I am a new in Machine Learning area & I am (trying to) implementing anomaly detection algorithms, one algorithm is Autoencoder implemented with help of keras from tensorflow library and the second one is IsolationForest implemented with help of sklearn library and I want to compare these algorithms with help of roc_auc_score ( function from Python), but I am not sure if I am doing it correct.
In documentation of roc_auc_score function I can see, that for input it should be like:
sklearn.metrics.roc_auc_score(y_true, y_score, average=’macro’, sample_weight=None, max_fpr=None
y_true :
True binary labels or binary label indicators.
y_score :
Target scores, can either be probability estimates of the positive class, confidence values, or non-thresholded measure of decisions (as returned by “decision_function” on some classifiers). For binary y_true, y_score is supposed to be the score of the class with greater label.
For AE I am computing roc_auc_score like this:
model.fit(...) # model from https://www.tensorflow.org/api_docs/python/tf/keras/Sequential
pred = model.predict(x_test) # predict function from https://www.tensorflow.org/api_docs/python/tf/keras/Sequential#predict
metric = np.mean(np.power(x_test - pred, 2), axis=1) #MSE
print(roc_auc_score(y_test, metric) # where y_test is true binary labels 0/1
For IsolationForest I am computing roc_auc_score like this:
model.fit(...) # model from https://scikit-learn.org/stable/modules/generated/sklearn.ensemble.IsolationForest.html
metric = -(model.score_samples(x_test)) # https://scikit-learn.org/stable/modules/generated/sklearn.ensemble.IsolationForest.html#sklearn.ensemble.IsolationForest.score_samples
print(roc_auc_score(y_test, metric) #where y_test is true binary labels 0/1
I am just curious if returned roc_auc_score from both implementations of AE and IsolationForest are comparable (I mean, if I am computing them in the correct way)? Especially in AE model, where I am putting MSE into the roc_auc_score (if not, what should be the input as y_score to this function?)
Comparing AE and IsolationForest in the context of anomaly dection using sklearn.metrics.roc_auc_score based on scores coming from AE MSE loss and IF decision_function() respectively is okay. Varying range of the y_score when switching classifier isn't an issue, since this range is taken into account for each classifier when computing the AUC.
To understand that AUC isn't range dependent, remember that you travel along the decision function values to obtain the ROC points. Rescaling the decision function values will only change the decision function thresholds accordingly, defining similar points of the ROC since the new thresholds will lead each to the same TPR and FPR as they did before the rescaling.
Couldn't find a convincing code line in sklearn.metrics.roc_auc_score's implementation, but you can easily observe this comparison in published code associated with a research paper. For example, in the Deep One-Class Classification paper's code (I'm not an author, I know the paper's code because I'm reproducing their results), AE MSE loss and IF decision_function() are the roc_auc_score inputs (whose outputs the paper is comparing):
AE roc_auc_score computation
Found in this script on github.
from sklearn.metrics import roc_auc_score
(...)
scores = torch.sum((outputs - inputs) ** 2, dim=tuple(range(1, outputs.dim())))
(...)
auc = roc_auc_score(labels, scores)
IsolationForest roc_auc_score computation
Found in this script on github.
from sklearn.metrics import roc_auc_score
(...)
scores = (-1.0) * self.isoForest.decision_function(X.astype(np.float32)) # compute anomaly score
y_pred = (self.isoForest.predict(X.astype(np.float32)) == -1) * 1 # get prediction
(...)
auc = roc_auc_score(y, scores.flatten())
Note: The two scripts come from two different repositories but are actually the source of a single paper's results. The authors only chose to create an extra repository for their PyTorch implementation of an AD method requiring a neural network.
If I had 2 features x1 and x2 where I know that the pattern is:
if x1 < x2 then
class1
else
class2
Can any machine learning algorithm find such a pattern? What algorithm would that be?
I know that I could create a third feature x3 = x1-x2. Then feature x3 can easily be used by some machine learning algorithms. For example a decision tree can solve the problem 100% using x3 and just 3 nodes (1 decision and 2 leaf nodes).
