Построить массив из случайного распределения, пока последний столбец не превысит порог

Question

Я хочу построить двумерный массив из случайного распределения, чтобы каждое из значений в последнем столбце каждой строки превышало пороговое значение.

Вот рабочий код, который у меня сейчас есть.Есть ли более чистый способ создания массивов с произвольным условием?

def new_array(
        num_rows: int,
        dist: Callable[[int], np.ndarray],
        min_hours: int) -> np.ndarray:
    # Get the 40th percentile as a reasonable guess for how many samples we need.
    # Use a lower percentile to increase num_cols and avoid looping in most cases.
    p40_val = np.quantile(dist(20), 0.4)
    # Generate at least 10 columns each time.
    num_cols = max(int(min_hours / p40_val), 10)

    def create_starts() -> np.ndarray:
        return dist(num_rows * num_cols).reshape((num_rows, num_cols)).cumsum(axis=1)

    max_iters = 20
    starts = create_starts()
    for _ in range(max_iters):
        if np.min(starts[:, -1]) >= min_hours:
            # All the last columns exceed min_hours.
            break

        last_col_vals = starts[:, -1].repeat(num_cols).reshape(starts.shape)
        next_starts = create_starts() + last_col_vals
        starts = np.append(starts, next_starts, axis=1)
    else:
        # We didn't break out of the for loop, so we hit the max iterations.
        raise AssertionError('Failed to create enough samples to exceed '
                             'sim duration for all columns')

    # Only keep columns up to the column where each value > min_hours.
    mins_per_col = np.min(starts, axis=0)
    cols_exceeding_sim_duration = np.nonzero(mins_per_col > min_hours)[0]
    cols_to_keep = cols_exceeding_sim_duration[0]
    return np.delete(starts, np.s_[cols_to_keep:], axis=1)


new_array(5, lambda size: np.random.normal(3, size=size), 7)

# Example output
array([[1.47584632, 4.04034105, 7.19592256],
       [3.10804306, 6.46487043, 9.74177227],
       [1.03633165, 2.62430309, 6.92413189],
       [3.46100139, 6.53068143, 7.37990547],
       [2.70152742, 6.09488369, 9.58376664]])

Answer 1

Я упростил несколько вещей и заменил их логическим индексированием Numpy.Цикл for теперь работает, и нет необходимости обрабатывать ошибку, поскольку она просто выполняется, пока не будет достаточно строк.

Это все еще работает, как вы ожидаете?

def new_array(num_rows, dist, min_hours):

    # Get the 40th percentile as a reasonable guess for how many samples we need.
    # Use a lower percentile to increase num_cols and avoid looping in most cases.
    p40_val = np.quantile(dist(20), 0.4)
    # Generate at least 10 columns each time.
    num_cols = max(int(min_hours / p40_val), 10)

    # no need to reshape here, size can be a shape tuple
    def create_starts() -> np.ndarray:
        return dist((num_rows, num_cols)).cumsum(axis=1)

    # append to list, in the end stack it into a Numpy array once.
    # faster than numpy.append
    # due to Numpy's pre-allocation which will slow down things here.
    storage = []

    while True:

        starts = create_starts()

        # boolean / logical array
        is_larger = starts[:, -1] >= min_hours

        # Use Numpy boolean indexing instead to find the rows
        # fitting your condition
        good_rows = starts[is_larger, :]

        # can also be empty array if none found, but will
        # be skipped later
        storage.append(good_rows)

        # count what is in storage so far, empty arrays will be skipped
        # due to shape (0, x)
        number_of_good_rows = sum([_a.shape[0] for _a in storage])
        print('number_of_good_rows', number_of_good_rows)

        if number_of_good_rows >= num_rows:
            starts = np.vstack(storage)
            print(starts)
            break

    # Only keep columns up to the column where each value > min_hours.
    # also use logical indexing here
    is_something = np.logical_not(np.all(starts > min_hours, axis=0))

    return starts[:, is_something]