I have time series data consisting of a vector
v=(x_1,…, x_n)
of binary categorical variables and the probabilities for four outcomes
p_1, p_2, p_3, p_4.
Given a new vector of categorical variables I want to predict the probabilities
p_1,…,p_4
The probabilities are very unbalanced with
p_1>.99 and p_2, p_3, p_4 < .01.
For example
v_1= (1,0,0,0,1,0,0,0) , p_1=.99, p_2=.005, p_3=.0035, p_4= .0015
v_2=(0,0,1,0,0,0,0,1), p_1=.99, p_2=.006, p_3=.0035, p_4= .0005
v_3=(0,1,0,0,1,1,1,0), p_1=.99, p_2=.005, p_3=.003, p_4= .002
v_4=(0,0,1,0,1,0,0,1), p_1=.99, p_2=.0075, p_3=.002, p_4= .0005
Given a new vector
v_5= (0,0,1,0,1,1,0,0)
I want to predict
p_1, p_2, p_3, p_4.
I should also note that the new vector could be identical to one of the input vectors, i.e.,
v_5=(0,0,1,0,1,0,0,1)= v_4.
My initial approach is to turn this into 4 regression problems.
The first would predict p_1, the second would predict p_2, the third would predict p_3, and the fourth would predict p_4. The problem with this is that I need
p_1+p_2+p_3+p_4=1
I’m not classifying, but should I also be worried about the unbalanced probabilities. Any ideas would be welcome.
Your suggestion of considering this as a multiple problem + final normalization, has some sense, but it's known to be problematic in many cases (see, e.g., the problem of masking).
What you're describing here is multiclass (soft) classification, and there are many many known techniques for doing so. You didn't specify which language/tool/library you're using, or if you're planning on rolling your own (which only makes sense for didactic purposes). I'd suggest starting with Linear Discriminant Analysis which is very simple to understand and implement, and - despite its strong assumptions - is known to often work well in practice (see the classical book by Hastie & Tibshirani).
Irrespective of the underlying algorithm you use for soft binary classification (e.g., LDA or not), It is not very difficult to transform aggregate input into labeled input.
Consider for example the instance
v_1= (1,0,0,0,1,0,0,0) , p_1=.99, p_2=.005, p_3=.0035, p_4= .0015
If your classifier supports instance weights, feed it 4 instances, labeled 1, 2, ..., with weights given by p_1, p_2, ..., respectively.
If it does not support instance weights, simply simulate what the law of large numbers says would happen: generate some large n instance from this input; for each such new input, choose a label randomly proportionally to its probability.
Related
I have a large dataset (~20,000 samples x 2,000 features-- each sample w/ a corresponding y-value) that I'm constructing a regression ML model for.
The input vectors are bitvectors with either 1s or 0s at each position.
Interestingly, I have noticed that when I 'randomly' select N samples such that their y-values are between two arbitrary values A and B (such that B-A is much smaller than the total range of values in y), the subsequent model is much better at predicting other values with the A-->B range not used in the training of the model.
However, the overall similarity of the input X vectors for these values are in no way more similar than any random selection of X values across the whole dataset.
Is there an available method to transform the input X-vectors such that those with more similar y-values are "closer" (I'm not particular the methodology, but it could be something like cosine similarity), and those with not similar y-values are separated?
After more thought, I believe this question can be re-framed as a supervised clustering problem. What might be able to accomplish this might be as simple as:
import umap
print(df.shape)
>> (23,312, 2149)
print(len(target))
>> 23,312
embedding = umap.UMAP().fit_transform(df, y=target)
I'm trying to teach myself machine learning and I have a similar question to this.
Is this correct:
For example, if I have an input matrix, where X1, X2 and X3 are three numerical features (e.g. say they are petal length, stem length, flower length, and I'm trying to label whether the sample is a particular flower species or not):
x1 x2 x3 label
5 1 2 yes
3 9 8 no
1 2 3 yes
9 9 9 no
That you take the vector of the first ROW (not column) of the table above to be inputted into the network like this:
i.e. there would be three neurons (1 for each value of the first table row), and then w1,w2 and w3 are randomly selected, then to calculate the first neuron in the next column, you do the multiplication I have described, and you add a randomly selected bias term. This gives the value of that node.
This is done for a set of nodes (i.e. each column actually will have four nodes (three + a bias), for simplicity, i removed the other three nodes from the second column), and then in the last node before the output, there is an activation function to transform the sum into a value (e.g. 0-1 for sigmoid) and that value tells you whether the classification is yes or no.
I'm sorry for how basic this is, I want to really understand the process, and I'm doing it from free resources. So therefore generally, you should select the number of nodes in your network to be a multiple of the number of features, e.g. in this case, it would make sense to write:
from keras.models import Sequential
from keras.models import Dense
model = Sequential()
model.add(Dense(6,input_dim=3,activation='relu'))
model.add(Dense(6,input_dim=3,activation='relu'))
model.add(Dense(3,activation='softmax'))
What I don't understand is why the keras model has an activation function in each layer of the network and not just at the end, which is why I'm wondering if my understanding is correct/why I added the picture.
Edit 1: Just a note I saw that in the bias neuron, I put on the edge 'b=1', that might be confusing, I know the bias doesn't have a weight, so that was just a reminder to myself that the weight of the bias node is 1.
Several issues here apart from the question in your title, but since this is not the time & place for full tutorials, I'll limit the discussion to some of your points, taking also into account that at least one more answer already exists.
So therefore generally, you should select the number of nodes in your network to be a multiple of the number of features,
No.
The number of features is passed in the input_dim argument, which is set only for the first layer of the model; the number of inputs for every layer except the first one is simply the number of outputs of the previous one. The Keras model you have written is not valid, and it will produce an error, since for your 2nd layer you ask for input_dim=3, while the previous one has clearly 6 outputs (nodes).
Beyond this input_dim argument, there is no other relationship whatsoever between the number of data features and the number of network nodes; and since it seems you have in mind the iris data (4 features), here is a simple reproducible example of applying a Keras model to them.
What is somewhat hidden in the Keras sequential API (which you use here) is that there is in fact an implicit input layer, and the number of its nodes is the dimensionality of the input; see own answer in Keras Sequential model input layer for details.
So, the model you have drawn in your pad actually corresponds to the following Keras model written using the sequential API:
model = Sequential()
model.add(Dense(1,input_dim=3,activation='linear'))
where in the functional API it would be written as:
inputs = Input(shape=(3,))
outputs = Dense(1, activation='linear')(inputs)
model = Model(inputs, outputs)
and that's all, i.e. it is actually just linear regression.
I know the bias doesn't have a weight
The bias does have a weight. Again, the useful analogy is with the constant term of linear (or logistic) regression: the bias "input" itself is always 1, and its corresponding coefficient (weight) is learned through the fitting process.
why the keras model has an activation function in each layer of the network and not just at the end
I trust this has been covered sufficiently in the other answer.
I'm sorry for how basic this is, I want to really understand the process, and I'm doing it from free resources.
We all did; no excuse though to not benefit from Andrew Ng's free & excellent Machine Learning MOOC at Coursera.
It seems your question is why there is a activation function for each layer instead of just the last layer. The simple answer is, if there are no non-linear activations in the middle, no matter how deep your network is, it can be boiled down to a single linear equation. Therefore, non-linear activation is one of the big enablers that enable deep networks to be actually "deep" and learn high-level features.
Take the following example, say you have 3 layer neural network without any non-linear activations in the middle, but a final softmax layer. The weights and biases for these layers are (W1, b1), (W2, b2) and (W3, b3). Then you can write the network's final output as follows.
h1 = W1.x + b1
h2 = W2.h1 + b2
h3 = Softmax(W3.h2 + b3)
Let's do some manipulations. We'll simply replace h3 as a function of x,
h3 = Softmax(W3.(W2.(W1.x + b1) + b2) + b3)
h3 = Softmax((W3.W2.W1) x + (W3.W2.b1 + W3.b2 + b3))
In other words, h3 is in the following format.
h3 = Softmax(W.x + b)
So, without the non-linear activations, our 3-layer networks has been squashed to a single layer network. That's is why non-linear activations are important.
Imagine, you have an activation layer only in the last layer (In your case, sigmoid. It can be something else too.. say softmax). The purpose of this is to convert real values to a 0 to 1 range for a classification sort of answer. But, the activation in the inner layers (hidden layers) has a different purpose altogether. This is to introduce nonlinearity. Without the activation (say ReLu, tanh etc.), what you get is a linear function. And how many ever, hidden layers you have, you still end up with a linear function. And finally, you convert this into a nonlinear function in the last layer. This might work in some simple nonlinear problems, but will not be able to capture a complex nonlinear function.
Each hidden unit (in each layer) comprises of activation function to incorporate nonlinearity.
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 am new to SVM. I am using jlibsvm for a multi-class classification problem. Basically, I am doing a sentence classification problem. There are 3 Classes. What I understood is I am doing One-against-all classification. I have a comparatively small train set. A total of 75 sentences, In which 25 sentences belongs to each class.
I am making 3 SVMs (so 3 different models), where, while training, in SVM_A, sentences belong to CLASS A will have a true label, i.e., 1 and other sentences will have a -1 label. Correspondingly done for SVM_B, and SVM_C.
While testing, to get the true label of a sentence, I am giving the sentence to 3 models and I am taking the prediction probability returned by these 3 models. Which one returns the highest will be the class the sentence belong to.
This is how I am doing. But I am getting the same prediction probability for every sentence in the test set for all models.
A predicted:0.012820514
B predicted:0.012820514
C predicted:0.012820514
These values repeat for all sentences in the training set.
