- get the `datetime` of the first or last order within a pixel - unify some fixtures in `tests.forecasts.timify.conftest`
54 lines
1.5 KiB
Python
54 lines
1.5 KiB
Python
"""Fixture for testing the `urban_meal_delivery.forecast.timify` module."""
|
|
|
|
import pandas as pd
|
|
import pytest
|
|
|
|
from tests import config as test_config
|
|
from urban_meal_delivery import config
|
|
from urban_meal_delivery.forecasts import timify
|
|
|
|
|
|
@pytest.fixture
|
|
def good_pixel_id(pixel):
|
|
"""A `pixel_id` that is on the `grid`."""
|
|
return pixel.id # `== 1`
|
|
|
|
|
|
@pytest.fixture
|
|
def order_totals(good_pixel_id):
|
|
"""A mock for `OrderHistory.totals`.
|
|
|
|
To be a bit more realistic, we sample two pixels on the `grid`.
|
|
|
|
Uses the LONG_TIME_STEP as the length of a time step.
|
|
"""
|
|
pixel_ids = [good_pixel_id, good_pixel_id + 1]
|
|
|
|
gen = (
|
|
(pixel_id, start_at)
|
|
for pixel_id in pixel_ids
|
|
for start_at in pd.date_range(
|
|
test_config.START, test_config.END, freq=f'{test_config.LONG_TIME_STEP}T',
|
|
)
|
|
if config.SERVICE_START <= start_at.hour < config.SERVICE_END
|
|
)
|
|
|
|
# Re-index `data` filling in `0`s where there is no demand.
|
|
index = pd.MultiIndex.from_tuples(gen)
|
|
index.names = ['pixel_id', 'start_at']
|
|
|
|
df = pd.DataFrame(data={'n_orders': 1}, index=index)
|
|
|
|
# Sanity check: n_pixels * n_time_steps_per_day * n_weekdays * n_weeks.
|
|
assert len(df) == 2 * 12 * (7 * 2 + 1)
|
|
|
|
return df
|
|
|
|
|
|
@pytest.fixture
|
|
def order_history(order_totals, grid):
|
|
"""An `OrderHistory` object that does not need the database."""
|
|
oh = timify.OrderHistory(grid=grid, time_step=test_config.LONG_TIME_STEP)
|
|
oh._data = order_totals
|
|
|
|
return oh
|