我正在嘗試將 R 代碼轉(zhuǎn)換為 Python。R 代碼使用lsoda函數(shù)，它是 FORTRANDOE求解器的包裝器。Python 對應(yīng)物似乎solve_ivp是 FORTRAN 的包裝器ODEPACK。我method='LSODA'在 Python 中使用它應(yīng)該是 R 使用的等效方法。但是，我的結(jié)果有高達(dá) 1% 的誤差。我的代碼中沒有任何東西是隨機(jī)的，所以我相信我應(yīng)該能夠完全復(fù)制結(jié)果。任何的想法？！這是 R 代碼的一部分（之前的代碼只是計(jì)算參數(shù)的值：val = c("A1" = 1, "A2" = 1, "A3" = 1, "A4" = 1, "A5" = 1, "A6" = 1, "A7" = 1)hamberg_ode <- function(t,val,p) {     dA1 = p["ktr1"]*(1 - ((p["E_MAX"] * p["C_s_gamma"])/(p["EC_50_gamma"] + p["C_s_gamma"]))) - p["ktr1"]*val["A1"]    dA2 = p["ktr1"]*val["A1"] - p["ktr1"]*val["A2"]    dA3 = p["ktr1"]*val["A2"] - p["ktr1"]*val["A3"]    dA4 = p["ktr1"]*val["A3"] - p["ktr1"]*val["A4"]    dA5 = p["ktr1"]*val["A4"] - p["ktr1"]*val["A5"]    dA6 = p["ktr1"]*val["A5"] - p["ktr1"]*val["A6"]    dA7 = p["ktr2"]*(1 - ((p["E_MAX"] * p["C_s_gamma"])/(p["EC_50_gamma"] + p["C_s_gamma"]))) - p["ktr2"]*val["A7"]    cat(val["A1"], dA1, '\n')    list(c(dA1, dA2, dA3, dA4, dA5, dA6, dA7))}out = lsoda(val, times, hamberg_ode, p)Python代碼：val = [1]*7class hamberg_ode:    def __init__(self, p):        self.p = p        return (dA1, dA2, dA3, dA4, dA5, dA6, dA7)h_function = hamberg_ode(p).fout = solve_ivp(h_function, (0, maxTime), val, t_eval=times, method='LSODA')作為數(shù)字如何發(fā)散的示例，以下是兩個代碼的 A1 和 dA1 的幾個第一個值： R1 -0.22891511 -0.22891510.9997726 -0.22879750.9997727 -0.22879760.9995454 -0.228680.9995455 -0.22868010.9901534 -0.22382210.9901523 -0.22382150.9809609 -0.21906730.9809587 -0.21906620.9719626 -0.2144130.9719604 -0.21441190.9493722 -0.20272840.9493668 -0.20272550.927996 -0.19167170.9280039 -0.19167580.9078033 -0.18122720.9078049 -0.1812280.8887056 -0.17134910.8887071 -0.1713499