The following is how I set parameters for training:
C_SVC svm = new C_SVC();
MutableBinaryClassificationProblemImpl problem;
ImmutableSvmParameterGrid.Builder builder = ImmutableSvmParameterGrid.builder();
// create training parameters ------------
HashSet<Float> cSet;
HashSet<LinearKernel> kernelSet;
cSet = new HashSet<Float>();
cSet.add(1.0f);
kernelSet = new HashSet<LinearKernel>();
kernelSet.add(new LinearKernel());
// configure finetuning parameters
builder.eps = 0.001f; // epsilon
builder.Cset = cSet; // C values used
builder.kernelSet = kernelSet; //Kernel used
builder.probability=true; // To get the prediction probability
ImmutableSvmParameter params = builder.build();
What am I doing wrong?
Is there any other better way to do multi-class classification other than this?
You are getting the same output, because you generate the same model three times.
The reason for this is, that jlibsvm is able to perform multiclass classification out of the box based on the provided data (LIBSVM itself supports this too). If it detects, that more than two class labes are provided in the given data, it automatically performs multiclass classification. So there is no need for a manually 1vsN approach. Just supply the data with class-labels for each category.
However, jlibsvm is still in beta and relies on a rather old version of LIBSVM (2.88). A lot has changed. For a more intiuitive Java binding (in comparison to the default LIBSVM version), you can take a look at zlibsvm, which is available via Maven Central and based on the latest LIBSVM version.
Any idea on the recommended parameters for OpenCV RTrees? I have read the documentation and I'm trying to apply it to MNIST dataset, i.e. 60000 training images, with 10000 testing images. I'm trying to optimize MaxDepth, MinSampleCount, setMaxCategories, and setPriors? e.g.
Ptr<RTrees> model = RTrees::create();
/* Depth of the tree.
A low value will likely underfit and conversely
a high value will likely overfit.
The optimal value can be obtained using cross validation
or other suitable methods.
*/
model->setMaxDepth(?); // letter_recog.cpp uses 10
/* minimum samples required at a leaf node for it to be split.
A reasonable value is a small percentage of the total data e.g. 1%.
MNIST 70000 * 0.01 = 700
*/
model->setMinSampleCount(700?); letter_recog.cpp uses 10
/* regression_accuracy – Termination criteria for regression trees.
If all absolute differences between an estimated value in a node and
values of train samples in this node are less than this parameter
then the node will not be split. */
model->setRegressionAccuracy(0); // I think this is already correct
/*
use_surrogates – If true then surrogate splits will be built.
These splits allow to work with missing data and compute variable importance correctly.'
To compute variable importance correctly, the surrogate splits must be enabled in
the training parameters, even if there is no missing data.
*/
model->setUseSurrogates(true); // I think this is already correct
/*
Cluster possible values of a categorical variable into K \leq max_categories clusters
to find a suboptimal split. If a discrete variable, on which the training procedure
tries to make a split, takes more than max_categories values, the precise best subset
estimation may take a very long time because the algorithm is exponential.
Instead, many decision trees engines (including ML) try to find sub-optimal split
in this case by clustering all the samples into max_categories clusters that is
some categories are merged together. The clustering is applied only in n>2-class
classification problems for categorical variables with N > max_categories possible values.
In case of regression and 2-class classification the optimal split can be found
efficiently without employing clustering, thus the parameter is not used in these cases.
*/
model->setMaxCategories(?); letter_recog.cpp uses 15
/*
priors – The array of a priori class probabilities, sorted by the class label value.
The parameter can be used to tune the decision tree preferences toward a certain class.
For example, if you want to detect some rare anomaly occurrence, the training base will
likely contain much more normal cases than anomalies, so a very good classification
performance will be achieved just by considering every case as normal.
To avoid this, the priors can be specified, where the anomaly probability is
artificially increased (up to 0.5 or even greater), so the weight of the misclassified
anomalies becomes much bigger, and the tree is adjusted properly. You can also think about
this parameter as weights of prediction categories which determine relative weights that
you give to misclassification. That is, if the weight of the first category is 1 and
the weight of the second category is 10, then each mistake in predicting the
second category is equivalent to making 10 mistakes in predicting the first category.
*/
model->setPriors(Mat()); // ?
/* If true then variable importance will be calculated and
then it can be retrieved by CvRTrees::get_var_importance().
*/
model->setCalculateVarImportance(true); // I think this is already correct
/*
The size of the randomly selected subset of features at each tree node and
that are used to find the best split(s). If you set it to 0 then the size
will be set to the square root of the total number of features.
*/
model->setActiveVarCount(0); // I think this is already correct
/*
CV_TERMCRIT_ITER Terminate learning by the max_num_of_trees_in_the_forest;
CV_TERMCRIT_EPS Terminate learning by the forest_accuracy;
CV_TERMCRIT_ITER | CV_TERMCRIT_EPS Use both termination criteria.
*/
model->setTermCriteria(TC(100,0.01f)); // I think this is already correct