求曲线曲线的最大值 [英] Finding the maximum of a curve scipy

问题描述

我已将曲线拟合到一组数据点.我想知道如何找到曲线的最大点，然后我想注释该点(我不想使用数据中的最大 y 值来执行此操作).我不能完全编写我的代码，但这里是我的代码的基本布局.

I have fitted curve to a set of data points. I would like to know how to find the maximum point of my curve and then I would like to annotate that point (I don't want to use by largest y value from my data to do this). I cannot exactly write my code but here is the basic layout of my code.

import matplotlib.pyplot as plt
from scipy.optimize import curve_fit

x = [1,2,3,4,5]
y = [1,4,16,4,1]

def f(x, p1, p2, p3):
    return p3*(p1/((x-p2)**2 + (p1/2)**2))   

p0 = (8, 16, 0.1) # guess perameters 
plt.plot(x,y,"ro")
popt, pcov = curve_fit(f, x, y, p0)
plt.plot(x, f(x, *popt)) 

还有找到峰宽的方法吗?

Also is there a way to find the peak width?

我是否缺少一个可以执行此操作的简单内置函数?我可以区分函数并找到它为零的点吗?如果是这样怎么办?

Am I missing a simple built in function that could do this? Could I differentiate the function and find the point at which it is zero? If so how?

推荐答案

在您拟合找到最佳参数以最大化您的函数后，您可以使用 minimize_scalar(或其他方法之一)找到峰值 scipy.optimize 中的方法).

After you fit to find the best parameters to maximize your function, you can find the peak using minimize_scalar (or one of the other methods from scipy.optimize).

请注意，在下面，我已经移动了 x[2]=3.2 以便曲线的峰值不会落在数据点上，我们可以确定我们正在找到达到曲线的峰值，而不是数据.

Note that in below, I've shifted x[2]=3.2 so that the peak of the curve doesn't land on a data point and we can be sure we're finding the peak to the curve, not the data.

import numpy as np
import matplotlib.pyplot as plt
from scipy.optimize import curve_fit, minimize_scalar

x = [1,2,3.2,4,5]
y = [1,4,16,4,1]

def f(x, p1, p2, p3):
    return p3*(p1/((x-p2)**2 + (p1/2)**2))   

p0 = (8, 16, 0.1) # guess perameters 
plt.plot(x,y,"ro")
popt, pcov = curve_fit(f, x, y, p0)

# find the peak
fm = lambda x: -f(x, *popt)
r = minimize_scalar(fm, bounds=(1, 5))
print "maximum:", r["x"], f(r["x"], *popt)  #maximum: 2.99846874275 18.3928199902

x_curve = np.linspace(1, 5, 100)
plt.plot(x_curve, f(x_curve, *popt))
plt.plot(r['x'], f(r['x'], *popt), 'ko')
plt.show()

当然，除了优化功能外，我们还可以针对一堆x值计算并获得接近值:

Of course, rather than optimizing the function, we could just calculate it for a bunch of x-values and get close:

x = np.linspace(1, 5, 10000)
y = f(x, *popt)
imax = np.argmax(y)
print imax, x[imax]     # 4996 2.99859985999

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