python - 任意长度的 Numpy 分段

Question

我需要构建一个具有任意数量的间隔和函数的分段函数，能够对输入的 numpy 数组进行操作。

我可以使用 for 循环和指示符数组来做到这一点，如下面的代码片段所示，但是有没有一种更 Pythonic 的方法来做到这一点？

我尝试使用 numpy.piecewise，但据我所知，段和函数的数量需要在源代码中静态定义。

import numpy as np
import matplotlib.pyplot as plt 

# inputs:
#    -xs: points in which to compute the piecewise function
#    -segments: the extremes of the piecewise intervals (as a list)
#    -funcs: the functions (as a list; len(funcs)==len(segments)-1 )
def calc_piecewise(xs, segments, funcs):
    # prepare indicators and results arrays
    indaseg = np.zeros(len(xs), np.bool)
    ys = np.zeros_like(xs)

    # loop through intervals and compute the ys
    for ii in range(len(funcs)):
        indaseg = np.logical_and(xs>=segments[ii], xs<=segments[ii+1])
        ys[indaseg] = funcs[ii](xs[indaseg])

    return ys

def test_calc_piecewise():
    segments = [0.0, 1.0, 2.5, 4.0, 5.0]
    def f0(xs):
        return xs
    def f1(xs):
        return xs*xs
    def f2(xs):
        return 12.5-xs*xs
    def f3(xs):
        return 4.0*xs-19.5
    funcs = [f0, f1, f2, f3]

    xs = np.linspace(0.0, 5.0, 500)
    ys = calc_piecewise(xs, segments, funcs)

    plt.figure()
    title = "calc_piecewise"
    plt.title(title)
    plt.plot(xs, ys, 'r-')
    plt.show()

    return 


test_calc_piecewise()

Answer 1

您可以使用np.piecewise来做到这一点如下(抱歉格式错误!):

ys = np.piecewise(
        xs,
        [(xs >= segments[i]) & (xs <= segments[i+1]) for i in range(len(segments)-1)],
        funcs)

结果是一样的。

本质上你的循环和测试相当于你的行 indaseg = np.logical_and(xs>=segments[ii], xs<=segments[ii+1])被移动到调用代码中。