python scipy.signal.peak_widths->绝对高度吗? (fft -3dB阻尼)

时间:2018-12-14 11:15:54

标签: python scipy

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我认为链接函数只能计算相对高度的峰宽。有谁知道是否有一个函数可以对所有峰以固定值(peak_amplitude-x)计算宽度?

当前,我正在尝试更改原始内部函数“ _peak_widths”。 cimport已经失败。这里仅部分了解源代码。我在代码中进行了修改。

 with nogil:
    for p in range(peaks.shape[0]):
        i_min = left_bases[p]
        i_max = right_bases[p]
        peak = peaks[p]
        # Validate bounds and order
        if not 0 <= i_min <= peak <= i_max < x.shape[0]:
            with gil:
                raise ValueError("prominence data is invalid for peak {}"
                                 .format(peak))
        height = width_heights[p] = x[peak] - prominences[p] * rel_height 

更改为x [peak]-3

        # Find intersection point on left side
        i = peak
        while i_min < i and height < x[i]:
            i -= 1
        left_ip = <np.float64_t>i
        if x[i] < height:
            # Interpolate if true intersection height is between samples
            left_ip += (height - x[i]) / (x[i + 1] - x[i])

        # Find intersection point on right side
        i = peak
        while i < i_max and height < x[i]:
            i += 1
        right_ip = <np.float64_t>i
        if  x[i] < height:
            # Interpolate if true intersection height is between samples
            right_ip -= (height - x[i]) / (x[i - 1] - x[i])

        widths[p] = right_ip - left_ip
        if widths[p] == 0:
            show_warning = True
        left_ips[p] = left_ip
        right_ips[p] = right_ip

2 个答案:

答案 0 :(得分:1)

如果这仍然与您有关,则可以按原样使用scipy.signal.peak_widths,通过传入修改后的prominence_data来实现所需的功能。根据您自己的answer

import numpy as np
from scipy.signal import find_peaks, peak_prominences, peak_widths

# Create sample data
x = np.linspace(0, 6 * np.pi, 1000)
x = np.sin(x) + 0.6 * np.sin(2.6 * x)

# Find peaks
peaks, _ = find_peaks(x)
prominences, left_bases, right_bases = peak_prominences(x, peaks)

peak_widths的文档中所述,测量宽度的高度计算为 h_eval = h_peak - prominence * relative_height

我们可以通过参数prominence_datarel_height控制后两个变量。因此,我们可以创建一个数组,其中所有值都相同,并使用其创建绝对高度:

prominence

enter image description here

您可以看到,以相同的恒定偏移量1评估每个峰的宽度。通过使用# Create constant offset as a replacement for prominences offset = np.ones_like(prominences) # Calculate widths at x[peaks] - offset * rel_height widths, h_eval, left_ips, right_ips = peak_widths( x, peaks, rel_height=1, prominence_data=(offset, left_bases, right_bases) ) # Check that h_eval is 1 everywhere np.testing.assert_equal(x[peaks] - h_eval, 1) # Visualize result import matplotlib.pyplot as plt plt.plot(x) plt.plot(peaks, x[peaks], "x") plt.hlines(h_eval, left_ips, right_ips, color="C2") plt.show() 提供的原始left_basesright_bases,我们限制了最大值测量宽度(例如,见299和533的峰值)。如果要消除该限制,则必须自己创建这些数组。

答案 1 :(得分:0)

我刚刚删除了c内容。那就是我的解决办法:

def gauss(x, p): # p[0]==mean, p[1]==stdev
    return 1.0/(p[1]*np.sqrt(2*np.pi))*np.exp(-(x-p[0])**2/(2*p[1]**2))

def _peak_widths(x,peaks,prop,val=3):

    i_min = prop['left_bases']
    i_max = prop['right_bases']
    peak = peaks[0]
    # Validate bounds and order
    height = x[peak] - val

    # Find intersection point on left side
    i = peak
    while i_min < i and height < x[i]:
        i -= 1
    left_ip = i
    if x[i] < height:
        # Interpolate if true intersection height is between samples
        left_ip += (height - x[i]) / (x[i + 1] - x[i])

    # Find intersection point on right side
    i = peak
    while i < i_max and height < x[i]:
        i += 1
    right_ip = i
    if  x[i] < height:
        # Interpolate if true intersection height is between samples
        right_ip -= (height - x[i]) / (x[i - 1] - x[i])

    widths = right_ip - left_ip
    left_ips = left_ip
    right_ips = right_ip

    return [height, widths, int(left_ips), int(right_ips)]

if __name__ == '__main__':

    # Create some sample data
    known_param = np.array([2.0, 0.07])
    xmin,xmax = -1.0, 5.0
    N = 1000
    X = np.linspace(xmin,xmax,N)
    Y = gauss(X, known_param)
    fig, ax= plt.subplots()
    ax.plot(X,Y)

    #find peaks
    peaks, prop = signal.find_peaks(Y, prominence = 3.1)
    ax.scatter(X[peaks],Y[peaks], color='r')

    #calculate peak width
    y, widths, x1, x2 = _peak_widths(Y,peaks, prop)

    print(f'width = { X[x1] - X[x2]}')

    l = mlines.Line2D([X[x1],X[x2]], [y,y], color='r')
    ax.add_line(l)
    plt.show()