The image on the left shows a standard ROC curve formed by sweeping a single threshold and recording the corresponding True Positive Rate (TPR) and False Positive Rate (FPR).
The image on the right shows my problem setup where there are 3 parameters, and for each, we have only 2 choices. Together, it produces 8 points as depicted on the graph. In practice, I intend to have thousands of possible combinations of 100s of parameters, but the concept remains the same in this down-scaled case.
I intend to find 2 things here:
Determine the optimum parameter(s) for the given data
Provide an overall performance score for all combinations of parameters
In the case of the ROC curve on the left, this is done easily using the following methods:
Optimal parameter: Maximal difference of TPR and FPR with a cost component (I believe it is called the J-statistic?)
Overall performance: Area under the curve (the shaded portion in the graph)
However, for my case in the image on the right, I do not know if the methods I have chosen are the standard principled methods that are normally used.
Optimal parameter set: Same maximal difference of TPR and FPR
Parameter score = TPR - FPR * cost_ratio
Overall performance: Average of all "parameter scores"
I have found a lot of reference material for the ROC curve with a single threshold and while there are other techniques available to determine the performance, the ones mentioned in this question is definitely considered a standard approach. I found no such reading material for the scenario presented on the right.
Bottomline, the question here is two-fold: (1) Provide methods to evaluate the optimal parameter set and overall performance in my problem scenario, (2) Provide reference that claims the suggested methods to be a standard approach for the given scenario.
P.S.: I had first posted this question on the "Cross Validated" forum, but didn't get any responses, in fact, got only 7 views in 15 hours.
I'm going to expand a little on aberger's previous answer on a Grid Search. As with any tuning of a model it's best to optimise hyper-parameters using one portion of the data and evaluate those parameters using another proportion of the data, so GridSearchCV is best for this purpose.
First I'll create some data and split it into training and test
import numpy as np
from sklearn import model_selection, ensemble, metrics
np.random.seed(42)
X = np.random.random((5000, 10))
y = np.random.randint(0, 2, 5000)
X_train, X_test, y_train, y_test = model_selection.train_test_split(X, y, test_size=0.3)
This gives us a classification problem, which is what I think you're describing, though the same would apply to regression problems too.
Now it's helpful to think about what parameters you may want to optimise. A cross-validated grid search is a computational expensive process, so the smaller the search space the quicker it gets done. I will show an example for a RandomForestClassifier because it's my go to model.
clf = ensemble.RandomForestClassifier()
parameters = {'n_estimators': [10, 20, 30],
'max_features': [5, 8, 10],
'max_depth': [None, 10, 20]}
So now I have my base estimator and a list of parameters that I want to optimise. Now I just have to think about how I want to evaluate each of the models that I'm going to build. It seems from your question that you're interested in the ROC AUC, so that's what I'll use for this example. Though you can chose from many default metrics in scikit or even define your own.
gs = model_selection.GridSearchCV(clf, param_grid=parameters,
scoring='roc_auc', cv=5)
gs.fit(X_train, y_train)
This will fit a model for all possible combinations of parameters that I have given it, using 5-fold cross-validation evaluate how well those parameters performed using the ROC AUC. Once that's been fit, we can look at the best parameters and pull out the best performing model.
print gs.best_params_
clf = gs.best_estimator_
Outputs:
{'max_features': 5, 'n_estimators': 30, 'max_depth': 20}
Now at this point you may want to retrain your classifier on all of the training data, as currently it's been trained using cross-validation. Some people prefer not to, but I'm a retrainer!
clf.fit(X_train, y_train)
So now we can evaluate how well the model performs on both our training and test set.
print metrics.classification_report(y_train, clf.predict(X_train))
print metrics.classification_report(y_test, clf.predict(X_test))
Outputs:
precision recall f1-score support
0 1.00 1.00 1.00 1707
1 1.00 1.00 1.00 1793
avg / total 1.00 1.00 1.00 3500
precision recall f1-score support
0 0.51 0.46 0.48 780
1 0.47 0.52 0.50 720
avg / total 0.49 0.49 0.49 1500
We can see that this model has overtrained by the poor score on the test set. But this is not surprising as the data is just random noise! Hopefully when performing these methods on data with a signal you will end up with a well-tuned model.
