我从表面光度仪收集数据,并尝试将收集的数据拟合到理想曲线。目的是获得曲率半径。
我编写的代码将适合所有原始数据的曲线。
适合整个范围
问题是要指定一个范围来进行拟合。即:所有轮廓仪数据都不适合拟合。在下面的例子中,我希望从x = 180到x = 380
适合指定范围
当然,点击并指定x_start和x_end会更有帮助,但我无法从5.3 docs中提取数据
奖励:如果我可以在改变适合范围时显示R²,那就太棒了!
感谢您的帮助
以下是我提出的代码:
#importing modules
import tkinter.filedialog as tk
import numpy as np
import matplotlib.pyplot as plt
import scipy
from scipy.optimize import curve_fit
from matplotlib.widgets import Slider
#Graphic properties
fig, ax = plt.subplots()
plt.subplots_adjust(left=0.15, bottom=0.45)
#Search and open desired file:
def openfile():
opennm = tk.askopenfile()
f = open(opennm.name,"r")
data=np.genfromtxt(f,
skip_header=6,
names=True,
dtype=None,
delimiter=',')
x=[]
y=[]
for i in range(0,len(data)-1):
x.append(data[i][0])
y.append(data[i][1])
return x,y
#after opening the file, x and y data are plotted
x,y=openfile()
k,= plt.plot(x,y)
#Define the fitting function
def func(x,ROC,x_shift,y_shift):
return ((ROC*1000)-((ROC*1000)**2-(x-x_shift)**2)**0.5-y_shift)*1000
popt, pcov = curve_fit(func, x, y, p0=[45,250,0.8],bounds=((1,100,0),(100,300,2))) #, bounds=((-np.inf,10**-8,-np.inf,-np.inf),(np.inf,3.,np.inf,np.inf))
#The fitted curve will be overlaid on raw data
l, = plt.plot(x,func(x,*popt),'r--')
list_of_para=['ROC','X shift','Y shift']
list_of_units=['mm','a.u.','um']
for i in range (0,3):
print (list_of_para[i],'= ',"%.2f" %popt[i],' ', list_of_units[i])
#definig initial values
ROC0 = popt[0]
x_shift0 = popt[1]
y_shift0 = popt[2]
#defining sliders
axcolor = 'lightgoldenrodyellow'
axROC= plt.axes([0.15, 0.05, 0.75, 0.02], facecolor=axcolor)
axx_shift= plt.axes([0.15, 0.15, 0.75, 0.02], facecolor=axcolor)
axy_shift= plt.axes([0.15, 0.25, 0.75, 0.02], facecolor=axcolor)
#define slider limites
sROC = Slider(axROC, 'ROC', popt[0]-np.absolute(popt[0]),popt[0]+np.absolute(popt[0]), valinit=ROC0)
sx_shift = Slider(axx_shift, 'X Shift', popt[1]-np.absolute(popt[1]), popt[1]+np.absolute(popt[1]), valinit=x_shift0)
sy_shift = Slider(axy_shift, 'Y shift', popt[2]-np.absolute(popt[2]), popt[2]+np.absolute(popt[2]), valinit=y_shift0)
#define slider update values
def update(val):
ROC = sROC.val
x_shift = sx_shift.val
y_shift = sy_shift.val
l.set_ydata(((ROC*1000)-((ROC*1000)**2-(x-x_shift)**2)**0.5-y_shift)*1000)
fig.canvas.draw_idle()
sROC.on_changed(update)
sx_shift.on_changed(update)
sy_shift.on_changed(update)
plt.show()
注意:您可以从此处获取轮廓测量仪数据的示例:onclick
答案 0 :(得分:0)
如果您无法轻松截断数据,则需要对所有数据进行建模。我没有尝试过您的数据(似乎没有发布 - 可能会这样做?)但似乎数据可以建模为
Parabola * Rectangle_with_soft_boundaries
使用抛物线模型给出您正在寻找的曲率半径,矩形给出“截断背景”。
您可能会发现lmfit(https://github.com/lmfit/lmfit-py)对此很有用。它为曲线拟合提供了相当高级的界面,并且有许多可以组合的内置模型。
例如,您可以执行以下操作:
from lmfit.models import RectangleModel
#Define the fitting function
def func(x, roc, x_shift, y_shift):
return ((roc*1000)-((roc*1000)**2-(x-x_shift)**2)**0.5-y_shift)*1000
# turn model function into model,
# multiply by rectangle with error-function for sides:
model = Model(func) * RectangleModel(prefix='r_', form='erf')
