Hive 蜂巢如何实现计数（不同？

在GenericUDAFCount.java中：

@Description(name = "count",
value = "_FUNC_(*) - Returns the total number of retrieved rows, including "
      +        "rows containing NULL values.\n"

      + "_FUNC_(expr) - Returns the number of rows for which the supplied "
      +        "expression is non-NULL.\n"

      + "_FUNC_(DISTINCT expr[, expr...]) - Returns the number of rows for "
      +        "which the supplied expression(s) are unique and non-NULL.")

但是我没有看到任何处理“distinct”表达式的代码

public static class GenericUDAFCountEvaluator extends GenericUDAFEvaluator {
private boolean countAllColumns = false;
private LongObjectInspector partialCountAggOI;
private LongWritable result;

@Override
public ObjectInspector init(Mode m, ObjectInspector[] parameters)
throws HiveException {
  super.init(m, parameters);
  partialCountAggOI =
    PrimitiveObjectInspectorFactory.writableLongObjectInspector;
  result = new LongWritable(0);
  return PrimitiveObjectInspectorFactory.writableLongObjectInspector;
}

private GenericUDAFCountEvaluator setCountAllColumns(boolean countAllCols) {
  countAllColumns = countAllCols;
  return this;
}

/** class for storing count value. */
static class CountAgg implements AggregationBuffer {
  long value;
}

@Override
public AggregationBuffer getNewAggregationBuffer() throws HiveException {
  CountAgg buffer = new CountAgg();
  reset(buffer);
  return buffer;
}

@Override
public void reset(AggregationBuffer agg) throws HiveException {
  ((CountAgg) agg).value = 0;
}

@Override
public void iterate(AggregationBuffer agg, Object[] parameters)
  throws HiveException {
  // parameters == null means the input table/split is empty
  if (parameters == null) {
    return;
  }
  if (countAllColumns) {
    assert parameters.length == 0;
    ((CountAgg) agg).value++;
  } else {
    assert parameters.length > 0;
    boolean countThisRow = true;
    for (Object nextParam : parameters) {
      if (nextParam == null) {
        countThisRow = false;
        break;
      }
    }
    if (countThisRow) {
      ((CountAgg) agg).value++;
    }
  }
}

@Override
public void merge(AggregationBuffer agg, Object partial)
  throws HiveException {
  if (partial != null) {
    long p = partialCountAggOI.get(partial);
    ((CountAgg) agg).value += p;
  }
}

@Override
public Object terminate(AggregationBuffer agg) throws HiveException {
  result.set(((CountAgg) agg).value);
  return result;
}

@Override
public Object terminatePartial(AggregationBuffer agg) throws HiveException {
  return terminate(agg);
}

}

hive是如何实现计数的？当任务运行时，它确实花费了很多时间。

---

它在源代码中的什么位置？

因为您可以从表1中运行SELECT DISTINCT column1，DISTINCT expression不是一个标志或选项，它是独立计算的

说：

绑定到参数类型的数据的实际筛选 实现由框架而不是计数UDAF处理 实施

如果您想深入了解源代码的详细信息，请查看