自动减少分段函数组件 - Pyomo答案

【问题标题】：Automatically reduce piecewise function components - Pyomo自动减少分段函数组件 - Pyomo
【发布时间】：2019-07-10 14:49:20
【问题描述】：

在 pyomo 中，我有一个通过 pyomo.environ.Piecewise 定义的分段线性约束。我不断收到类似

的警告

Piecewise component '<component name>' has detected slopes of consecutive piecewise segments to be within <tolerance> of one another. Refer to the Piecewise help documentation for information on how to disable this warning.

我知道我可以增加容差并消除警告，但我想知道如果两个连续斜率低于给定值，是否有一种通用方法（通过 Pyomo 或 numpy）来减少“段”的数量容忍度。

我显然可以自己实现这一点，但我想避免重新发明轮子。

【问题讨论】：

标签： python numpy pyomo piecewise

【解决方案1】：

好的，这就是我想出的。绝对没有针对性能进行优化，但我的情况取决于几点。它还缺少对输入的更多验证（例如，x 已排序且唯一）。

def reduce_piecewise(x, y, abs_tol):
    """
    Remove unnecessary points from piece-wise curve.

    Points are remove if the slopes of consecutive segments
    differ by less than `abs_tol`.

    x points must be sorted and unique.
    Consecutive y points can be the same though!

    Parameters
    ----------
    x : List[float]
        Points along x-axis.
    y : List[float]
    abs_tol : float
        Tolerance between consecutive segments.

    Returns
    -------
    (np.array, np.array)
        x and y points - reduced.
    """
    if not len(x) == len(y):
        raise ValueError("x and y must have same shape")

    x_reduced = [x[0]]
    y_reduced = [y[0]]

    for i in range(1, len(x) - 1):
        left_slope  = (y[i] - y_reduced[-1])/(x[i] - x_reduced[-1])
        right_slope = (y[i+1] - y[i])/(x[i+1] - x[i])
        if abs(right_slope - left_slope) > abs_tol:
            x_reduced.append(x[i])
            y_reduced.append(y[i])

    x_reduced.append(x[-1])
    y_reduced.append(y[-1])

    return np.array(x_reduced), np.array(y_reduced)

这里有一些例子：

>>> x = np.array([0, 1, 2, 3])
>>> y = np.array([0, 1, 2, 3])
>>> reduce_piecewise(x, y, 0.01)
(array([0, 3]), array([0, 3]))

>>> x = np.array([0, 1, 2, 3, 4, 5])
>>> y = np.array([0, 2, -1, 3, 4.001, 5]) # 4.001 should be removed
>>> reduce_piecewise(x, y, 0.01)
(array([0, 1, 2, 3, 5]), array([ 0.,  2., -1.,  3.,  5.]))

【讨论】：