params = model.make_params(r_amplitude=-600, r_center1=150,
r_center2=375, r_sigma1=10, r_sigma2=10,
roc=45, x_shift=200, y_shift=1)
# you can apply bounds to any of the parameters here:
params['roc'].min = 1
params['roc'].max = 100
params['r_amplitude'].max = 0 # make sure rectangle points down
params['r_center1'].min = 50
params['r_center1'].max = 250
# then perform fit
result = model.fit(y, params, x=x)
# print report of statistics, parameter values
print(result.fit_report())
# plot results
plt.plot(x, y)
plt.plot(x, result.best_fit)
plt.show()
当然,你仍然需要知道关于矩形背景打开和关闭的位置的一些信息 - 我怀疑这是来自某些光圈,你可能知道一些事情(或者可以搞清楚!)。
答案 1 :(得分:0)
感谢大家的所有意见和回答
我可以想出另一种解决方法(解释可能很长,所以我更喜欢发帖作为答案,而不是简单地编辑问题)
请耐心等待我仅编码3个月,我很高兴得到您的支持:
因此,我没有使用滑块来调整拟合曲线,而是使用了一个点 - 并且 - 点击来确定我想要执行拟合的范围(我最后会有两个问题)强>
import tkinter.filedialog as tk
import numpy as np
import matplotlib.pyplot as plt
import scipy
from scipy.optimize import curve_fit
fig, ax = plt.subplots()
def openfile(): #define function to open file and extract x and y
opennm = tk.askopenfile()
f = open(opennm.name,"r")
data=np.genfromtxt(f,
skip_header=6,
names=True,
dtype=None,
delimiter=',')
x=[]
y=[]
for i in range(0,len(data)-1):
x.append(data[i][0])
y.append(data[i][1])
return x,y
x,y=openfile()
k,= plt.plot(x,y)
coords = [] #I define the coordinates from where to where i want to run the fitting
def onclick(event): #this function to point and click on graph and define fitting area
global ix, iy
ix, iy = event.xdata , event.ydata
print ('x = %d, y = %d'%(
ix, iy))
global coords
coords.append((ix, iy))
if len(coords) == 2:
fig.canvas.mpl_disconnect(cid)
global l
for i in coords:
l,=ax.plot (*i,'ro' )
fig.canvas.draw_idle()
global m
def find_index(array,value): #this functions is to find index of my 2 point-and-click data
global limit
limit=(np.abs(array-value)).argmin()
return limit
limit1=find_index(x,coords[0][0]) #I extract index of Xstart
limit2=find_index(x,coords[1][0]) #I extraxt index of Xend
fitx=x[limit1:limit2] #From initial x data I extract the range where the fitting will be done
fity=y[limit1:limit2] #From initial y data I extract the range where the fitting will be done
def func(x,ROC,x_shift,y_shift): #This si the function I want to fit
return ((ROC*1000)-((ROC*1000)**2-(x-x_shift)**2)**0.5-y_shift)*1000
popt, pcov = curve_fit(func, fitx, fity, p0=[45,250,0.8],bounds=((0.4,100,0),(500,300,2))) #I run the fitting
m, = plt.plot(x,func(x,*popt),'r--')
print('ROC= ',round(popt[0],0),'mm| X Shift= ', round(popt[1],0),'um| Y Shift= ',round(popt[2],3),'um')
return coords, l,m
cid = fig.canvas.mpl_connect('button_press_event', onclick)
plt.show()
问题1: 在运行时:我选择拟合范围的下限,在选择上限之前,我收到以下信息:
limit2 = find_index(x,coords [1] [0])IndexError:列表索引超出范围
这很自然,因为我没有选择上限YET ...如何避免这个烦人的错误信息?
问题2: 运行拟合后,图形自动缩放...如何修正y轴的上限并保持自动缩放的下限?
我查看了我在互联网上找到的所有建议解决方案(plt.gca()。set_ylim(top = 200)...等),但每次下限自动固定为0
除了这两个恼人的错误,程序运行正常
再一次,我将链接发布到我的个人测量仪的测试数据中: