符號計算插值函數的輸入？

Question

我有一個相當複雜的功能H(x)，我試圖解決的價值x，使H(x) = constant。我想用離散間隔生成的插值對象和H（間隔）的相應輸出來做到這一點，其中其他輸入保持不變。我表示插值對象f。

我的問題是插值對象的調用函數接受一個array_like，所以傳遞一個符號到f(x)來使用sage的求解器方法是不成問題的。任何想法如何解決這個問題？我有插補功能f。我想求解方程f(x) == sageconstant for X .

from scipy.interpolate import InterpolatedUnivariateSpline as IUspline 
    import numpy as np 

    #Generating my interpolation object 
    xint = srange(30,200,step=.1) 
    val = [H(i,1,.1,0,.2,.005,40) for i in srange(30,299,step=.1)] 
    f = IUspline(xint,val,k=4) 

    #This will yield a sage constant 
    eq_G(x) = freeB - x 

    #relation that I would like to solve 
    eq_m(x) = eq_G(39.9) == f(x) 
    m = solve(eq_m(x),x) 

上面的代碼（f(x)更具體）生成

"TypeError: Cannot cast array data from dtype('0') to dtype('float64') according to the rule 'safe'.

編輯：任何功能H(x)將導致同樣的錯誤，因此，它不不管什麼H(x)是。爲了簡單起見（當我說H很複雜時，我不是在開玩笑），請嘗試H(x) = x。然後該塊將顯示如下：

from scipy.interpolate import InterpolatedUnivariateSpline as IUspline 
    import numpy as np 

    #Generating my interpolation object 
    xint = srange(30,200,step=.1) 
    H(x) = x 
    val = [H(i) for i in srange(30,299,step=.1)] 
    f = IUspline(xint,val,k=4) 

    #This will yield a sage constant 
    eq_G(x) = freeB - x 

    #relation that I would like to solve 
    eq_m(x) = eq_G(39.9) == f(x) 
    m = solve(eq_m(x),x) 

Answer 1

在使用numpy和scipy時，首選Python類型爲Sage類型。

而不是Sage Integers和Reals，使用Python ints和float。

也許你可以像這樣修復你的代碼。

from scipy.interpolate import InterpolatedUnivariateSpline as IUspline 
import numpy as np 

# Generate interpolation object 
xint = srange(30,200,step=.1) 
xint = [float(x) for x in xint] 
val = [float(H(i,1,.1,0,.2,.005,40)) for i in srange(30,299,step=.1)] 
f = IUspline(xint,val,k=4) 

# This will yield a Sage constant 
eq_G(x) = freeB - x 

# relation that I would like to solve 
eq_m(x) = eq_G(39.9) == f(x) 
m = solve(eq_m(x),x)