I am trying to use the 'is_unbalance' parameter in my model training for a binary classification problem where the positive class is approximately 3%. If I set the parameter 'is_unbalance', I observe that the binary log loss drops in the first iteration but then keeps on increasing. I'm noticing this behavior only if I enable this parameter 'is_unbalance'. Otherwise, there is a steady drop in log_loss. Appreciate your help on this. Thanks.
When you do not balance the sets for such an unbalanced dataset, then obviously the objective value will always drop - and will probably reach the point of classifying all the predictions to the majority class, while having a fantastic objective value.
Balancing the classes is necessary, but it doesn't mean that you should stop on is_unbalanced - you can use sample_pos_weight, have customized metric, or apply weights to your samples, like following:
WEIGHTS = y_train.value_counts(normalize = True).min() / y_train.value_counts(normalize = True)
TRAIN_WEIGHTS = pd.DataFrame(y_train.rename('old_target')).merge(WEIGHTS, how = 'left', left_on = 'old_target', right_on = WEIGHTS.index).target.values
train_data = lgb.Dataset(X_train, label=y_train, weight = TRAIN_WEIGHTS)
Also, optimizing other hyperparameters should solve the issue of increasing log_loss.
When you set Is_unbalace: True, the algorithm will try to Automatically balance the weight of the dominated label (with the pos/neg fraction in train set).
If you want change scale_pos_weight (it is by default 1 which mean assume both positive and negative label are equal) in case of unbalance dataset you can use following formula(based on this issue on lightgbm repository) to set it correctly.
sample_pos_weight = number of negative samples / number of positive samples
Related
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)
Here is my code for SVC classifier.
vectorizer = TfidfVectorizer(lowercase=False)
train_vectors = vectorizer.fit_transform(training_data)
classifier_linear = svm.LinearSVC()
clf = CalibratedClassifierCV(classifier_linear)
linear_svc_model = clf.fit(train_vectors, train_labels)
training_data here is a list of english sentences and train_lables are the labels associated. I do the usual stopwords removal and some preprocessing before creating final version of training_data. Here is how my testing code:
test_lables = ["no"]
test_vectors = vectorizer.transform(test_lables)
prediction_linear = clf.predict_proba(test_vectors)
counter = 0
class_probability = {}
lables = []
for item in train_labels:
if item in lables:
continue
else:
lables.append(item)
for val in np.nditer(prediction_linear):
new_val = val.item(0)
class_probability[lables[counter]] = new_val
counter = counter + 1
sorted_class_probability = sorted(class_probability.items(), key=operator.itemgetter(1), reverse=True)
print(sorted_class_probability)
Now when I run the code with a phrase that is already there in the training set (a word 'no' in this case), it identifies properly, but the confidence score is even below .9. The output is as follows:
[('no', 0.8474342514152964), ('hi', 0.06830103628879058), ('thanks', 0.03070201906552546), ('confused', 0.02647134535600733), ('ok', 0.015857384248465656), ('yes', 0.005961945963546264), ('bye', 0.005272017662368208)]
When I am studying online, I have seen that usually confidence score for data already in the training set is closer to 1 or almost 1 and rest of them are really negligible. What can I do to get better confidence score? Should I be worried that if I add more classes to it, the confidence score will further dip and it will be difficult for me to surely point out one standout class?
As long as your scores help you classify your inputs correctly, you shouldn't worry at all. If anything, if your confidence on the input already in your training data is too high, that probably means your method has overfit to the data, and cannot generalize to the unseen data.
However, you can tune the complexity of your method by changing the penalization parameters. In the case of a LinearSVC, you have both the penalty and the C parameter. Try different values of those two and observe the effect. Make sure you also observe the effect on an unseen test set.
Just a not that the values of C should be in exponential space, eg. [0.001, 0.01, 0.1, 1, 10, 100, 1000] for you to see meaningful effects.
The SGDClassifier may be relevant to your case if you're interested in such linear models and tuning your parameters.
I have a set of inputs that has 5000ish features with values that vary from 0.005 to 9000000. Each of the features has similar values (a feature with a value of 10ish will not also have a value of 0.1ish)
I am trying to apply linear regression to this data set, however, the wide range of input values is inhibiting effective gradient descent.
What is the best way to handle this variance? If normalization is best, please include details on the best way to implement this normalization.
Thanks!
Simply perform it as a pre-processing step. You can do it as following:
1) Calculate mean values for each of the features in the training set and store it. Be careful, do not mess up feature mean and sample mean, so you will have a vector of size [number_of_features (5000ish)].
2) Calculate std. for each feature in the training set and store it. Size of [number_of_feature] as well
3) Update each training and testing entry as:
updated = (original_vector - mean_vector)/ std_vector
That's it!
