问题描述
这个问题特别解决了在油漆，颜料等颜色混合的情况下曲线拟合的问题.
This question addresses in particular the question of curve fitting in the context of color mixing of paints, pigments, etc.
我试图猜测两种涂料的比例，比如说棕色"(B)和白色"(W)，以达到给定的亮度值L.
I am trying to guess the required proportions of two paints, let's say "Brown" (B) and "white" (W) to get to a given lightness value L.
我以一种将比尔-兰伯特定律"应用到化学中的方式制作了校准曲线".但是，曲线不是线性的，所以我不能使用比尔-朗伯定律.
I have made a "calibration curve" in the same fashion as one does so for applying the Beer-lambert law in chemistry. However, the curve is not linear so I cannot use the Beer-Lambert law.
这就是我所做的:
(1)
-
我已经测量了这些比例的混合物的油漆样品的光谱，标记为a，b，c，d，...等.
I have measured the spectrum of paint samples for these proportions of mixture, labeled a, b, c, d, ... etc.
a >>> W = 1，B = 0(纯白色)
a >>> W = 1, B = 0 (pure white)
b >>> W = 63/64，B = 1/64
b >>> W = 63/64, B = 1/64
c >>> W = 31/32，B = 1/32
c >>> W = 31/32, B = 1/32
d >>> W = 15/16，B = 1/16
d >>> W = 15/16, B = 1/16
e >>> W = 7/8，B = 1/8
e >>> W = 7/8, B = 1/8
f >>> W = 3/4，B = 1/4
f >>> W = 3/4, B = 1/4
g >>> W = 1/2，B = 1/2
g >>> W = 1/2, B = 1/2
h >>> W = 0，B = 1(纯棕色)
h >>> W = 0, B = 1 (pure brown)
这些是我得到的光谱反射率曲线:
And these are the spectral reflectance curves that I got :
如果我在给定波长(例如，波长)下获得一个反射率值， 500 nm，我得到了一条漂亮的曲线，其中x轴代表混合物中白色涂料的比例，y轴代表500 nm处的反射光:
If I pick-up one reflectance value at a given wavelength, e.g. 500 nm, I get this nice curve, where the x axis represents the proportion of white paint in the mix, and the y axis the reflected light at 500 nm :
我想通过内插法猜测达到一定数量的反射光需要多少白色.
I'd like to guess by interpolation how much white I need to arrive at a certain amount of reflected light.
(2)
我尝试使用scipy.optimize.curve_fit将指数曲线拟合到数据，但拟合度很差:
I have tried to fit an exponential curve to the data with scipy.optimize.curve_fit but the fit is pretty poor:
什么样的功能可以使数据紧密匹配?
What kind of function would fit the data closely?
推荐答案
由于没有人回答，我将扩展我的评论.
I'll expand my comment since nobody has answered.
根据我在图中看到的，有一个模式.最好的方法是拟合一条适合该图案整体的曲线.您可以使用Eureqa进行任何数学运算(免费试用就足够了): http://www .nutonian.com/products/eureqa/
From what I see in the figure, there is a pattern. The best way would be to fit a curve that fits that pattern as a whole. You can do this without any math using Eureqa (the free trial should be enough): http://www.nutonian.com/products/eureqa/
如果要保留在python中并适合指数分布，可以执行以下操作:如何做指数和对数曲线拟合在Python?我发现只有多项式拟合
If you want to remain in python and fit an exponential distribution, you can do the following:How to do exponential and logarithmic curve fitting in Python? I found only polynomial fitting
因此，假设您对于500nm波长具有以下值:
So imagine you have for the wavelength 500nm the following values:
y = [10,20,30,30,50,60,70,80,90,100]
x = [0.,0.3,0.5,0.6,0.72,0.77,0.84,0.9,0.95,1]
然后，拟合指数曲线的代码将是:
Then the code to fit the exponential curve would be:
import numpy as np
import matplotlib.pyplot as plt
from scipy.optimize import curve_fit
def func(x, a, b, c):
return a * np.exp(-b * x) + c
popt, pcov = curve_fit(func, x, y)
在这种情况下，我们得出a，b和c为:
In this case we get that a,b, and c are:
popt = array([ 7.1907744 , -2.62804994, 2.45029842])
因此，要获取特定x处的反射光值(例如0.2)，可以执行以下操作:
So to get the value of reflected light at a certain x (for instance 0.2), you can do:
func(0.2, 7.1907744 , -2.62804994, 2.45029842)
哪个是14.61
但是您说这很不合适，如果不需要模型，可以执行以下操作:如果您不太在意模型，则可以使用以下方法: https://docs.scipy.org/doc/scipy/reference/Generated/scipy.interpolate.interp1d.html#scipy.interpolate.interp1d
But you say it's a bad fit, if you don't need a model, you can do the following: If you don't really care about having a model you can use this: https://docs.scipy.org/doc/scipy/reference/generated/scipy.interpolate.interp1d.html#scipy.interpolate.interp1d
from scipy import interpolate
f = interpolate.interp1d(x, y, kind="quadratic") #you can try different kinds of interpolation
然后找到一个值(例如x = 0.2):
And then to find a value (for instance x=0.2):
ynew = f(0.2)
哪个是6.549
或者具有许多值，以便您可以绘制它们: ynew = f(np.linspace(0,1,1000)
Or to have many values so you can plot them: ynew = f(np.linspace(0,1,1000)
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