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Version: Python

sorted_last

agg.sorted_last returns an aggregator that sorts a table in descending order and then computes the first value, within an aggregation group, for each input column.

note

agg.sorted_last will produce the same results as a sort operation followed by agg.last.

Syntax​

sorted_last(order_by: str, cols: Union[str, list[str]]) -> Aggregation

Parameters​

ParameterTypeDescription
order_bystr

The column to sort by.

colsUnion[str, list[str]]

The source column(s) for the calculations.

  • ["X"] will output the last value in the X column for each group.
  • ["Y = X"] will output the last value in the X column for each group and rename it to Y.
  • ["X, A = B"] will output the last value in the X column for each group and the last value 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.agg_by([agg.sum_(cols=[“X”]), agg.avg(cols=["X"]), both the sum and the average aggregators produce column X, which results in an error.

Returns​

An aggregator that sorts the table in descending order and then computes the first value, within an aggregation group, for each input column.

Examples​

In this example, agg.sorted_last returns the last Y value as sorted by Z and grouped by X.

from deephaven import new_table
from deephaven.column import string_col, int_col, double_col
from deephaven import agg as agg

source = new_table(
    [
        string_col("X", ["A", "B", "A", "C", "B", "A", "B", "B", "C"]),
        string_col("Y", ["N", "O", "P", "N", "P", "N", "", "Q", "O"]),
        int_col("Z", [3, 2, 1, 1, 3, 1, 4, 1, 2]),
    ]
)

result = source.agg_by([agg.sorted_last(order_by="Z", cols=["Y"])], by=["X"])

In this example, agg.sorted_last returns the last Y value (renamed to Z), as sorted by Z and grouped by X.

from deephaven import new_table
from deephaven.column import string_col, int_col, double_col
from deephaven import agg as agg

source = new_table(
    [
        string_col("X", ["A", "B", "A", "C", "B", "A", "B", "B", "C"]),
        string_col("Y", ["N", "O", "P", "N", "P", "N", "", "Q", "O"]),
        int_col("Z", [3, 2, 1, 1, 3, 1, 4, 1, 2]),
    ]
)

result = source.agg_by([agg.sorted_last(order_by="Z", cols=["Z = Y"])], by=["X"])

In this example, agg.sorted_last returns the last Y string and last Z integer, as sorted by Z and grouped by X.

from deephaven import new_table
from deephaven.column import string_col, int_col, double_col
from deephaven import agg as agg

source = new_table(
    [
        string_col("X", ["A", "B", "A", "C", "B", "A", "B", "B", "C"]),
        string_col("Y", ["N", "O", "P", "N", "P", "N", "", "Q", "O"]),
        int_col("Z", [3, 2, 1, 1, 3, 1, 4, 1, 2]),
    ]
)

result = source.agg_by([agg.sorted_last(order_by="Z", cols=["Y", "Z"])], by=["X"])

In this example, agg.sorted_last returns the last Z integer, as sorted by Z and grouped by X and Y.

from deephaven import new_table
from deephaven.column import string_col, int_col, double_col
from deephaven import agg as agg

source = new_table(
    [
        string_col("X", ["A", "B", "A", "C", "B", "A", "B", "B", "C"]),
        string_col("Y", ["N", "O", "P", "N", "P", "N", "", "Q", "O"]),
        int_col("Z", [3, 2, 1, 1, 3, 1, 4, 1, 2]),
    ]
)

result = source.agg_by([agg.sorted_last(order_by="Z", cols=["Z"])], by=["X", "Y"])

In this example, agg.sorted_last returns the last Y string, and agg.max_ returns the maximum Z integer, as sorted by Z and grouped by X.

from deephaven import new_table
from deephaven.column import string_col, int_col, double_col
from deephaven import agg as agg

source = new_table(
    [
        string_col("X", ["A", "B", "A", "C", "B", "A", "B", "B", "C"]),
        string_col("Y", ["N", "O", "P", "N", "P", "N", "", "Q", "O"]),
        int_col("Z", [3, 2, 1, 1, 3, 1, 4, 1, 2]),
    ]
)

result = source.agg_by(
    [agg.sorted_last(order_by="Z", cols=["SortedLastY = Y"]), agg.max_("Z")], by=["X"]
)