EDIT
This is one of those situations where 'everyone does it' but there's no real clear reference to say this is the best way to do it. I would suggest looking for an example close to the classification problem that you're working on. For example using Google Scholar to search for "grid search" "SVM" "gene expression"
I feeeeel like we're talking about Grid Search in scikit-learn. It (1), provides methods to evaluate optimal (hyper)parameters and (2), is implemented in a massively popular and well referenced statistical software package.
Related
I have 38 variables, like oxygen, temperature, pressure, etc and have a task to determine the total yield produced every day from these variables. When I calculate the regression coefficients and intercept value, they seem to be abnormal and very high (Impractical). For example, if 'temperature' coefficient was found to be +375.456, I could not give a meaning to them saying an increase in one unit in temperature would increase yield by 375.456g. That's impractical in my scenario. However, the prediction accuracy seems right. I would like to know, how to interpret these huge intercept( -5341.27355) and huge beta values shown below. One other important point is that I removed multicolinear columns and also, I am not scaling the variables/normalizing them because I need beta coefficients to have meaning such that I could say, increase in temperature by one unit increases yield by 10g or so. Your inputs are highly appreciated!
modl.intercept_
Out[375]: -5341.27354961415
modl.coef_
Out[376]:
array([ 1.38096017e+00, -7.62388829e+00, 5.64611255e+00, 2.26124164e-01,
4.21908571e-01, 4.50695302e-01, -8.15167717e-01, 1.82390184e+00,
-3.32849969e+02, 3.31942553e+02, 3.58830763e+02, -2.05076898e-01,
-3.06404757e+02, 7.86012402e+00, 3.21339318e+02, -7.00817205e-01,
-1.09676321e+04, 1.91481734e+00, 6.02929848e+01, 8.33731416e+00,
-6.23433431e+01, -1.88442804e+00, 6.86526274e+00, -6.76103795e+01,
-1.11406021e+02, 2.48270706e+02, 2.94836048e+01, 1.00279016e+02,
1.42906659e-02, -2.13019683e-03, -6.71427100e+02, -2.03158515e+02,
9.32094007e-03, 5.56457014e+01, -2.91724945e+00, 4.78691176e-01,
8.78121854e+00, -4.93696073e+00])
It's very unlikely that all of these variables are linearly correlated, so I would suggest that you have a look at simple non-linear regression techniques, such as Decision Trees or Kernel Ridge Regression. These are however more difficult to interpret.
Going back to your issue, these high weights might well be due to there being some high amount of correlation between the variables, or that you simply don't have very much training data.
If you instead of linear regression use Lasso Regression, the solution is biased away from high regression coefficients, and the fit will likely improve as well.
A small example on how to do this in scikit-learn, including cross validation of the regularization hyper-parameter:
from sklearn.linear_model LassoCV
# Make up some data
n_samples = 100
n_features = 5
X = np.random.random((n_samples, n_features))
# Make y linear dependent on the features
y = np.sum(np.random.random((1,n_features)) * X, axis=1)
model = LassoCV(cv=5, n_alphas=100, fit_intercept=True)
model.fit(X,y)
print(model.intercept_)
If you have a linear regression, the formula looks like this (y= target, x= features inputs):
y= x1*b1 +x2*b2 + x3*b3 + x4*b4...+ c
where b1,b2,b3,b4... are your modl.coef_. AS you already realized one of your bigges number is 3.319+02 = 331 and the intercept is also quite big with -5431.
As you already mentioned the coeffiecient variables means how much the target variable changes, if the coeffiecient feature changes with 1 unit and all others features are constant.
so for your interpretation, the higher the absoult coeffienct, the higher the influence of your analysis. But it is important to note that the model is using a lot of high coefficient, that means your model is not depending only of one variable
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.
I have been trying to get into more details of resampling methods and implemented them on a small data set of 1000 rows. The data was split into 800 training set and 200 validation set. I used K-fold cross validation and repeated K-fold cross validation to train the KNN using the training set. Based on my understanding I have done some interpretations of the results - however, I have certain doubts about them (see questions below):
Results :
10 Fold Cv
Resampling: Cross-Validated (10 fold)
Summary of sample sizes: 720, 720, 720, 720, 720, 720, ...
Resampling results across tuning parameters:
k Accuracy Kappa
5 0.6600 0.07010791
7 0.6775 0.09432414
9 0.6800 0.07054371
Accuracy was used to select the optimal model using the largest value.
The final value used for the model was k = 9.
Repeated 10 fold with 10 repeats
Resampling results across tuning parameters:
k Accuracy Kappa
5 0.670250 0.10436607
7 0.676875 0.09288219
9 0.683125 0.08062622
Accuracy was used to select the optimal model using the largest value.
The final value used for the model was k = 9.
10 fold, 1000 repeats
k Accuracy Kappa
5 0.6680438 0.09473128
7 0.6753375 0.08810406
9 0.6831800 0.07907891
Accuracy was used to select the optimal model using the largest value.
The final value used for the model was k = 9.
10 fold with 2000 repeats
k Accuracy Kappa
5 0.6677981 0.09467347
7 0.6750369 0.08713170
9 0.6826894 0.07772184
Doubts:
While selecting the parameter, K=9 is the optimal value for highest accuracy. However, I don't understand how to take Kappa into consideration while finally choosing parameter value?
Repeat number has to be increased until we get stabilised result, the accuracy changes when the repeats are increased from 10 to 1000. However,the results are similar for 1000 repeats and 2000 repeats. Will it be right to consider the results of 1000/2000 repeats to be stabilised performance estimate?
Any thumb rule for the repeat number?
Finally,should I train the model on my complete training data (800 rows) now test the accuracy on the validation set ?
Accuracy and Kappa are just different classification performance metrics. In a nutshell, their difference is that Accuracy does not take possible class imbalance into account when calculating the metrics, while Kappa does. Therefore, with imbalanced classes, you might be better off using Kappa. With R caret you can do so via the train::metric parameter.
You could see a similar effect of slightly different performance results when running e.g. the 10CV with 10 repeats multiple times - you will just get slightly different results for those as well. Something you should look out for is the variance of classification performance over your partitions and repeats. In case you obtain a small variance you can derive that you by training on all your data, you likely obtain a model that will give you similar (hence stable) results on new data. But, in case you obtain a huge variance, you can derive that just by chance (being lucky or unlucky) you might instead obtain a model that either gives you rather good or rather bad performance on new data. BTW: the prediction performance variance is something e.g. R caret::train will give you automatically, hence I'd advice on using it.
See above: look at the variance and increase the repeats until you can e.g. repeat the whole process and obtain a similar average performance and variance of performance.
Yes, CV and resampling methods exist to give you information about how well your model will perform on new data. So, after performing CV and resampling and obtaining this information, you will usually use all your data to train a final model that you use in your e.g. application scenario (this includes both train and test partition!).
I'm using an example extracted from the book "Mastering Machine Learning with scikit learn".
It uses a decision tree to predict whether each of the images on a web page is an
advertisement or article content. Images that are classified as being advertisements could then be hidden using Cascading Style Sheets. The data is publicly available from the Internet Advertisements Data Set: http://archive.ics.uci.edu/ml/datasets/Internet+Advertisements, which contains data for 3,279 images.