But, is it possible to solve this without creating new features? This seems like a problem that should be easily solved 100% if a machine learning algorithm could only find such a pattern.
I tried MLP and SVM with different kernels, including svg kernel and the results are not great. As an example of what I tried, here is the scikit-learn code where the SVM could only get a score of 0.992:
import numpy as np
from sklearn.svm import SVC
# Generate 1000 samples with 2 features with random values
X_train = np.random.rand(1000,2)
# Label each sample. If feature "x1" is less than feature "x2" then label as 1, otherwise label is 0.
y_train = X_train[:,0] < X_train[:,1]
y_train = y_train.astype(int) # convert boolean to 0 and 1
svc = SVC(kernel = "rbf", C = 0.9) # tried all kernels and C values from 0.1 to 1.0
svc.fit(X_train, y_train)
print("SVC score: %f" % svc.score(X_train, y_train))
Output running the code:
SVC score: 0.992000
This is an oversimplification of my problem. The real problem may have hundreds of features and different patterns, not just x1 < x2. However, to start with it would help a lot to know how to solve for this simple pattern.
To understand this, you must go into the settings of all the parameters provided by sklearn, and C in particular. It also helps to understand how the value of C influences the classifier's training procedure.
If you look at the equation in the User Guide for SVC, there are two main parts to the equation - the first part tries to find a small set of weights that solves the problem, and the second part tries to minimize the classification errors.
C is the penalty multiplier associated with misclassifications. If you decrease C, then you reduce the penalty (lower training accuracy but better generalization to test) and vice versa.
Try setting C to 1e+6. You will see that you almost always get 100% accuracy. The classifier has learnt the pattern x1 < x2. But it figures that a 99.2% accuracy is enough when you look at another parameter called tol. This controls how much error is negligible for you and by default it is set to 1e-3. If you reduce the tolerance, you can also expect to get similar results.
In general, I would suggest you to use something like GridSearchCV (link) to find the optimal values of hyper parameters like C as this internally splits the dataset into train and validation. This helps you to ensure that you are not just tweaking the hyperparameters to get a good training accuracy but you are also making sure that the classifier will do well in practice.
I am using the LogisticRegression() method in scikit-learn on a highly unbalanced data set. I have even turned the class_weight feature to auto.
I know that in Logistic Regression it should be possible to know what is the threshold value for a particular pair of classes.
Is it possible to know what the threshold value is in each of the One-vs-All classes the LogisticRegression() method designs?
I did not find anything in the documentation page.
Does it by default apply the 0.5 value as threshold for all the classes regardless of the parameter values?
There is a little trick that I use, instead of using model.predict(test_data) use model.predict_proba(test_data). Then use a range of values for thresholds to analyze the effects on the prediction;
pred_proba_df = pd.DataFrame(model.predict_proba(x_test))
threshold_list = [0.05,0.1,0.15,0.2,0.25,0.3,0.35,0.4,0.45,0.5,0.55,0.6,0.65,.7,.75,.8,.85,.9,.95,.99]
for i in threshold_list:
print ('\n******** For i = {} ******'.format(i))
Y_test_pred = pred_proba_df.applymap(lambda x: 1 if x>i else 0)
test_accuracy = metrics.accuracy_score(Y_test.as_matrix().reshape(Y_test.as_matrix().size,1),
Y_test_pred.iloc[:,1].as_matrix().reshape(Y_test_pred.iloc[:,1].as_matrix().size,1))
print('Our testing accuracy is {}'.format(test_accuracy))
print(confusion_matrix(Y_test.as_matrix().reshape(Y_test.as_matrix().size,1),
Y_test_pred.iloc[:,1].as_matrix().reshape(Y_test_pred.iloc[:,1].as_matrix().size,1)))
Best!
