本文整理汇总了Java中org.apache.lucene.util.automaton.Transition类的典型用法代码示例。如果您正苦于以下问题:Java Transition类的具体用法?Java Transition怎么用?Java Transition使用的例子?那么恭喜您, 这里精选的类代码示例或许可以为您提供帮助。
Transition类属于org.apache.lucene.util.automaton包,在下文中一共展示了Transition类的6个代码示例,这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞,您的评价将有助于我们的系统推荐出更棒的Java代码示例。
示例1: topoSortStates
import org.apache.lucene.util.automaton.Transition; //导入依赖的package包/类
private int[] topoSortStates(Automaton a) {
int[] states = new int[a.getNumStates()];
final Set<Integer> visited = new HashSet<>();
final LinkedList<Integer> worklist = new LinkedList<>();
worklist.add(0);
visited.add(0);
int upto = 0;
states[upto] = 0;
upto++;
Transition t = new Transition();
while (worklist.size() > 0) {
int s = worklist.removeFirst();
int count = a.initTransition(s, t);
for (int i=0;i<count;i++) {
a.getNextTransition(t);
if (!visited.contains(t.dest)) {
visited.add(t.dest);
worklist.add(t.dest);
states[upto++] = t.dest;
}
}
}
return states;
}
开发者ID:justor,项目名称:elasticsearch_my,代码行数:25,代码来源:XAnalyzingSuggester.java
示例2: setUp
import org.apache.lucene.util.automaton.Transition; //导入依赖的package包/类
@Override
public void setUp() throws Exception {
super.setUp();
// build an automaton matching this jvm's letter definition
State initial = new State();
State accept = new State();
accept.setAccept(true);
for (int i = 0; i <= 0x10FFFF; i++) {
if (Character.isLetter(i)) {
initial.addTransition(new Transition(i, i, accept));
}
}
Automaton single = new Automaton(initial);
single.reduce();
Automaton repeat = BasicOperations.repeat(single);
jvmLetter = new CharacterRunAutomaton(repeat);
}
开发者ID:pkarmstr,项目名称:NYBC,代码行数:18,代码来源:TestDuelingAnalyzers.java
示例3: copyDestTransitions
import org.apache.lucene.util.automaton.Transition; //导入依赖的package包/类
private void copyDestTransitions(State from, State to, List<Transition> transitions) {
if (to.isAccept()) {
from.setAccept(true);
}
for(Transition t : to.getTransitions()) {
transitions.add(t);
}
}
开发者ID:pkarmstr,项目名称:NYBC,代码行数:9,代码来源:AnalyzingSuggester.java
示例4: addHoles
import org.apache.lucene.util.automaton.Transition; //导入依赖的package包/类
private static void addHoles(State startState, RollingBuffer<Position> positions, int pos) {
Position posData = positions.get(pos);
Position prevPosData = positions.get(pos-1);
while(posData.arriving == null || prevPosData.leaving == null) {
if (posData.arriving == null) {
posData.arriving = new State();
posData.arriving.addTransition(new Transition(POS_SEP, posData.leaving));
}
if (prevPosData.leaving == null) {
if (pos == 1) {
prevPosData.leaving = startState;
} else {
prevPosData.leaving = new State();
}
if (prevPosData.arriving != null) {
prevPosData.arriving.addTransition(new Transition(POS_SEP, prevPosData.leaving));
}
}
prevPosData.leaving.addTransition(new Transition(HOLE, posData.arriving));
pos--;
if (pos <= 0) {
break;
}
posData = prevPosData;
prevPosData = positions.get(pos-1);
}
}
开发者ID:pkarmstr,项目名称:NYBC,代码行数:29,代码来源:TokenStreamToAutomaton.java
示例5: finish
import org.apache.lucene.util.automaton.Transition; //导入依赖的package包/类
/**
* Call this once you are done adding states/transitions.
* @param maxDeterminizedStates Maximum number of states created when
* determinizing the automaton. Higher numbers allow this operation to
* consume more memory but allow more complex automatons.