The following is the complete code for completing the classification task:
import pandas as pd
from sklearn.tree import DecisionTreeClassifier
from sklearn.cross_validation import train_test_split
from sklearn.metrics import classification_report
from sklearn.pipeline import Pipeline
from sklearn.grid_search import GridSearchCV
import sys,random
def main(argv):
df = pd.read_csv('ad-dataset/ad.data', header=None)
explanatory_variable_columns = set(df.columns.values)
response_variable_column = df[len(df.columns.values)-1]
explanatory_variable_columns.remove(len(df.columns.values)-1)
y = [1 if e == 'ad.' else 0 for e in response_variable_column]
X = df[list(explanatory_variable_columns)]
X.replace(to_replace=' *\?', value=-1, regex=True, inplace=True)
X_train, X_test, y_train, y_test = train_test_split(X, y,random_state=100000)
pipeline = Pipeline([('clf',DecisionTreeClassifier(criterion='entropy',random_state=20000))])
parameters = {
'clf__max_depth': (150, 155, 160),
'clf__min_samples_split': (1, 2, 3),
'clf__min_samples_leaf': (1, 2, 3)
}
grid_search = GridSearchCV(pipeline, parameters, n_jobs=-1,verbose=1, scoring='f1')
grid_search.fit(X_train, y_train)
print 'Best score: %0.3f' % grid_search.best_score_
print 'Best parameters set:'
best_parameters = grid_search.best_estimator_.get_params()
for param_name in sorted(parameters.keys()):
print '\t%s: %r' % (param_name, best_parameters[param_name])
predictions = grid_search.predict(X_test)
print classification_report(y_test, predictions)
if __name__ == '__main__':
main(sys.argv[1:])
The RESULTS of using scoring='f1' in GridSearchCV as in the example is:
The RESULTS of using scoring=None (by default Accuracy measure) is the same as using F1 score:
If I'm not wrong optimizing the parameter search by different scoring functions should yield different results. The following case shows that different results are obtained when scoring='precision' is used.
The RESULTS of using scoring='precision' is DIFFERENT than the other two cases. The same would be true for 'recall', etc:
WHY 'F1' AND None, BY DEFAULT ACCURACY, GIVE THE SAME RESULT??
EDITED
I agree with both answers by Fabian & Sebastian. The problem should be the small param_grid. But I just wanted to clarify that the problem surged when I was working with a totally different (not the one in the example here) highly imbalance dataset 100:1 (which should affect the accuracy) and using Logistic Regression. In this case also 'F1' and accuracy gave the same result.
The param_grid that I used, in this case, was the following:
parameters = {"penalty": ("l1", "l2"),
"C": (0.001, 0.01, 0.1, 1, 10, 100),
"solver": ("newton-cg", "lbfgs", "liblinear"),
"class_weight":[{0:4}],
}
I guess that the parameter selection is also too small.
I think that the author didn't choose this example very well. I may be missing something here, but min_samples_split=1 doesn't make sense to me: Isn't it the same as setting min_samples_split=2 since you can't split 1 sample -- essentially, it's a waste of computational time.
From the documentation: min_samples_split: "The minimum number of samples required to split an internal node."
Btw. this is a very small grid and there is not much choice anyways, which may explain why accuracy and f1 give you the same parameter combinations and hence the same scoring tables.
Like mentioned above, the dataset may be well balanced which is why F1 and accuracy scores may prefer the same parameter combinations. So, looking further at your GridSearch results using (a) F1 score and (b) Accuracy, I conclude that in both cases a depth of 150 works best. Since this is the lower boundary, it gives you a slight hind that lower "depth" values may work even better. However, I suspect that the tree doesn't even go that deep on this dataset (you can end up with "pure" leaves even well before reaching the max depth).
So, let's repeat the experiment with a little bit more sensible values using the following parameter grid
parameters = {
'clf__max_depth': list(range(2, 30)),
'clf__min_samples_split': (2,),
'clf__min_samples_leaf': (1,)
}
The optimal "depth" for the best F1 score seems to be around 15.
Best score: 0.878
Best parameters set:
clf__max_depth: 15
clf__min_samples_leaf: 1
clf__min_samples_split: 2
precision recall f1-score support
0 0.98 0.99 0.99 716
1 0.92 0.89 0.91 104
avg / total 0.98 0.98 0.98 820
Next, let's try it using "accuracy" (or None) as our scoring metric:
> Best score: 0.967
Best parameters set:
clf__max_depth: 6
clf__min_samples_leaf: 1
clf__min_samples_split: 2
precision recall f1-score support
0 0.98 0.99 0.98 716
1 0.93 0.85 0.88 104
avg / total 0.97 0.97 0.97 820
As you can see, you get different results now, and the "optimal" depth is different if you use "accuracy."
I don't agree that optimizing the parameter search by different scoring functions should yield necessarily different results necessarily. If your dataset is balanced (roughly same number of samples in each class), I would expect that model selection by accuracy and F1 would yield very similar results.