Logistic regression chooses the class that has the biggest probability. In case of 2 classes, the threshold is 0.5: if P(Y=0) > 0.5 then obviously P(Y=0) > P(Y=1). The same stands for the multiclass setting: again, it chooses the class with the biggest probability (see e.g. Ng's lectures, the bottom lines).
Introducing special thresholds only affects in the proportion of false positives/false negatives (and thus in precision/recall tradeoff), but it is not the parameter of the LR model. See also the similar question.
Yes, Sci-Kit learn is using a threshold of P>=0.5 for binary classifications. I am going to build on some of the answers already posted with two options to check this:
One simple option is to extract the probabilities of each classification using the output from model.predict_proba(test_x) segment of the code below along with class predictions (output from model.predict(test_x) segment of code below). Then, append class predictions and their probabilities to your test dataframe as a check.
As another option, one can graphically view precision vs. recall at various thresholds using the following code.
### Predict test_y values and probabilities based on fitted logistic
regression model
pred_y=log.predict(test_x)
probs_y=log.predict_proba(test_x)
# probs_y is a 2-D array of probability of being labeled as 0 (first
column of
array) vs 1 (2nd column in array)
from sklearn.metrics import precision_recall_curve
precision, recall, thresholds = precision_recall_curve(test_y, probs_y[:,
1])
#retrieve probability of being 1(in second column of probs_y)
pr_auc = metrics.auc(recall, precision)
plt.title("Precision-Recall vs Threshold Chart")
plt.plot(thresholds, precision[: -1], "b--", label="Precision")
plt.plot(thresholds, recall[: -1], "r--", label="Recall")
plt.ylabel("Precision, Recall")
plt.xlabel("Threshold")
plt.legend(loc="lower left")
plt.ylim([0,1])
we can use a wrapper as follows:
model = LogisticRegression()
model.fit(X, y)
def custom_predict(X, threshold):
probs = model.predict_proba(X)
return (probs[:, 1] > threshold).astype(int)
new_preds = custom_predict(X=X, threshold=0.4)
I'm trying to train a CNN to categorize text by topic. When I use binary cross-entropy I get ~80% accuracy, with categorical cross-entropy I get ~50% accuracy.
I don't understand why this is. It's a multiclass problem, doesn't that mean that I have to use categorical cross-entropy and that the results with binary cross-entropy are meaningless?
model.add(embedding_layer)
model.add(Dropout(0.25))
# convolution layers
model.add(Conv1D(nb_filter=32,
filter_length=4,
border_mode='valid',
activation='relu'))
model.add(MaxPooling1D(pool_length=2))
# dense layers
model.add(Flatten())
model.add(Dense(256))
model.add(Dropout(0.25))
model.add(Activation('relu'))
# output layer
model.add(Dense(len(class_id_index)))
model.add(Activation('softmax'))
Then I compile it either it like this using categorical_crossentropy as the loss function:
model.compile(loss='categorical_crossentropy', optimizer='adam', metrics=['accuracy'])
or
model.compile(loss='binary_crossentropy', optimizer='adam', metrics=['accuracy'])
Intuitively it makes sense why I'd want to use categorical cross-entropy, I don't understand why I get good results with binary, and poor results with categorical.
The reason for this apparent performance discrepancy between categorical & binary cross entropy is what user xtof54 has already reported in his answer below, i.e.:
the accuracy computed with the Keras method evaluate is just plain
wrong when using binary_crossentropy with more than 2 labels
I would like to elaborate more on this, demonstrate the actual underlying issue, explain it, and offer a remedy.
This behavior is not a bug; the underlying reason is a rather subtle & undocumented issue at how Keras actually guesses which accuracy to use, depending on the loss function you have selected, when you include simply metrics=['accuracy'] in your model compilation. In other words, while your first compilation option
model.compile(loss='categorical_crossentropy', optimizer='adam', metrics=['accuracy'])
is valid, your second one:
model.compile(loss='binary_crossentropy', optimizer='adam', metrics=['accuracy'])
will not produce what you expect, but the reason is not the use of binary cross entropy (which, at least in principle, is an absolutely valid loss function).
