python - 如何使用 python 获得正确的正面区域？

Question

我很难获得积极的区域（高于 y=0）。从数学上讲，面积应该是 1.125。

x=np.arange(1,5,1)

y=np.array([-1.5,-0.5,0.5,1.5])

但是下面的函数都没有给我答案。第一个是 0。第二个给我 1.25，它不会插值来计算面积。我需要得到答案 1.125。任何人都可以帮忙吗？谢谢！

np.trapz(y,x)

np.trapz(y.clip(min=0),x)

标签： pythonnumpyscipy

Answer 1

为此，您必须找到数据的线性插值与 x 轴相交的位置。此功能是我在另一个答案（数字化模拟信号）中包含的功能的变体：

def find_zero_crossings(t, y):
    """
    Given the input signal `y` with samples at times `t`,
    find the times where the linearly interpolated graph
    crosses 0.

    `t` and `y` must be 1-D numpy arrays.
    """
    transition_indices = np.where((np.sign(y[:-1]) * np.sign(y[1:])) == -1)[0]

    # Linearly interpolate the time values where the transition occurs.
    t0 = t[transition_indices]
    t1 = t[transition_indices + 1]
    y0 = y[transition_indices]
    y1 = y[transition_indices + 1]
    slope = (y1 - y0) / (t1 - t0)
    transition_times = t0 - y0 / slope

    return transition_times

您可以在如下脚本中将其用于您的示例：

xx = np.arange(1, 5, 1)
yy = np.array([-1.5, -0.5, 0.5, 1.5])

xz = find_zero_crossings(xx, yy)
print("xz:", xz)

# Insert the x intercepts into the data.
xx2 = np.append(xx, xz)
yy2 = np.append(yy, np.zeros_like(xz))

# Restore the order of xx2, and order yy2 in the same way.
k = xx2.argsort()
xx2 = xx2[k]
yy2 = yy2[k]

print("xx2:", xx2)
print("yy2:", yy2)

# pos_yy2 is the clipped version of yy2.
pos_yy2 = np.maximum(yy2, 0.0)

print("pos_yy2:", pos_yy2)

# Compute the area with `trapz`.
pos_area = np.trapz(pos_yy2, xx2)
print("pos_area:", pos_area)

输出：

xz: [2.5]
xx2: [1.  2.  2.5 3.  4. ]
yy2: [-1.5 -0.5  0.   0.5  1.5]
pos_yy2: [0.  0.  0.  0.5 1.5]
pos_area: 1.125

完成所有这些的功能是：

def pos_trapz(y, x=None, dx=1.0):
    if x is None:
        x = np.arange(len(y))*dx
    xz = find_zero_crossings(x, y)

    # Insert the x intercepts into the data.
    x2 = np.append(x, xz)
    y2 = np.append(y, np.zeros_like(xz))

    # Restore the order of x2, and order y2 in the same way.
    # (The function assumes that the input `x` is in ascending order.)
    k = x2.argsort()
    x2 = x2[k]
    y2 = y2[k]

    # pos_y2 is the clipped version of y2.
    pos_y2 = np.maximum(y2, 0.0)

    # Compute the area with `trapz`.
    return np.trapz(pos_y2, x2)

在 ipython 会话中：

In [107]: xx = np.arange(1, 5, 1)

In [108]: yy = np.array([-1.5, -0.5, 0.5, 1.5])

In [109]: pos_trapz(yy, xx)                        
Out[109]: 1.125