# Copyright 2016 Google Inc.
#
# Licensed under the Apache License, Version 2.0 (the "License");
# you may not use this file except in compliance with the License.
# You may obtain a copy of the License at
#
#     http://www.apache.org/licenses/LICENSE-2.0
#
# Unless required by applicable law or agreed to in writing, software
# distributed under the License is distributed on an "AS IS" BASIS,
# WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
# See the License for the specific language governing permissions and
# limitations under the License.

"""Time series as :mod:`pandas` dataframes."""

import itertools

TOP_RESOURCE_LABELS = (
    'project_id',
    'aws_account',
    'location',
    'region',
    'zone',
)


def _build_dataframe(time_series_iterable,
                     label=None, labels=None):  # pragma: NO COVER
    """Build a :mod:`pandas` dataframe out of time series.

    :type time_series_iterable:
        iterable over :class:`~google.cloud.monitoring.timeseries.TimeSeries`
    :param time_series_iterable:
        An iterable (e.g., a query object) yielding time series.

    :type label: str
    :param label:
        (Optional) The label name to use for the dataframe header. This can be
        the name of a resource label or metric label (e.g.,
        ``"instance_name"``), or the string ``"resource_type"``.

    :type labels: list of strings, or None
    :param labels:
        A list or tuple of label names to use for the dataframe header.
        If more than one label name is provided, the resulting dataframe
        will have a multi-level column header.

        Specifying neither ``label`` or ``labels`` results in a dataframe
        with a multi-level column header including the resource type and
        all available resource and metric labels.

        Specifying both ``label`` and ``labels`` is an error.

    :rtype: :class:`pandas.DataFrame`
    :returns: A dataframe where each column represents one time series.
    """
    import pandas   # pylint: disable=import-error

    if labels is not None:
        if label is not None:
            raise ValueError('Cannot specify both "label" and "labels".')
        elif not labels:
            raise ValueError('"labels" must be non-empty or None.')

    columns = []
    headers = []
    for time_series in time_series_iterable:
        pandas_series = pandas.Series(
            data=[point.value for point in time_series.points],
            index=[point.end_time for point in time_series.points],
        )
        columns.append(pandas_series)
        headers.append(time_series.header())

    # Implement a smart default of using all available labels.
    if label is None and labels is None:
        resource_labels = set(itertools.chain.from_iterable(
            header.resource.labels for header in headers))
        metric_labels = set(itertools.chain.from_iterable(
            header.metric.labels for header in headers))
        labels = (['resource_type'] +
                  _sorted_resource_labels(resource_labels) +
                  sorted(metric_labels))

    # Assemble the columns into a DataFrame.
    dataframe = pandas.DataFrame.from_records(columns).T

    # Convert the timestamp strings into a DatetimeIndex.
    dataframe.index = pandas.to_datetime(dataframe.index)

    # Build a multi-level stack of column headers. Some labels may
    # be undefined for some time series.
    levels = []
    for key in labels or [label]:
        level = [header.labels.get(key, '') for header in headers]
        levels.append(level)

    # Build a column Index or MultiIndex. Do not include level names
    # in the column header if the user requested a single-level header
    # by specifying "label".
    dataframe.columns = pandas.MultiIndex.from_arrays(
        levels,
        names=labels or None)

    # Sort the rows just in case (since the API doesn't guarantee the
    # ordering), and sort the columns lexicographically.
    return dataframe.sort_index(axis=0).sort_index(axis=1)


def _sorted_resource_labels(labels):
    """Sort label names, putting well-known resource labels first."""
    head = [label for label in TOP_RESOURCE_LABELS if label in labels]
    tail = sorted(label for label in labels
                  if label not in TOP_RESOURCE_LABELS)
    return head + tail
