Let's assume that we have a few data points that can be used as the training set. Each row is consisted of 4 say columns (features) that take boolean values. The 5th column expresses the class and it also takes boolean values. Here is an example (they are almost random):
1,1,1,0,1
0,1,1,0,1
1,1,0,0,1
0,0,0,0,0
1,0,0,1,0
0,0,0,0,0
Now, what I want to do is to build a model such that for any given input (new line) the system does not return the class itself (like in the case of a regular classification problem) but instead the probability this particular input belongs to class 0 or class 1. How can I do that? What's more, how can I generate a confidence interval or error rate associated with that computation?
Not all classification algorithms return probabilities, because not all of them have an underlying probabilistic model. For example, a classification tree is just a set of rules that you follow to assign each new input to a particular class.
An example of a classification algorithm that does have an underlying probabilistic model is logistic regression. In this algorithm, the probability that a particular input x is in the class is
prob = 1 / (1 + exp( -theta * x ))
where theta is a vector of coefficients with the same number of dimensions as x. Generally to move from probabilities to classifications, you simply threshold, e.g.
if prob < 0.5
return 0;
else
return 1;
end
Other classification algorithms may have probabilistic interpretations, for example random forests are essentially a voting algorithm with multiple classification trees. If 80% of the trees vote for class 1 and 20% vote for class 2, then you could output an 80% probability of being in class 1. But this is a side effect of how the model works, rather than an explicit underlying probability model.
Related
I have a classification task. The training data has 50 different labels. The customer wants to differentiate the low probability predictions, meaning that, I have to classify some test data as Unclassified / Other depending on the probability (certainty?) of the model.
When I test my code, the prediction result is a numpy array (I'm using different models, this is one is pre-trained BertTransformer). The prediction array doesn't contain probabilities such as in Keras predict_proba() method. These are numbers generated by prediction method of pretrained BertTransformer model.
[[-1.7862008 -0.7037363 0.09885322 1.5318055 2.1137428 -0.2216074
0.18905772 -0.32575375 1.0748093 -0.06001111 0.01083148 0.47495762
0.27160102 0.13852511 -0.68440574 0.6773654 -2.2712054 -0.2864312
-0.8428862 -2.1132915 -1.0157436 -1.0340284 -0.35126117 -1.0333195
9.149789 -0.21288703 0.11455813 -0.32903734 0.10503325 -0.3004114
-1.3854568 -0.01692022 -0.4388664 -0.42163098 -0.09182278 -0.28269592
-0.33082992 -1.147654 -0.6703184 0.33038092 -0.50087476 1.1643585
0.96983343 1.3400391 1.0692116 -0.7623776 -0.6083422 -0.91371405
0.10002492]]
I'm using numpy.argmax() to identify the correct label. The prediction works just fine. However, since these are not probabilities, I cannot compare the best result with a threshold value.
My question is, how can I define a threshold (say, 0.6), and then compare the probability of the argmax() element of the BertTransformer prediction array so that I can classify the prediction as "Other" if the probability is less than the threshold value?
Edit 1:
We are using 2 different models. One is Keras, and the other is BertTransformer. We have no problem in Keras since it gives the probabilities so I'm skipping Keras model.
The Bert model is pretrained. Here is how it is generated:
def model(self, data):
number_of_categories = len(data['encoded_categories'].unique())
model = BertForSequenceClassification.from_pretrained(
"dbmdz/bert-base-turkish-128k-uncased",
num_labels=number_of_categories,
output_attentions=False,
output_hidden_states=False,
)
# model.cuda()
return model
The output given above is the result of model.predict() method. We compare both models, Bert is slightly ahead, therefore we know that the prediction works just fine. However, we are not sure what those numbers signify or represent.
Here is the Bert documentation.
BertForSequenceClassification returns logits, i.e., the classification scores before normalization. You can normalize the scores by calling F.softmax(output, dim=-1) where torch.nn.functional was imported as F.
With thousands of labels, the normalization can be costly and you do not need it when you are only interested in argmax. This is probably why the models return the raw scores only.
Is it possible to include weights in a poisson point process model fitted to a logistic regression quadrature scheme? My data is a stratified sample and I would like to account for this sampling strategy in order to have valid population level predictions.
This is a question about the model-fitting function ppm in the R package spatstat.
Yes, you can include survey weights. The easiest way is to create a covariate surveyweight, which could be a function(x,y) or a pixel image or a column of data associated with your quadrature scheme. Then when fitting the model using ppm, add the model term +offset(log(surveyweight)).
The result of ppm will be a fitted model that describes the observed point pattern. You can do prediction, simulation etc from this model, but be aware that these will be predictions or simulations of the observed point process including the effect of non-constant survey effort.
