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<p>我试图用曲线拟合来优化函数科学优化. 这是我的密码。你知道吗</p>
<pre><code>import pandas as pd
import numpy as np
from scipy.optimize import curve_fit
xdata = [row[0] for row in pd.read_excel("C:\\Users\\310967\\Desktop\\Scholar\\Wound Chelation Draft\\ChelationFiles.xlsx", sheetname="Case2Data",skiprows=0).as_matrix()]
ydata = [row[1] for row in pd.read_excel("C:\\Users\\310967\\Desktop\\Scholar\\Wound Chelation Draft\\ChelationFiles.xlsx", sheetname="Case2Data",skiprows=0).as_matrix()]
SF = [row[4] for row in pd.read_excel("C:\\Users\\310967\\Desktop\\Scholar\\Wound Chelation Draft\\ChelationFiles.xlsx", sheetname="Case2Data",skiprows=0).as_matrix()]
uncertainty = [(np.sqrt(np.exp(np.log(a)**2)-1))*b for a,b in zip(SF, ydata)]
Tau = [0,1,5,7]
def func(x, I, E, ic1, ic2, ih1, ih2):
def iu(t):
return ((0.01295*np.exp(-0.645974*t))+(4.3688e-4*np.exp(-0.04251*t))+(5.642452e-5*np.exp(-0.00160863*t)))
def ic(t,tj):
if t > tj:
return ic1*np.exp(-0.693/ih1*(t-tj))+ic1*np.exp(-0.693/ih1*(t-tj))
else:
return 0
def listofic(t):
list1 = []
for tj in Tau:
list1.append(ic(t,tj))
return list1
def Kj(tj):
return iu(tj+1)*(E-1)/(ic(1,0)-iu(tj+1))
def listofKj():
list2 = []
for tj in Tau:
list2.append(Kj(tj))
return list2
Kjs = listofKj()
def listofOneMinusKj(t):
list3 = []
for a in Tau:
if t > a:
value = 1-Kj(a)
else:
value = 1
list3.append(value)
return list3
return (iu(x)*np.prod(listofOneMinusKj(x))+sum([a*b for a,b in zip(Kjs,listofic(x))]))*I
popt, pcov = curve_fit(func, xdata, ydata, sigma=uncertainty)
print(popt)
</code></pre>
<p>当我运行上面的代码时,我得到一个错误,指出“一个数组中有多个元素的真值是不明确的。使用a.any()或a.all()。这是指函数列表ofoneminuskj(t)中的“if t>;a”部分。你知道吗</p>
<p>但是,如果我运行下面的代码,尽管有一个“if t>;a”,代码的运行方式与我期望的一样。我想知道上面的代码有什么问题。你知道吗</p>
<pre><code>import numpy as np
Tau = [0,1,5,7]
def func(x, I, E, ic1, ic2, ih1, ih2):
def iu(t):
return ((0.01295*np.exp(-0.645974*t))+(4.3688e-4*np.exp(-0.04251*t))+(5.642452e-5*np.exp(-0.00160863*t)))
def ic(t,tj):
if t > tj:
return ic1*np.exp(-0.693/ih1*(t-tj))+ic1*np.exp(-0.693/ih1*(t-tj))
else:
return 0
def listofic(t):
list1 = []
for tj in Tau:
list1.append(ic(t,tj))
return list1
def Kj(tj):
return iu(tj+1)*(E-1)/(ic(1,0)-iu(tj+1))
def listofKj():
list2 = []
for tj in Tau:
list2.append(Kj(tj))
return list2
Kjs = listofKj()
def listofOneMinusKj(t):
list3 = []
for a in Tau:
if t > a:
value = 1-Kj(a)
else:
value = 1
list3.append(value)
return list3
return (iu(x)*np.prod(listofOneMinusKj(x))+sum([a*b for a,b in zip(Kjs,listofic(x))]))*I
print(func(1,400,12.5,0.99,0.01,0.55,10))
</code></pre>