Why is that? If you check the metrics source code, Keras does not define a single accuracy metric, but several different ones, among them binary_accuracy and categorical_accuracy. What happens under the hood is that, since you have selected binary cross entropy as your loss function and have not specified a particular accuracy metric, Keras (wrongly...) infers that you are interested in the binary_accuracy, and this is what it returns - while in fact you are interested in the categorical_accuracy.
Let's verify that this is the case, using the MNIST CNN example in Keras, with the following modification:
model.compile(loss='binary_crossentropy', optimizer='adam', metrics=['accuracy']) # WRONG way
model.fit(x_train, y_train,
batch_size=batch_size,
epochs=2, # only 2 epochs, for demonstration purposes
verbose=1,
validation_data=(x_test, y_test))
# Keras reported accuracy:
score = model.evaluate(x_test, y_test, verbose=0)
score[1]
# 0.9975801164627075
# Actual accuracy calculated manually:
import numpy as np
y_pred = model.predict(x_test)
acc = sum([np.argmax(y_test[i])==np.argmax(y_pred[i]) for i in range(10000)])/10000
acc
# 0.98780000000000001
score[1]==acc
# False
To remedy this, i.e. to use indeed binary cross entropy as your loss function (as I said, nothing wrong with this, at least in principle) while still getting the categorical accuracy required by the problem at hand, you should ask explicitly for categorical_accuracy in the model compilation as follows:
from keras.metrics import categorical_accuracy
model.compile(loss='binary_crossentropy', optimizer='adam', metrics=[categorical_accuracy])
In the MNIST example, after training, scoring, and predicting the test set as I show above, the two metrics now are the same, as they should be:
# Keras reported accuracy:
score = model.evaluate(x_test, y_test, verbose=0)
score[1]
# 0.98580000000000001
# Actual accuracy calculated manually:
y_pred = model.predict(x_test)
acc = sum([np.argmax(y_test[i])==np.argmax(y_pred[i]) for i in range(10000)])/10000
acc
# 0.98580000000000001
score[1]==acc
# True
System setup:
Python version 3.5.3
Tensorflow version 1.2.1
Keras version 2.0.4
UPDATE: After my post, I discovered that this issue had already been identified in this answer.
It all depends on the type of classification problem you are dealing with. There are three main categories
binary classification (two target classes),
multi-class classification (more than two exclusive targets),
multi-label classification (more than two non exclusive targets), in which multiple target classes can be on at the same time.
In the first case, binary cross-entropy should be used and targets should be encoded as one-hot vectors.
In the second case, categorical cross-entropy should be used and targets should be encoded as one-hot vectors.
In the last case, binary cross-entropy should be used and targets should be encoded as one-hot vectors. Each output neuron (or unit) is considered as a separate random binary variable, and the loss for the entire vector of outputs is the product of the loss of single binary variables. Therefore it is the product of binary cross-entropy for each single output unit.
The binary cross-entropy is defined as
and categorical cross-entropy is defined as
where c is the index running over the number of classes C.
I came across an "inverted" issue — I was getting good results with categorical_crossentropy (with 2 classes) and poor with binary_crossentropy. It seems that problem was with wrong activation function. The correct settings were:
for binary_crossentropy: sigmoid activation, scalar target
for categorical_crossentropy: softmax activation, one-hot encoded target
It's really interesting case. Actually in your setup the following statement is true:
binary_crossentropy = len(class_id_index) * categorical_crossentropy
This means that up to a constant multiplication factor your losses are equivalent. The weird behaviour that you are observing during a training phase might be an example of a following phenomenon:
At the beginning the most frequent class is dominating the loss - so network is learning to predict mostly this class for every example.
After it learnt the most frequent pattern it starts discriminating among less frequent classes. But when you are using adam - the learning rate has a much smaller value than it had at the beginning of training (it's because of the nature of this optimizer). It makes training slower and prevents your network from e.g. leaving a poor local minimum less possible.
That's why this constant factor might help in case of binary_crossentropy. After many epochs - the learning rate value is greater than in categorical_crossentropy case. I usually restart training (and learning phase) a few times when I notice such behaviour or/and adjusting a class weights using the following pattern:
class_weight = 1 / class_frequency
This makes loss from a less frequent classes balancing the influence of a dominant class loss at the beginning of a training and in a further part of an optimization process.
EDIT:
Actually - I checked that even though in case of maths:
binary_crossentropy = len(class_id_index) * categorical_crossentropy
should hold - in case of keras it's not true, because keras is automatically normalizing all outputs to sum up to 1. This is the actual reason behind this weird behaviour as in case of multiclassification such normalization harms a training.
After commenting #Marcin answer, I have more carefully checked one of my students code where I found the same weird behavior, even after only 2 epochs ! (So #Marcin's explanation was not very likely in my case).
And I found that the answer is actually very simple: the accuracy computed with the Keras method evaluate is just plain wrong when using binary_crossentropy with more than 2 labels. You can check that by recomputing the accuracy yourself (first call the Keras method "predict" and then compute the number of correct answers returned by predict): you get the true accuracy, which is much lower than the Keras "evaluate" one.
a simple example under a multi-class setting to illustrate
suppose you have 4 classes (onehot encoded) and below is just one prediction
true_label = [0,1,0,0]
predicted_label = [0,0,1,0]
when using categorical_crossentropy, the accuracy is just 0 , it only cares about if you get the concerned class right.
however when using binary_crossentropy, the accuracy is calculated for all classes, it would be 50% for this prediction. and the final result will be the mean of the individual accuracies for both cases.
it is recommended to use categorical_crossentropy for multi-class(classes are mutually exclusive) problem but binary_crossentropy for multi-label problem.
As it is a multi-class problem, you have to use the categorical_crossentropy, the binary cross entropy will produce bogus results, most likely will only evaluate the first two classes only.
50% for a multi-class problem can be quite good, depending on the number of classes. If you have n classes, then 100/n is the minimum performance you can get by outputting a random class.
You are passing a target array of shape (x-dim, y-dim) while using as loss categorical_crossentropy. categorical_crossentropy expects targets to be binary matrices (1s and 0s) of shape (samples, classes). If your targets are integer classes, you can convert them to the expected format via:
from keras.utils import to_categorical
y_binary = to_categorical(y_int)
Alternatively, you can use the loss function sparse_categorical_crossentropy instead, which does expect integer targets.
model.compile(loss='sparse_categorical_crossentropy', optimizer='adam', metrics=['accuracy'])
when using the categorical_crossentropy loss, your targets should be in categorical format (e.g. if you have 10 classes, the target for each sample should be a 10-dimensional vector that is all-zeros except for a 1 at the index corresponding to the class of the sample).
Take a look at the equation you can find that binary cross entropy not only punish those label = 1, predicted =0, but also label = 0, predicted = 1.
However categorical cross entropy only punish those label = 1 but predicted = 1.That's why we make assumption that there is only ONE label positive.
The main point is answered satisfactorily with the brilliant piece of sleuthing by desernaut. However there are occasions when BCE (binary cross entropy) could throw different results than CCE (categorical cross entropy) and may be the preferred choice. While the thumb rules shared above (which loss to select) work fine for 99% of the cases, I would like to add a few new dimensions to this discussion.
The OP had a softmax activation and this throws a probability distribution as the predicted value. It is a multi-class problem. The preferred loss is categorical CE. Essentially this boils down to -ln(p) where 'p' is the predicted probability of the lone positive class in the sample. This means that the negative predictions dont have a role to play in calculating CE. This is by intention.
