本文整理汇总了Java中beast.evolution.tree.Node.isDirty方法的典型用法代码示例。如果您正苦于以下问题：Java Node.isDirty方法的具体用法？Java Node.isDirty怎么用？Java Node.isDirty使用的例子？那么恭喜您, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类beast.evolution.tree.Node的用法示例。
在下文中一共展示了Node.isDirty方法的1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，您的评价将有助于系统推荐出更棒的Java代码示例。
示例1: isDirtyBranch
import beast.evolution.tree.Node; //导入方法依赖的package包/类
@Override
public boolean isDirtyBranch(Node speciesNode) {
if (needsUpdate) {
final RealParameter tipPopSizes = tipPopSizesInput.get();
final RealParameter topPopSizes = topPopSizesInput.get();
Arrays.fill(speciesBranchStatus, false);
Node[] speciesNodes = speciesTree.getNodesAsArray();
// non-root nodes (linear population sizes)
for (int nodeI = 0; nodeI < speciesNodes.length; nodeI++) {
// if the "top" population is dirty, no need to check the tip
if (nodeI < rootNodeNumber && topPopSizes.isDirty(nodeI)) { // not the root node
speciesBranchStatus[nodeI] = true;
continue;
}
if (nodeI < leafNodeCount) { // is a leaf node
speciesBranchStatus[nodeI] = tipPopSizes.isDirty(nodeI);
} else { // is an internal node
final Node leftChild = speciesNodes[nodeI].getLeft();
final Node rightChild = speciesNodes[nodeI].getRight();
final int leftChildTopI = leftChild.getNr();
final int rightChildTopI = rightChild.getNr();
speciesBranchStatus[nodeI] = topPopSizes.isDirty(leftChildTopI) ||
topPopSizes.isDirty(rightChildTopI) ||
leftChild.isDirty() != Tree.IS_CLEAN ||
rightChild.isDirty() != Tree.IS_CLEAN;
}
}
//Arrays.fill(speciesBranchStatus, true);
needsUpdate = false;
}
return speciesBranchStatus[speciesNode.getNr()];
}