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Version: Java (Groovy)

AggCountDistinct

AggCountDistinct returns an aggregator that computes the number of distinct values, within an aggregation group, for each input column.

Syntax​

AggCountDistinct(countNulls, pairs...)
AggCountDistinct(pairs...)

Parameters​

ParameterTypeDescription
countNullsboolean

If true, null values are counted as a distinct value; otherwise null values are ignored. The default value is false.

pairsString...

The source column(s) for the calculations.

  • "X" will output the number of distinct values in the X column for each group.
  • "Y = X" will output the number of distinct values in the X column for each group and rename it to Y.
  • "X, A = B" will output the number of distinct values in the X column for each group and the number of distinct values in the B column while renaming it to A.
caution

If an aggregation does not rename the resulting column, the aggregation column will appear in the output table, not the input column. If multiple aggregations on the same column do not rename the resulting columns, an error will result, because the aggregations are trying to create multiple columns with the same name. For example, in table.aggBy([agg.AggSum(“X”), agg.AggAvg(“X”)]), both the sum and the average aggregators produce column X, which results in an error.

Returns​

An aggregator that computes the number of distinct values, within an aggregation group, for each input column.

Examples​

In this example, AggCountDistinct returns the number of distinct values of Y as grouped by X

import static io.deephaven.api.agg.Aggregation.AggCountDistinct

source = newTable(
    stringCol("X", "A", "B", "A", "C", "B", "A", "B", "B", "A", "A", "B", "A", "C", "B", "A", "B", "B", "C"),
    stringCol("Y", "M", "N", "M", "N", "N", "M", "O", "P", "N", "M", "N", "M", "N", "N", "M", "O", "P", "N"),
    intCol("Number", 55, 76, 55, 130, 230, 50, 76, 137, 214, 55, 76, 55, 130, 230, 50, 76, 137, 214),
)

result = source.aggBy([AggCountDistinct("Y")], "X")

In this example, AggCountDistinct returns the number of distinct values of Y (renamed to Z), as grouped by X.

import static io.deephaven.api.agg.Aggregation.AggCountDistinct

source = newTable(
    stringCol("X", "A", "B", "A", "C", "B", "A", "B", "B", "A", "A", "B", "A", "C", "B", "A", "B", "B", "C"),
    stringCol("Y", "M", "N", "M", "N", "N", "M", "O", "P", "N", "M", "N", "M", "N", "N", "M", "O", "P", "N"),
    intCol("Number", 55, 76, 55, 130, 230, 50, 76, 137, 214, 55, 76, 55, 130, 230, 50, 76, 137, 214),
)

result = source.aggBy([AggCountDistinct("Z = Y")], "X")

In this example, AggCountDistinct returns the number of distinct values of Y and the number of distinct values of Number, as grouped by X.

import static io.deephaven.api.agg.Aggregation.AggCountDistinct

source = newTable(
    stringCol("X", "A", "B", "A", "C", "B", "A", "B", "B", "A", "A", "B", "A", "C", "B", "A", "B", "B", "C"),
    stringCol("Y", "M", "N", "M", "N", "N", "M", "O", "P", "N", "M", "N", "M", "N", "N", "M", "O", "P", "N"),
    intCol("Number", 55, 76, 55, 130, 230, 50, 76, 137, 214, 55, 76, 55, 130, 230, 50, 76, 137, 214),
)

result = source.aggBy([AggCountDistinct("Y", "Number")], "X")

In this example, AggCountDistinct returns the number of distinct values of Number, as grouped by X and Y.

import static io.deephaven.api.agg.Aggregation.AggCountDistinct

source = newTable(
    stringCol("X", "A", "B", "A", "C", "B", "A", "B", "B", "A", "A", "B", "A", "C", "B", "A", "B", "B", "C"),
    stringCol("Y", "M", "N", "M", "N", "N", "M", "O", "P", "N", "M", "N", "M", "N", "N", "M", "O", "P", "N"),
    intCol("Number", 55, 76, 55, 130, 230, 50, 76, 137, 214, 55, 76, 55, 130, 230, 50, 76, 137, 214),
)

result = source.aggBy([AggCountDistinct("Number")], "X", "Y")

In this example, AggCountDistinct returns the number of distinct values of Number, and AggLast returns the last Number integer, as grouped by X.

import static io.deephaven.api.agg.Aggregation.AggCountDistinct
import static io.deephaven.api.agg.Aggregation.AggLast

source = newTable(
    stringCol("X", "A", "B", "A", "C", "B", "A", "B", "B", "A", "A", "B", "A", "C", "B", "A", "B", "B", "C"),
    stringCol("Y", "M", "N", "M", "N", "N", "M", "O", "P", "N", "M", "N", "M", "N", "N", "M", "O", "P", "N"),
    intCol("Number", 55, 76, 55, 130, 230, 50, 76, 137, 214, 55, 76, 55, 130, 230, 50, 76, 137, 214),
)

result = source.aggBy([AggCountDistinct("FirstNumber = Number"),AggLast("LastNumber = Number")], "X")