Question

我正在尝试在java中实现C4.5算法。为了初步了解C4.5算法，我从这个link中获取了一个python代码作为参考。在这个项目中有一个名为mine.py的文件，其中包含以下函数。

def mine_c45(table, result):
    """ An entry point for C45 algorithm.
        _table_ - a dict representing data table in the following format:
        {
            '<column name>': [<column values>],
            '<column name>': [<column values>],
            ...
        }
        _result_: a string representing a name of column indicating a result.
    """
    col = max([(k, gain(table, k, result)) for k in table.keys() if k != result],
              key=lambda x: x[1])[0]
    tree = []
    for subt in get_subtables(table, col):
        v = subt[col][0]
        if is_mono(subt[result]):
            tree.append(['%s=%s' % (col, v),
                         '%s=%s' % (result, subt[result][0])])
        else:
            del subt[col]
            tree.append(['%s=%s' % (col, v)] + mine_c45(subt, result))
    return tree

通过使用链接中的代码，我尝试在Java中转换此代码并对其进行一些修改。我成功获得了我想要的输出，但问题是我无法以递归方式构建树。

这里是用java。

转换的代码

public void mineC45(Map<String, Attribute> table, String result) {
    int maxGain = 0;
    double[] gains = new double[table.size()];
    int counter = 0;
    SplitPoints point = null;
    for (Entry<String, Attribute> entry : table.entrySet()) {
        if (!entry.getKey().equals(result)) {
            boolean nominal = entry.getValue().isNominal();
            if (nominal)
                gains[counter++] = Utils
                        .gain(table, entry.getKey(), result);
            else {
                point = Utils.numericGain(table, entry.getKey(), result);
                gains[counter++] = point.getGain();
            }
        }
    }

    // calculate maximum gain column index
    maxGain = Utils.getMax(gains);
    List<String> keys = new ArrayList<String>(table.keySet());
    String column = keys.get(maxGain);
    if (table.get(column).isNominal()) {
        for (Map<String, Attribute> subTable : Utils.createSubTables(table,
                column)) {
            String value = subTable.get(column).getValues().get(0);
            if (Utils.isMono(subTable.get(result))) {
                System.out.println("\t" + column + " = " + value + " "
                        + result + " = "
                        + subTable.get(result).getValues().get(0));
            } else {
                subTable.remove(column);
                System.out.println(column + " = " + value + " ");
                mineC45(subTable, result);
            }
        }
    } else {
        boolean first = true;
        for (Map<String, Attribute> subTable : Utils.createNSubtables(
                table, column, result, point.getSplitValue())) {
            String sign = "";
            sign = first ? "<=" : ">";
            first = false;
            if (Utils.isMono(subTable.get(result))) {
                System.out.println("\t" + column + " "
                        + point.getSplitValue().toString() + " " + result
                        + " = " + subTable.get(result).getValues().get(0));
            } else {
                subTable.remove(column);
                System.out.println(column + " "
                        + point.getSplitValue().toString() + " ");
                mineC45(subTable, result);
            }
        }
    }
}

我在这里创建了代表表格的Map<String, Attribute>。字符串键表示列名称，属性存储值列表。如果任何人可以解释我如何将输出转换为树，以便我可以形成规则。

在递归函数中创建树

0 个答案: