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If the character is not found most of these routines return an invalid index value – -1 where indexes are 0-based, 0 where they are 1-based – or some value to be interpreted as Boolean FALSE. This can be accomplished as a special case of #Find , with a string of one character; but it may be simpler or more efficient in many languages to ...
Word2vec is a technique in natural language processing (NLP) for obtaining vector representations of words. These vectors capture information about the meaning of the word based on the surrounding words.
In Java, lazy evaluation can be done by using objects that have a method to evaluate them when the value is needed. The body of this method must contain the code required to perform this evaluation. The body of this method must contain the code required to perform this evaluation.
Given a database in which each entry has multiple attributes (viewed as a 0–1 matrix with a row per database entry and a column per attribute) they use MinHash-based approximations to the Jaccard index to identify candidate pairs of attributes that frequently co-occur, and then compute the exact value of the index for only those pairs to ...
The most common superscript digits (1, 2, and 3) were included in ISO-8859-1 and were therefore carried over into those code points in the Latin-1 range of Unicode. The remainder were placed along with basic arithmetical symbols, and later some Latin subscripts, in a dedicated block at U+2070 to U+209F. The table below shows these characters ...
From Java 8 onwards, the default keyword can be used to allow an interface to provide an implementation of a method. do The do keyword is used in conjunction with while to create a do-while loop , which executes a block of statements associated with the loop and then tests a boolean expression associated with the while .
A simple and inefficient way to see where one string occurs inside another is to check at each index, one by one. First, we see if there is a copy of the needle starting at the first character of the haystack; if not, we look to see if there's a copy of the needle starting at the second character of the haystack, and so forth.
Given a DFA, the problem of determining if it has a synchronizing word can be solved in polynomial time [2] using a theorem due to Ján Černý. A simple approach considers the power set of states of the DFA, and builds a directed graph where nodes belong to the power set, and a directed edge describes the action of the transition function.