本文整理汇总了Java中pal.tree.Node.getBranchLength方法的典型用法代码示例。如果您正苦于以下问题：Java Node.getBranchLength方法的具体用法？Java Node.getBranchLength怎么用？Java Node.getBranchLength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pal.tree.Node的用法示例。
在下文中一共展示了Node.getBranchLength方法的3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: nodeFromRoot
import pal.tree.Node; //导入方法依赖的package包/类
private Pair<Integer,Double> nodeFromRoot(Node node) {
Node parent = node;
int step = 0;
double height = 0;
while (parent != null && !parent.isRoot()) {
step++;
height += parent.getBranchLength();
parent = parent.getParent();
}
return Pair.of(step,height);
}
示例2: printNodeInASCII
import pal.tree.Node; //导入方法依赖的package包/类
private static void printNodeInASCII(PrintWriter out, Node node, int level, int m, int maxm,
int numExternalNodes, boolean[] umbrella, int[] position, double proportion, int minLength) {
position[level] = (int) (node.getBranchLength() * proportion);
if (position[level] < minLength) {
position[level] = minLength;
}
if (node.isLeaf()) // external branch
{
if (m == maxm - 1) {
umbrella[level - 1] = true;
}
printlnNodeWithNumberAndLabel(out, node, level, numExternalNodes, umbrella, position);
if (m == 0) {
umbrella[level - 1] = false;
}
} else // internal branch
{
for (int n = node.getChildCount() - 1; n > -1; n--) {
printNodeInASCII(out, node.getChild(n), level + 1, n, node.getChildCount(),
numExternalNodes, umbrella, position, proportion, minLength);
if (m == maxm - 1 && n == node.getChildCount() / 2) {
umbrella[level - 1] = true;
}
if (n != 0) {
if (n == node.getChildCount() / 2) {
printlnNodeWithNumberAndLabel(out, node, level, numExternalNodes, umbrella, position);
} else {
for (int i = 0; i < level + 1; i++) {
if (umbrella[i]) {
putCharAtLevel(out, i, '|', position);
} else {
putCharAtLevel(out, i, ' ', position);
}
}
out.println();
}
}
if (m == 0 && n == node.getChildCount() / 2) {
umbrella[level - 1] = false;
}
}
}
}
示例3: getTreeBranchLengths
import pal.tree.Node; //导入方法依赖的package包/类
private double[] getTreeBranchLengths(List<Node> nodes) {
double[] branchLengths = new double[nodes.size()];
int i = 0;
for (Node n : nodes) branchLengths[i++] = n.getBranchLength();
return branchLengths;
}