=n_folds, penalty='l1',
refit=True, scoring='roc_auc',
solver='liblinear', random_state=0,
fit_intercept=False)
clf.fit(X_train_scaled, y_train)
########################
# Get and plot coefficients using method 1
########################
coefs1 = clf.coef_
coefs1_series = pd.Series(coefs1.ravel(), index=cancer['feature_names'])
coefs1_series.sort_values().plot(kind="barh")
########################
# Get and plot coefficients using method 2
########################
# mean of scores of class "1"
scores = clf.scores_[1]
mean_scores = np.mean(scores, axis=0)
# Get index of the C that has the highest average score across all folds
best_C_idx = np.where(mean_scores==np.max(mean_scores))[0][0]
# Get index (here: indices) of the folds with highest scores for the
# best C
best_folds_idx = np.where(scores[:, best_C_idx]==np.max(scores[:, best_C_idx]))[0]
paths = clf.coefs_paths_[1] # has shape (n_folds, len(C_values), n_features)
coefs2 = np.squeeze(paths[best_folds_idx, best_C_idx, :])
coefs2 = np.mean(coefs2, axis=0)
coefs2_series = pd.Series(coefs2.ravel(), index=cancer['feature_names'])
coefs2_series.sort_values().plot(kind="barh")
I think this article answers your question: https://orvindemsy.medium.com/understanding-grid-search-randomized-cvs-refit-true-120d783a5e94.
The key point is the refit parameter of LogisticRegressionCV.
According to sklearn (https://scikit-learn.org/stable/modules/generated/sklearn.linear_model.LogisticRegressionCV.html)
refitbool, default=True
If set to True, the scores are averaged across all folds, and the coefs and the C that corresponds to the best score is taken, and a final refit is done using these parameters. Otherwise the coefs, intercepts and C that correspond to the best scores across folds are averaged.
Best.
I have he following code to run a 10-fold cross validation in SkLearn:
cv = model_selection.KFold(n_splits=10, shuffle=True, random_state=0)
scores = model_selection.cross_val_score(MyEstimator(), x_data, y_data, cv=cv, scoring='mean_squared_error') * -1
For debugging purposes, while I am trying to make MyEstimator work, I would like to run only one fold of this cross-validation, instead of all 10. Is there an easy way to keep this code but just say to run the first fold and then exit?
I would still like that data is split into 10 parts, but that only one combination of that 10 parts is fitted and scored, instead of 10 combinations.
No, not with cross_val_score I suppose. You can set n_splits to minimum value of 2, but still that will be 50:50 split of train, test which you may not want.
If you want maintain a 90:10 ration and test other parts of code like MyEstimator(), then you can use a workaround.
You can use KFold.split() to get the first set of train and test indices and then break the loop after first iteration.
cv = model_selection.KFold(n_splits=10, shuffle=True, random_state=0)
for train_index, test_index in cv.split(x_data):
print("TRAIN:", train_index, "TEST:", test_index)
X_train, X_test = x_data[train_index], x_data[test_index]
y_train, y_test = y_data[train_index], y_data[test_index]
break
Now use this X_train, y_train to train the estimator and X_test, y_test to score it.
Instead of :
scores = model_selection.cross_val_score(MyEstimator(),
x_data, y_data,
cv=cv,
scoring='mean_squared_error')
Your code becomes:
myEstimator_fitted = MyEstimator().fit(X_train, y_train)
y_pred = myEstimator_fitted.predict(X_test)
from sklearn.metrics import mean_squared_error
# I am appending to a scores list object, because that will be output of cross_val_score.
scores = []
scores.append(mean_squared_error(y_test, y_pred))
Rest assured, cross_val_score will be doing this only internally, just some enhancements for parallel processing.
I'm comparing the results of a logistic regressor written in Keras to the default Sklearn Logreg. My input is one-dimensional. My output has two classes and I'm interested in the probability that the output belongs to the class 1.
I'm expecting the results to be almost identical, but they are not even close.
Here is how I generate my random data. Note that X_train, X_test are still vectors, I'm just using capital letters because I'm used to it. Also there is no need for scaling in this case.