The code will look like:
# train_data shape [train_length,5000]
# test_data [test_length, 5000]
mean = np.mean(train_data,1)
std = np.std(train_data,1)
normalized_train_data = (train_data - mean)/ std
normalized_test_data = (test_data - mean)/ std
I am trying to produce a mathematical operation selection nn model, which is based on the scalar input. The operation is selected based on the softmax result which is produce by the nn. Then this operation has to be applied to the scalar input in order to produce the final output. So far I’ve come up with applying argmax and onehot on the softmax output in order to produce a mask which then is applied on the concated values matrix from all the possible operations to be performed (as show in the pseudo code below). The issue is that neither argmax nor onehot appears to be differentiable. I am new to this, so any would be highly appreciated. Thanks in advance.
#perform softmax
logits = tf.matmul(current_input, W) + b
softmax = tf.nn.softmax(logits)
#perform all possible operations on the input
op_1_val = tf_op_1(current_input)
op_2_val = tf_op_2(current_input)
op_3_val = tf_op_2(current_input)
values = tf.concat([op_1_val, op_2_val, op_3_val], 1)
#create a mask
argmax = tf.argmax(softmax, 1)
mask = tf.one_hot(argmax, num_of_operations)
#produce the input, by masking out those operation results which have not been selected
output = values * mask
I believe that this is not possible. This is similar to Hard Attention described in this paper. Hard attention is used in Image captioning to allow the model to focus only on a certain part of the image at each step. Hard attention is not differentiable but there are 2 ways to go around this:
1- Use Reinforcement Learning (RL): RL is made to train models that makes decisions. Even though, the loss function won't back-propagate any gradients to the softmax used for the decision, you can use RL techniques to optimize the decision. For a simplified example, you can consider the loss as penalty, and send to the node, with the maximum value in the softmax layer, a policy gradient proportional to the penalty in order to decrease the score of the decision if it was bad (results in a high loss).
2- Use something like soft attention: instead of picking only one operation, mix them with weights based on the softmax. so instead of:
output = values * mask
Use:
output = values * softmax
Now, the operations will converge down to zero based on how much the softmax will not select them. This is easier to train compared to RL but it won't work if you must completely remove the non-selected operations from the final result (set them to zero completely).
This is another answer that talks about Hard and Soft attention that you may find helpful: https://stackoverflow.com/a/35852153/6938290
I'm trying image inpainting using a NN with weights pretrained using denoising autoencoders. All according to https://papers.nips.cc/paper/4686-image-denoising-and-inpainting-with-deep-neural-networks.pdf
I have made the custom loss function they are using.
My set is a batch of overlapping patches (196x32x32) of an image. My input are the corrupted batches of the image, and the output should be the cleaned ones.
Part of my loss function is
dif_y = tf.subtract(y_xi,y_)
dif_norm = tf.norm(dif_y, ord = 'euclidean', axis = (1,2))
Where y_xi(196 x 1 x 3072) is the reconstructed clean image and y_ (196 x 1 x 3072) is the real clean image. So I actually I subtract all images from their corrupted version and I sum all those differences. I think it is normal to be a very big number.
train_step = tf.train.AdamOptimizer().minimize(loss)
The loss value begins at around 3*10^7 and is converging after 200 runs (I loop for 1000) at a close value. So my output image will be miles away from the original.
Edit: starts at 3.02391e+07 and converges to 3.02337e+07
Is there any way my loss value is correct? If so, how can I dramatically reduce it?
Thanks
Edit 2: My loss function
dif_y = tf.subtract(y,y_)
dif_norm = tf.norm(dif_y, ord = 'euclidean', axis = (1,2))
sqr_norm = tf.square(dif_norm)
prod = tf.multiply(sqr_norm,0.5)
sum_norm2 = tf.reduce_sum(prod,0)
error_1 = tf.divide(sum_norm2,196)
Just for the record if anyone else has a similar problem: Remember to normalize your data! I was actually subtracting values in range [0,1] from values in range [0,255]. Very noobish mistake, I learned it the hard way!
Input values / 255
Expected values / 255
Problem solved.
sum_norm2 = tf.reduce_sum(prod,0) - I don't think this is doing what you want it to do.
Say y and y_ have values for 500 images and you have 10 labels for a 500x10 matrix. When tf.reduce_sum(prod,0) processes that you will have 1 value that is the sum of 500 values each which will be the sum of all values in the 2nd rank.
I don't think that is what you want, the sum of the error across each label. Probably what you want is the average, at least in my experience that is what works wonders for me. Additionally, I don't want a whole bunch of losses, one for each image, but instead one loss for the batch.
My preference is to use something like
loss = tf.reduce_mean ( tf.reduce_mean( prod ) )
This has the additional upshot of making your Optimizer parameters simple. I haven't run into a situation yet where I have to use anything other than 1.0 for the learning_rate for GradientDescent, Adam, or MomentumOptimizer.
Now your loss will be independent of batch size or number of labels.