我正在嘗試解決一個(gè)具有 6x6 矩陣的矩陣作為它的條目（元素）我嘗試將 gen 的倒數(shù)乘以解決方案矩陣，但我不相信得到的答案的正確性。from sympy import Eq, solve_linear_system, Matrix,count_ops,Mul,hornerimport sympy as spa, b, c, d, e,f = sp.symbols('a b c d e f')ad = Matrix(([43.4,26.5,115,-40.5,52.4,0.921], [3.78,62.9,127,-67.6,110,4.80], [41.25,75.0,213,-88.9, 131, 5.88], [-10.6,-68.4,-120,64.6,-132,-8.49], [6.5,74.3,121,-72.8,179,29.7], [1.2,30.7,49.7,-28.7,91,29.9]))fb= Matrix(([1,0,0,0,0,0], [0,1,0,0,0,0], [0,0,1,0,0,0], [0,0,0,1,0,0], [0,0,0,0,1,0], [0,0,0,0,0,1]))ab = Matrix(([-0.0057], [0.0006], [-0.0037], [0.0009], [0.0025], [0.0042]))az = sp.symbols('az')bz = sp.symbols('bz')fz = sp.symbols('fz')gen = Matrix(([az, fz, 0, 0, 0, 0,bz], [fz,az,fz,0,0,0,bz], [0,fz,az,fz,0,0,bz], [0,0,fz,az,fz,0,bz], [0,0,0,fz,az,fz,bz], [0,0,0,0,fz,az,bz]))answer = solve_linear_system(gen,a,b,c,d,e,f)first_solution = answer[a]df = count_ops(first_solution)print(df,first_solution)disolved = zip(first_solution.simplify().as_numer_denom(),(1,-1))dft = Mul(*[horner(b)**e for b,e in disolved])dff = count_ops(dft)print(dff,dft)_1st_solution = dft.subs({az:ad,fz:fb,bz:ab},simultaneous = True).doit()print(_1st_solution)當(dāng)我運(yùn)行我的代碼時(shí)，它引發(fā)了 sympy.matrices.common.ShapeError
1 回答
慕姐4208626
TA貢獻(xiàn)1852條經(jīng)驗(yàn) 獲得超7個(gè)贊
horner在使用包含實(shí)際上不可交換的可交換符號(hào)的表達(dá)式時(shí)必須小心（在您的情況下，因?yàn)樗鼈儽硎揪仃嚕?。你的dft表情是
(az**2*bz - bz*fz**2)/(az*(az*(az + fz) - 2*fz**2) - fz**3)
但也許應(yīng)該是
(az**2 - fz**2)*(az*(az*(az + fz) - 2*fz**2) - fz**3)**(-1)*bz
如果您將符號(hào)創(chuàng)建為不可交換的，您將收到正確的表達(dá)式（如下所示）。
但是你不能使用horner非交換符號(hào)，所以我只是手動(dòng)重新排列了表達(dá)式；您將必須檢查訂單是否正確。作為手動(dòng)進(jìn)行分解的替代方法，您也可以嘗試使用factor_nc來(lái)幫助您——但它不會(huì)horner像表達(dá)式分解那樣處理：
>>> ax, bz, fz = symbols('az bz fz, commutative=False)
>>> (az**2*bz - fz**2*bz)
az**2*bz - fz**2*bz
>>> factor_nc(_)
(az**2 - fz**2)*bz
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