Python 如何找到滞后的过零点？

在numpy中，我想检测信号从（以前）低于某个阈值到高于某个其他阈值的交叉点。这适用于去抖动或存在噪声时的精确过零等情况

像这样：

import numpy

# set up little test problem
N = 1000
values = numpy.sin(numpy.linspace(0, 20, N))
values += 0.4 * numpy.random.random(N) - 0.2
v_high = 0.3
v_low = -0.3

# find transitions from below v_low to above v_high    
transitions = numpy.zeros_like(values, dtype=numpy.bool)

state = "high"

for i in range(N):
    if values[i] > v_high:
        # previous state was low, this is a low-to-high transition
        if state == "low":
            transitions[i] = True
        state = "high"
    if values[i] < v_low:
        state = "low"

导入numpy
#设置一个小测试问题
N=1000
值=numpy.sin（numpy.linspace（0,20,N））
值+=0.4*numpy.random.random（N）-0.2
v_高=0.3
v_低=-0.3
#查找从低于v_低到高于v_高的转换
transitions=numpy.zeros_like（值，dtype=numpy.bool）
state=“高”
对于范围（N）中的i：
如果值[i]>v_高：
#以前的状态为低，这是从低到高的转换
如果状态==“低”：
转换[i]=True
state=“高”
如果值[i]

---

我想要一种不显式循环数组的方法：但我想不出任何方法，因为每个状态值都依赖于前一个状态。可以不使用循环吗？

可以这样做：

def hyst(x, th_lo, th_hi, initial = False):
    hi = x >= th_hi
    lo_or_hi = (x <= th_lo) | hi
    ind = np.nonzero(lo_or_hi)[0]
    if not ind.size: # prevent index error if ind is empty
        return np.zeros_like(x, dtype=bool) | initial
    cnt = np.cumsum(lo_or_hi) # from 0 to len(x)
    return np.where(cnt, hi[ind[cnt-1]], initial)

结果:

---

我必须对我的工作进行修改，所有修改都基于上面的答案，以允许在使用标准阈值和反向阈值时检测阈值交叉

我不喜欢这个名字，也许现在应该改为

thu-hi2lo

和

thu-lo2hi

，而不是

thu-lo

和

thu-hi

？使用原始值，行为也同样艰难

def hyst(x, th_lo, th_hi, initial = False):
    """
    x : Numpy Array
        Series to apply hysteresis to.
    th_lo : float or int
        Below this threshold the value of hyst will be False (0).
    th_hi : float or int
        Above this threshold the value of hyst will be True (1).
    """        

    if th_lo > th_hi: # If thresholds are reversed, x must be reversed as well
        x = x[::-1]
        th_lo, th_hi = th_hi, th_lo
        rev = True
    else:
        rev = False

    hi = x >= th_hi
    lo_or_hi = (x <= th_lo) | hi

    ind = np.nonzero(lo_or_hi)[0]  # Index für alle darunter oder darüber
    if not ind.size:  # prevent index error if ind is empty
        x_hyst = np.zeros_like(x, dtype=bool) | initial
    else:
        cnt = np.cumsum(lo_or_hi)  # from 0 to len(x)
        x_hyst = np.where(cnt, hi[ind[cnt-1]], initial)

    if rev:
        x_hyst = x_hyst[::-1]

    return x_hyst

---

@SpaceDog您可以，但如果您使用C语言，则最好编写一个与原始问题类似的简单循环。我的答案中的技巧在Python中更快，因为它使用矢量化的numpy代码，而不是缓慢的Python循环。矢量化代码必须在数据上传递多次，而C中的一个简单循环可以在一次传递中完成所有操作。我问这个问题是因为我试图理解你的函数的功能，所以我可以用C编写代码。。。

def hyst(x, th_lo, th_hi, initial = False):
    """
    x : Numpy Array
        Series to apply hysteresis to.
    th_lo : float or int
        Below this threshold the value of hyst will be False (0).
    th_hi : float or int
        Above this threshold the value of hyst will be True (1).
    """        

    if th_lo > th_hi: # If thresholds are reversed, x must be reversed as well
        x = x[::-1]
        th_lo, th_hi = th_hi, th_lo
        rev = True
    else:
        rev = False

    hi = x >= th_hi
    lo_or_hi = (x <= th_lo) | hi

    ind = np.nonzero(lo_or_hi)[0]  # Index für alle darunter oder darüber
    if not ind.size:  # prevent index error if ind is empty
        x_hyst = np.zeros_like(x, dtype=bool) | initial
    else:
        cnt = np.cumsum(lo_or_hi)  # from 0 to len(x)
        x_hyst = np.where(cnt, hi[ind[cnt-1]], initial)

    if rev:
        x_hyst = x_hyst[::-1]

    return x_hyst

x = np.linspace(0,20, 1000)
y = np.sin(x)
h1 = hyst(y, -0.2, 0.2)
h2 = hyst(y, +0.5, -0.5)
plt.plot(x, y, x, -0.2 + h1*0.4, x, -0.5 + h2)
plt.legend(('input', 'output, classic, hyst(y, -0.2, +0.2)', 
            'output, reversed, hyst(y, +0.5, -0.5)'))
plt.title('Thresholding with hysteresis')
plt.show()