Python solve_ivp vs R lsoda

Question

I am trying to convert an R code to Python. R code uses lsoda function which is a wrapper for FORTRAN DOE solver. The Python counterpart seems to be solve_ivp which is a wrapper for FORTRAN ODEPACK . I use method='LSODA' in Python which should be an equivalent method of what R uses. However, my results are different by up to 1% error. Nothing is random in my code, so I believe I should be able to fully replicate the results.

Any idea?!

This is part of the R code (the code before that is just calculating values for the parameters:

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)

The Python code:

val = [1]*7

class hamberg_ode:
    def __init__(self, p):
        self.p = p

    def f(self, t, val, p=None):
        if p is None:
            p = self.p
        dA1=p["ktr1"]*(1 - ((p["E_MAX"] * p["C_s_gamma"]) /
                        (p["EC_50_gamma"] + p["C_s_gamma"]))) - p["ktr1"]*val[0]
        dA2=p["ktr1"]*val[0] - p["ktr1"]*val[1]
        dA3=p["ktr1"]*val[1] - p["ktr1"]*val[2]
        dA4=p["ktr1"]*val[2] - p["ktr1"]*val[3]
        dA5=p["ktr1"]*val[3] - p["ktr1"]*val[4]
        dA6=p["ktr1"]*val[4] - p["ktr1"]*val[5]
        dA7=p["ktr2"]*(1 - ((p["E_MAX"] * p["C_s_gamma"]) /
                        (p["EC_50_gamma"] + p["C_s_gamma"]))) - p["ktr2"]*val[6]
        print(val[0], dA1)

        return (dA1, dA2, dA3, dA4, dA5, dA6, dA7)


h_function = hamberg_ode(p).f
out = solve_ivp(h_function, (0, maxTime), val, t_eval=times, method='LSODA')

As an example for how numbers diverge, below are few first values of A1 and dA1 for the two code: R

1 -0.2289151

1 -0.2289151

0.9997726 -0.2287975

0.9997727 -0.2287976

0.9995454 -0.22868

0.9995455 -0.2286801

0.9901534 -0.2238221

0.9901523 -0.2238215

0.9809609 -0.2190673

0.9809587 -0.2190662

0.9719626 -0.214413

0.9719604 -0.2144119

0.9493722 -0.2027284

0.9493668 -0.2027255

0.927996 -0.1916717

0.9280039 -0.1916758

0.9078033 -0.1812272

0.9078049 -0.181228

0.8887056 -0.1713491

0.8887071 -0.1713499

Python

1.0 -0.22891514470392998

0.9868338969217406 -0.22210509138758888

0.9872255785819792 -0.22230768534978135

0.9744278526945864 -0.2156881719597506

0.9748085754105683 -0.2158850975024998

0.9069550726140441 -0.18078845812498728

0.906362742770375 -0.18048208061964116

0.8502494750308627 -0.15145797661644517

0.8491489787959607 -0.15088875442597866

0.8022897024620746 -0.1266511977015548

0.8013657199203642 -0.1261732756972218

0.7405758555885625 -0.094730242422152

0.7400188154524862 -0.09444211821383663

0.7145516960742005 -0.08126947025955095

0.7148846597052525 -0.08144169282733643

Answer 1

正如@astoeriko 所指出的，默认相对容差（ rtol ）在 scipy 的solve_ivp是 1e-3 而在 R 的lsoda 1e-6 。