X = np.linspace(0, 1, 10000)
y = np.random.sample(X.shape)
y = np.where(y<X, 1, 0)
Here's cumsum of y plotted over X. Doing a regression here is not rocket science.
I do a standard train-test-split:
X_train, X_test, y_train, y_test = train_test_split(X, y)
X_train = X_train.reshape(-1,1)
X_test = X_test.reshape(-1,1)
Next, I train a default logistic regressor:
from sklearn.linear_model import LogisticRegression
sk_lr = LogisticRegression()
sk_lr.fit(X_train, y_train)
sklearn_logreg_result = sk_lr.predict_proba(X_test)[:,1]
And a logistic regressor that I write in Keras:
from keras.models import Sequential
from keras.layers import Dense
keras_lr = Sequential()
keras_lr.add(Dense(1, activation='sigmoid', input_dim=1))
keras_lr.compile(loss='mse', optimizer='sgd', metrics=['accuracy'])
_ = keras_lr.fit(X_train, y_train, verbose=0)
keras_lr_result = keras_lr.predict(X_test)[:,0]
And a hand-made solution:
pearson_corr = np.corrcoef(X_train.reshape(X_train.shape[0],), y_train)[0,1]
b = pearson_corr * np.std(y_train) / np.std(X_train)
a = np.mean(y_train) - b * np.mean(X_train)
handmade_result = (a + b * X_test)[:,0]
I expect all three to deliver similar results, but here is what happens. This is a reliability diagram using 100 bins.
I have played around with loss functions and other parameters, but the Keras logreg stays roughly like this. What might be causing the problem here?
edit: Using binary crossentropy is not the solution here, as shown by this plot (note that the input data has changed between the two plots).
While both implementations are a form of Logistic Regression there's quite a few differences. While both solutions converge to a comparable minimum (0.75/0.76 ACC) they are not identical.
Optimizer - keras uses vanille SGD where sklearn's LR is based on
liblinear which implements trust region Newton method
Regularization - sklearn has built in L2 regularization
Weights -The weights are randomly initialized and probably sampled from a different distribution.
I'm using scikit-learn cross_validation(http://scikit-learn.org/stable/modules/cross_validation.html) and get for example 0.82 mean score(r2_scorer).
How could I know do I have over-fitting or under-fitting using scikit-learn functions?
Unfortunately I confirm that there is no built-in tool to compare train and test scores in a CV setup. The cross_val_score tool only reports test scores.
You can setup your own loop with the train_test_split function as in Ando's answer but you can also use any other CV scheme.
import numpy as np
from sklearn.cross_validation import KFold
from sklearn.metrics import SCORERS
scorer = SCORERS['r2']
cv = KFold(5)
train_scores, test_scores = [], []
for train, test in cv:
regressor.fit(X[train], y[train])
train_scores.append(scorer(regressor, X[train], y[train]))
test_scores.append(scorer(regressor, X[test], y[test]))
mean_train_score = np.mean(train_scores)
mean_test_score = np.mean(test_scores)
If you compute the mean train and test scores with cross validation you can then find out if you are:
Underfitting: the train score is far from the perfect score (which is 1.0 for r2)
Overfitting: the train and test scores are not close from on another (the mean test score is significantly lower than the mean train score).
Note: you can be both significantly underfitting and overfitting at the same time if your model is inadequate and your data is too noisy.
You should compare your scores when testing on training and testing data. If the scores are close to equal, you are likely underfitting. If they are far apart, you are likely overfitting (unless using a method such as random forest).
To compute the scores for both train and test data, you can use something along the following (assuming your data is in variables X and Y):
from sklearn import cross_validation
#do five iterations
for i in range(5):
X_train, X_test, y_train, y_test = cross_validation.train_test_split(X, Y, test_size=0.4)
#Your predictor, linear SVM in this example
clf = svm.SVC(kernel='linear', C=1).fit(X_train, y_train)
print "Test score", clf.score(X_test, y_test)
print "Train score", clf.score(X_train, y_train)