Also, have in mind that GridSearchCV optimizes over a discrete grid. Maybe using a thinner grid of parameters would yield the results that you are looking for.
On an unbalanced dataset use the "labels" parameter of the f1_score scorer to use only the f1 score of the class you are interested in. Or consider using "sample_weight".
I have crafted a little program for gender classification based on image of a face. I used Yale face databse (175 images for males and the same number for females), converted them to grayscale and equalized histograms, so after preprocessing images look like this:
I ran following code to test results (it uses SVM and linear kernel):
def run_gender_classifier():
Xm, Ym = mkdataset('gender/male', 1) # mkdataset just preprocesses images,
Xf, Yf = mkdataset('gender/female', 0) # flattens them and stacks into a matrix
X = np.vstack([Xm, Xf])
Y = np.hstack([Ym, Yf])
X_train, X_test, Y_train, Y_test = train_test_split(X, Y,
test_size=0.1,
random_state=100)
model = svm.SVC(kernel='linear')
model.fit(X_train, Y_train)
print("Results:\n%s\n" % (
metrics.classification_report(
Y_test, model.predict(X_test))))
And got 100% precision!
In [22]: run_gender_classifier()
Results:
precision recall f1-score support
0 1.00 1.00 1.00 16
1 1.00 1.00 1.00 19
avg / total 1.00 1.00 1.00 35
I could expect different results, but 100% correct for image classification look really suspicious to me.
Furthermore, when I changed kernel to RBF, results became totally bad:
In [24]: run_gender_classifier()
Results:
precision recall f1-score support
0 0.46 1.00 0.63 16
1 0.00 0.00 0.00 19
avg / total 0.21 0.46 0.29 35
Which seems even more strange for me.
So my questions are:
Is there any mistake in my approach or code?
If not, how can results for linear kernel be so good, and for RBF so bad?
Note, that I also got 100% correct results with logistic regression, and very poor results with deep belief networks, so it's not specific to SVM, but rather for linear and non-linear models.
Just for completeness, here's my code for preprocessing and making dataset:
import cv2
from sklearn import linear_model, svm, metrics
from sklearn.cross_validation import train_test_split
def preprocess(im):
im = cv2.cvtColor(im, cv2.COLOR_BGR2GRAY)
im = cv2.resize(im, (100, 100))
return cv2.equalizeHist(im)
def mkdataset(path, label):
images = (cv2.resize(cv2.imread(fname), (100, 100))
for fname in list_images(path))
images = (preprocess(im) for im in images)
X = np.vstack([im.flatten() for im in images])
Y = np.repeat(label, X.shape[0])
return X, Y
All of described models require tuning parameters:
Linear SVM : C
RBF SVM : C, gamma
DBN : Layers count, Neurons count, Output classifier, Training rate ...
And you simply omitted this element. So it is quite natural, that models with smallest number of tunable parameters behaved better - as simply there is bigger probability that default parameters actually worked.
100% score always looks suspicious and you should double check it "by hand" - phisically split data into train and test (put into different directories), train on one part, save your model to a file. Then in separate code - load a model, and test it on test files with displaying image + label from the model. This way you will make sure, that there is no implmenentation error (as you really don't care whether there is any processing error, if you have a physical proof that your model recognizes those faces, right?). This is purely "psychological method", which makes it obvious that there is no error in data splitting/sharing and further evaluation.
UPDATE
As suggested in the comment, I also checked your dataset, and as as it is stated on the official website:
The extended Yale Face Database B contains 16128 images of 28 human subjects under 9 poses and 64 illumination conditions.
So this is for sure a problem - this is not the dataset for the gender recognition. Your classifier simply memorizes these 28 subjects, which are easily splitted to male/female. It simply won't work on any image from other subjects. The only "valuable" part of this dataset is the set of 28 faces of distinctive individuals, which you can extract by hand, but 28 images seems at least row of magnitude too small to be useful.
Friend from what I understand your description of your questions and I think the explanation is simple, due to the problem with linear kernel is better than the RBF believe in your logic is correct however you should is using RBF somewhat wrong I think that it will serve to your problem, continue trying to develop a way to just use the linear kernel