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Support is a substantial measure because a rule that has very low support can appear easily by chance. A low support rule is also feasible to be tedious from a business viewpoint because it cannot be profitable to enhance items that users seldom purchase together.An association rule is an implication description of the form X→Y where X and Y are disjoint itemsets, i.e., $\mathrm{X\cap\:Y=\phi}$. The durability of an association rule can be computed in terms of its support and confidence. Support decides how to provide a rule that is accessible to a given data set, while confidence decides how frequently ... Read More
Sampling is a broadly used method for handling the class imbalance problem. The concept of sampling is to change the distribution of examples so that the rare class is well defined in the training set. There are various techniques for sampling such as undersampling, oversampling, and a hybrid of both approaches. For example, consider a data set that includes 100 positive examples and 1000 negative examples.In the method of undersampling, a random sample of 100 negative examples is selected to form the training set ahead with all the positive examples. One issue with this method is that some of the ... Read More
Random forest is a class of ensemble approaches particularly designed for decision tree classifiers. It integrates the predictions made by several decision trees, where each tree is created based on the values of a separate set of random vectors.The random vectors are produced from a constant probability distribution, unlike the adaptive methods used in AdaBoost, where the probability distribution is diverse to target instances that are difficult to classify.Bagging needs decision trees is a definite case of random forests, where randomness is inserted into the model-building procedure by randomly selecting N samples, with restoration, from the initial training set. Bagging ... Read More
The concept is to build multiple classifiers from the initial data and then aggregate their predictions when describing unknown examples. The ensemble of classifiers can be constructed in several methods which are as follows −By manipulating the training set − In this method, multiple training sets are generated by resampling the initial data as per some sampling distribution. The sampling distribution decides how likely it is that instances will be chosen for training, and it can change from one trial to another. A classifier is constructed from each training set using a specific learning algorithm. Bagging and boosting are instances ... Read More
In this problem, we are given an array arr[] of size N. It consists of integer values ranging from 1 to N. And one element x from the range is missing whereas one element y in the array occurs double. Our task is to find the repeating and the missing number using two equations.Let’s take an example to understand the problem, Inputarr[] = {1, 2 , 3, 3}Outputmissing = 4, double = 3Solution ApproachA method to solve the problem is using two equations for the two values x and y. Then solve the equation to get the value for x ... Read More
A classification approach that has received considerable scrutiny is the support vector machine (SVM). This approach has its roots in statistical learning theory and has displayed promising empirical outcomes in several practical applications, from handwritten digit identification to text classification.SVM also operates with high-dimensional data and prevents the curse of dimensionality issues. There is the second element of this approach is that it defines the decision boundary using a subset of the training instances, called the support vectors.SVM can be prepared to explicitly view this type of hyperplane in linearly separable data. It can achieve by displaying how the SVM ... Read More
In this problem, we are given an array arr[] of size N and an integer value M. Our task is to find the position of the last removed element from the array.The removal of values from the array is based on the operations −For an element in the array arr[i]. If arr[i] > M, pop the value and push arr[i] - M to the end of the array. Otherwise remove it from the array.Perform the operations till the array consists of elements.Let’s take an example to understand the problem, Inputarr[] = {5, 4, 8}, M = 3Output3ExplanationRemoving values using operations, ... Read More
An artificial neural network is a system placed on the functions of biological neural networks. It is a simulation of a biological neural system. The feature of artificial neural networks is that there are several structures, which required several approaches of algorithms, but regardless of being a complex system, a neural network is easy.These networks are between the specific signal-processing sciences in the director’s toolbox. The space is hugely interdisciplinary, but this technique will restrict the view to the engineering viewpoint.The input/output training data are essential in neural network technology because they send the essential record to “find” the optimal ... Read More
In this problem, we are given two arrays arr1[] and arr2[] consisting of unique values. Our task is to find the overlapping sum of two arrays.All elements of the arrays are distinct. And we need to return the sum of elements which are common for both arraysLet’s take an example to understand the problem, Inputarr1[] = {5, 4, 9, 2}, arr2[] = {6, 3, 9, 4}Output2ExplanationThe elements that are present in both arrays are 9 and 4. The sum is 9 + 9 + 4 + 4 = 26Solution ApproachA simple solution to the problem is traversing one array say ... Read More
In this problem, we are given the coordinates of two points of a line starting point A(xA, yA) and midpoint M(xM, yM) .Our task is to find the other end point of a line with given one end and mid.Let’s take an example to understand the problem, InputA = [1, 2], M = [3, 0]Output[5, -2]ExplanationThe line is −Solution ApproachTo solve the problem, we will be using the concepts of geometry we have learned in mathematics. If you remember there is a midpoint formula for every line which is, mid(x) = (x1 + x2) / 2 mid(y) = (y1 + ... Read More