使用python计算FWHM [英] FWHM calculation using python

问题描述

我正在尝试使用 python 计算光谱的 FWHM.光谱描述(我说的是物理学)对我来说有点复杂，我无法使用一些简单的高斯或洛伦兹分布来拟合数据.

I am trying to calculate the FWHM of spectra using python. The spectral description (I'm talking in terms of the physics) for me it's bit complicated and I can't fit the data using some simple Gaussian or Lorentizian profile.

到目前为止，我设法管理数据的插值并通过半最大值绘制一条平行于 X 轴的直线.

So far I managed to manage interpolation of the data and draw a straight line parallel to the X axis through the half maxima.

我怎样才能找到峰两侧的两条线的交点坐标?

How can I find the coordinates of the intersection of the two lines on both sides of the peak?

我知道如果我将光标放在那些点上，它会给我坐标，但我想自动化这个过程，以便它变得更加用户友好.我该怎么做?

I know if I take the cursor in those points it will give me the coordinates but I want to automate this process so that it becomes much more user friendly. How can I do that?

推荐答案

from matplotlib import pyplot as mp
import numpy as np

def peak(x, c):
    return np.exp(-np.power(x - c, 2) / 16.0)

def lin_interp(x, y, i, half):
    return x[i] + (x[i+1] - x[i]) * ((half - y[i]) / (y[i+1] - y[i]))

def half_max_x(x, y):
    half = max(y)/2.0
    signs = np.sign(np.add(y, -half))
    zero_crossings = (signs[0:-2] != signs[1:-1])
    zero_crossings_i = np.where(zero_crossings)[0]
    return [lin_interp(x, y, zero_crossings_i[0], half),
            lin_interp(x, y, zero_crossings_i[1], half)]

# make some fake data
x=np.linspace(0,20,21)
y=peak(x,10)

# find the two crossing points
hmx = half_max_x(x,y)

# print the answer
fwhm = hmx[1] - hmx[0]
print("FWHM:{:.3f}".format(fwhm))

# a convincing plot
half = max(y)/2.0
mp.plot(x,y)
mp.plot(hmx, [half, half])
mp.show()

两点的(x, y)坐标为(hmx[0], half)和(hmx[1], half).

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