*/
public void finish(int maxDeterminizedStates) {
Automaton automaton = builder.finish();
// System.out.println("before det:\n" + automaton.toDot());
Transition t = new Transition();
// TODO: should we add "eps back to initial node" for all states,
// and det that? then we don't need to revisit initial node at
// every position? but automaton could blow up? And, this makes it
// harder to skip useless positions at search time?
if (anyTermID != -1) {
// Make sure there are no leading or trailing ANY:
int count = automaton.initTransition(0, t);
for(int i=0;i<count;i++) {
automaton.getNextTransition(t);
if (anyTermID >= t.min && anyTermID <= t.max) {
throw new IllegalStateException("automaton cannot lead with an ANY transition");
}
}
int numStates = automaton.getNumStates();
for(int i=0;i<numStates;i++) {
count = automaton.initTransition(i, t);
for(int j=0;j<count;j++) {
automaton.getNextTransition(t);
if (automaton.isAccept(t.dest) && anyTermID >= t.min && anyTermID <= t.max) {
throw new IllegalStateException("automaton cannot end with an ANY transition");
}
}
}
int termCount = termToID.size();
// We have to carefully translate these transitions so automaton
// realizes they also match all other terms:
Automaton newAutomaton = new Automaton();
for(int i=0;i<numStates;i++) {
newAutomaton.createState();
newAutomaton.setAccept(i, automaton.isAccept(i));
}
for(int i=0;i<numStates;i++) {
count = automaton.initTransition(i, t);
for(int j=0;j<count;j++) {
automaton.getNextTransition(t);
int min, max;
if (t.min <= anyTermID && anyTermID <= t.max) {
// Match any term
min = 0;
max = termCount-1;
} else {
min = t.min;
max = t.max;
}
newAutomaton.addTransition(t.source, t.dest, min, max);
}
}
newAutomaton.finishState();
automaton = newAutomaton;
}
det = Operations.removeDeadStates(Operations.determinize(automaton,
maxDeterminizedStates));
}
开发者ID:europeana,项目名称:search,代码行数:74,代码来源:TermAutomatonQuery.java
示例6: nextString
import org.apache.lucene.util.automaton.Transition; //导入依赖的package包/类
/**
* Returns the next String in lexicographic order that will not put
* the machine into a reject state.
*
* This method traverses the DFA from the given position in the String,
* starting at the given state.
*
* If this cannot satisfy the machine, returns false. This method will
* walk the minimal path, in lexicographic order, as long as possible.
*
* If this method returns false, then there might still be more solutions,
* it is necessary to backtrack to find out.
*
* @param state current non-reject state
* @param position useful portion of the string
* @return true if more possible solutions exist for the DFA from this
* position
*/
private boolean nextString(int state, int position) {
/*
* the next lexicographic character must be greater than the existing
* character, if it exists.
*/
int c = 0;
if (position < seekBytesRef.length) {
c = seekBytesRef.bytes[position] & 0xff;
// if the next byte is 0xff and is not part of the useful portion,
// then by definition it puts us in a reject state, and therefore this
// path is dead. there cannot be any higher transitions. backtrack.
if (c++ == 0xff)
return false;
}
seekBytesRef.length = position;
visited[state] = curGen;
Transition transitions[] = allTransitions[state];
// find the minimal path (lexicographic order) that is >= c
for (int i = 0; i < transitions.length; i++) {
Transition transition = transitions[i];
if (transition.getMax() >= c) {
int nextChar = Math.max(c, transition.getMin());
// append either the next sequential char, or the minimum transition
seekBytesRef.grow(seekBytesRef.length + 1);
seekBytesRef.length++;
seekBytesRef.bytes[seekBytesRef.length - 1] = (byte) nextChar;
state = transition.getDest().getNumber();
/*
* as long as is possible, continue down the minimal path in
* lexicographic order. if a loop or accept state is encountered, stop.
*/
while (visited[state] != curGen && !runAutomaton.isAccept(state)) {
visited[state] = curGen;
/*
* Note: we work with a DFA with no transitions to dead states.
* so the below is ok, if it is not an accept state,
* then there MUST be at least one transition.
*/
transition = allTransitions[state][0];
state = transition.getDest().getNumber();
// append the minimum transition
seekBytesRef.grow(seekBytesRef.length + 1);
seekBytesRef.length++;
seekBytesRef.bytes[seekBytesRef.length - 1] = (byte) transition.getMin();
// we found a loop, record it for faster enumeration
if (!finite && !linear && visited[state] == curGen) {
setLinear(seekBytesRef.length-1);
}
}
return true;
}
}
return false;
}
开发者ID:pkarmstr,项目名称:NYBC,代码行数:79,代码来源:AutomatonTermsEnum.java
注:本文中的org.apache.lucene.util.automaton.Transition类示例整理自Github/MSDocs等源码及文档管理平台,相关代码片段筛选自各路编程大神贡献的开源项目,源码版权归原作者所有,传播和使用请参考对应项目的License;未经允许,请勿转载。 |
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