On a rare occasion, it may be needed to make the -ve voices count. This can be done by treating the above sample as a series of binary predictions. So if expected is [1 0 0 0 0] and predicted is [0.1 0.5 0.1 0.1 0.2], this is further broken down into:
expected = [1,0], [0,1], [0,1], [0,1], [0,1]
predicted = [0.1, 0.9], [.5, .5], [.1, .9], [.1, .9], [.2, .8]
Now we proceed to compute 5 different cross entropies - one for each of the above 5 expected/predicted combo and sum them up. Then:
CE = -[ ln(.1) + ln(0.5) + ln(0.9) + ln(0.9) + ln(0.8)]
The CE has a different scale but continues to be a measure of the difference between the expected and predicted values. The only difference is that in this scheme, the -ve values are also penalized/rewarded along with the +ve values. In case your problem is such that you are going to use the output probabilities (both +ve and -ves) instead of using the max() to predict just the 1 +ve label, then you may want to consider this version of CE.
How about a multi-label situation where expected = [1 0 0 0 1]? Conventional approach is to use one sigmoid per output neuron instead of an overall softmax. This ensures that the output probabilities are independent of each other. So we get something like:
expected = [1 0 0 0 1]
predicted is = [0.1 0.5 0.1 0.1 0.9]
By definition, CE measures the difference between 2 probability distributions. But the above two lists are not probability distributions. Probability distributions should always add up to 1. So conventional solution is to use same loss approach as before - break the expected and predicted values into 5 individual probability distributions, proceed to calculate 5 cross entropies and sum them up. Then:
CE = -[ ln(.1) + ln(0.5) + ln(0.9) + ln(0.9) + ln(0.9)] = 3.3
The challenge happens when the number of classes may be very high - say a 1000 and there may be only couple of them present in each sample. So the expected is something like: [1,0,0,0,0,0,1,0,0,0.....990 zeroes]. The predicted could be something like: [.8, .1, .1, .1, .1, .1, .8, .1, .1, .1.....990 0.1's]
In this case the CE =
- [ ln(.8) + ln(.8) for the 2 +ve classes and 998 * ln(0.9) for the 998 -ve classes]
= 0.44 (for the +ve classes) + 105 (for the negative classes)
You can see how the -ve classes are beginning to create a nuisance value when calculating the loss. The voice of the +ve samples (which may be all that we care about) is getting drowned out. What do we do? We can't use categorical CE (the version where only +ve samples are considered in calculation). This is because, we are forced to break up the probability distributions into multiple binary probability distributions because otherwise it would not be a probability distribution in the first place. Once we break it into multiple binary probability distributions, we have no choice but to use binary CE and this of course gives weightage to -ve classes.
One option is to drown the voice of the -ve classes by a multiplier. So we multiply all -ve losses by a value gamma where gamma < 1. Say in above case, gamma can be .0001. Now the loss comes to:
= 0.44 (for the +ve classes) + 0.105 (for the negative classes)
The nuisance value has come down. 2 years back Facebook did that and much more in a paper they came up with where they also multiplied the -ve losses by p to the power of x. 'p' is the probability of the output being a +ve and x is a constant>1. This penalized -ve losses even further especially the ones where the model is pretty confident (where 1-p is close to 1). This combined effect of punishing negative class losses combined with harsher punishment for the easily classified cases (which accounted for majority of the -ve cases) worked beautifully for Facebook and they called it focal loss.
So in response to OP's question of whether binary CE makes any sense at all in his case, the answer is - it depends. In 99% of the cases the conventional thumb rules work but there could be occasions when these rules could be bent or even broken to suit the problem at hand.
For a more in-depth treatment, you can refer to: https://towardsdatascience.com/cross-entropy-classification-losses-no-math-few-stories-lots-of-intuition-d56f8c7f06b0
The binary_crossentropy(y_target, y_predict) doesn't need to apply to binary classification problem.
In the source code of binary_crossentropy(), the nn.sigmoid_cross_entropy_with_logits(labels=target, logits=output) of tensorflow was actually used.
And, in the documentation, it says that:
Measures the probability error in discrete classification tasks in which each class is independent and not mutually exclusive. For instance, one could perform multilabel classification where a picture can contain both an elephant and a dog at the same time.