To get a prediction or simulation of the original point process (i.e. after removing the effect of non-constant survey effort) you need to replace the original covariate surveyweight by another covariate that is constant and equal to 1, then pass this to predict.ppm in the argument newdata.
Here are a few lines to elaborate on the answer by #adrian-baddeley.
If you have the setup of your related question and we imagine you have the weights and two covariates in a data.frame in the same order as the points of your quadscheme:
library(spatstat)
X <- split(chorley)$larynx
D <- split(chorley)$lung
Q <- quadscheme.logi(X,D)
covar <- data.frame(weights = runif(npoints(chorley)),
covar1 = rnorm(npoints(chorley)),
covar2 = rnorm(npoints(chorley)))
fit <- ppm(Q ~ offset(log(weights)) + covar1 + covar2, data = covar)
Say for example, a dataset contains 60% instances for "Yes" class and 30% instances for "NO" class.
In this scenario, Precision, Recall for the random classifier are
Precision =60%
Recall =50%
Then, what will be the accuracy for random classifier in this scenario?
Some caution is required here, since the very definition of a random classifier is somewhat ambiguous; this is best illustrated in cases of imbalanced data.
By definition, the accuracy of a binary classifier is
acc = P(class=0) * P(prediction=0) + P(class=1) * P(prediction=1)
where P stands for probability.
Indeed, if we stick to the intuitive definition of a random binary classifier as giving
P(prediction=0) = P(prediction=1) = 0.5
then the accuracy computed by the above formula is always 0.5, irrespectively of the class distribution (i.e. the values of P(class=0) and P(class=1)).
However, in this definition, there is an implicit assumption, i.e. that our classes are balanced, each one consisting of 50% of our dataset.
This assumption (and the corresponding intuition) breaks down in cases of class imbalance: if we have a dataset where, say, 90% of samples are of class 0 (i.e. P(class=0)=0.9), then it doesn't make much sense to use the above definition of a random binary classifier; instead, we should use the percentages of the class distributions themselves as the probabilities of our random classifier, i.e.:
P(prediction=0) = P(class=0) = 0.9
P(prediction=1) = P(class=1) = 0.1
Now, plugging these values to the formula defining the accuracy, we get:
acc = P(class=0) * P(prediction=0) + P(class=1) * P(prediction=1)
= (0.9 * 0.9) + (0.1 * 0.1)
= 0.82
which is nowhere close to the naive value of 0.5...
As I already said, AFAIK there are no clear-cut definitions of a random classifier in the literature. Sometimes the "naive" random classifier (always flip a fair coin) is referred to as a "random guess" classifier, while what I have described is referred to as a "weighted guess" one, but still this is far from being accepted as a standard...
The bottom line here is the following: since the main reason for using a random classifier is as a baseline, it makes sense to do so only in relatively balanced datasets. In your case of a 60-40 balance, the result turns out to be 0.52, which is admittedly not far from the naive one of 0.5; but for highly imbalanced datasets (e.g. 90-10), the usefulness itself of the random classifier as a baseline ceases to exist, since the correct baseline has become "always predict the majority class", which here would give an accuracy of 90%, in contrast to the random classifier accuracy of just 82% (let alone the 50% accuracy of the naive approach)...
As #desertnaut mentioned, if you're after a naïve benchmark for your model you're always better using "always predict the majority class" as your benchmark, achieving accuracy of %of_samples_in_majority_class (which is always better than either a random guess or a weighted guess).
In Deepchecks (a package I maintain) we have a check that automatically compares the performance of your model to a simple model (either weighted random, majority class or simple decision tree).
from deepchecks.checks import SimpleModelComparison
from deepchecks import Dataset
SimpleModelComparison().run(Dataset(train_df, label='target'), Dataset(test_df, label='target'), model)
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)
As a basic proof of concept, in a network that classifies K classes with input x, bias b, output y,S samples, weights v and t teacher signal in which t(k) equals 1 if the matching sample is under k class.
Variables
Let x_(is) represent the i_(th) input feature in the s_(th) sample.
v_(ks) represents the vector that holds the weights of connection to k_(th) output from all inputs within the s_(th) sample.
t_(s) represents the teacher signal for s_(th) sample.
If we extend the above variables to consider multiple samples, the changes below has to be applied while declaring the variable z_(k), the activation function f(.) and using the corss entropy as a cost function:
Derivation
Typically in learning rule, delta ( t_(k) - y_(k) ) is always included, why Delta doesnt show up in this equation? have i missed something or the delta rule showing up isnt a must?
I managed to find the solution, it's clear when we consider the Kronecker delta in which Where (δck = 1 if a class matches the classifier and δck otherwise). which means the derivation takes this shape:
Derivation
which leads to the delta rule.