我使用附加的代码来集成Fitzhugh-Nagumo模型的版本：
@Override
public void onTextChanged(CharSequence charSequence, int i, int i1, int i2) {
str=charSequence.toString();
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adapter.notifyDataSetChanged();
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Toast.makeText(getActivity(),e.getMessage(),Toast.LENGTH_SHORT).show();
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adapter.notifyDataSetChanged();
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Toast.makeText(getActivity(),e.getMessage(),Toast.LENGTH_SHORT).show();
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比较我的from scipy.integrate import odeint
import numpy as np
import time
P = {'epsilon':0.1,
'a1':1.0,
'a2':1.0,
'b':2.0,
'c':0.2}
def fhn_rhs(V,t,P):
u,v = V[0],V[1]
u_t = u - u**3 - v
v_t = P['epsilon']*(u - P['b']*v - P['c'])
return np.stack((u_t,v_t))
def integrate(func,V0,t,args,step='RK4'):
start = time.clock()
P = args[0]
result=[V0]
for i,tmp in enumerate(t[1:]):
result.append(RK4step(func,result[i],tmp,P,(t[i+1]-t[i])))
print "Integration took ",time.clock() - start, " s"
return np.array(result)
def RK4step(rhs,V,t,P,dt):
k_1 = dt*rhs(V,t,P)
k_2 = dt*rhs((V+(1.0/2.0)*k_1),t,P)
k_3 = dt*rhs((V+(1.0/2.0)*k_2),t,P)
k_4 = dt*rhs((V+k_3),t,P)
return V+(1.0/6.0)*k_1+(1.0/3.0)*k_2+(1.0/3.0)*k_3+(1.0/6.0)*k_4
和integrate会产生以下结果：
scipy.integrate.odeint有关如何让我的代码运行得更快的任何建议？我能以某种方式使用In [8]: import cProfile
In [9]: %timeit integrate(fhn_rhs,np.stack((0.1,0.2)),np.linspace(0,100,1000),args=(P,))
10 loops, best of 3: 36.4 ms per loop
In [10]: %timeit odeint(fhn_rhs,np.stack((0.1,0.2)),np.linspace(0,100,1000),args=(P,))
100 loops, best of 3: 3.45 ms per loop
In [11]: cProfile.run('integrate(fhn_rhs,np.stack((0.1,0.2)),np.linspace(0,100,1000),args=(P,))')
45972 function calls in 0.098 seconds
Ordered by: standard name
ncalls tottime percall cumtime percall filename:lineno(function)
1 0.000 0.000 0.098 0.098 <string>:1(<module>)
3996 0.011 0.000 0.078 0.000 fhn.py:20(fhn_rhs)
1 0.002 0.002 0.097 0.097 fhn.py:42(integrate)
999 0.016 0.000 0.094 0.000 fhn.py:52(RK4step)
1 0.000 0.000 0.000 0.000 function_base.py:9(linspace)
7994 0.011 0.000 0.021 0.000 numeric.py:484(asanyarray)
3997 0.029 0.000 0.067 0.000 shape_base.py:282(stack)
11991 0.008 0.000 0.008 0.000 shape_base.py:337(<genexpr>)
3997 0.002 0.000 0.002 0.000 {len}
999 0.001 0.000 0.001 0.000 {method 'append' of 'list' objects}
1 0.000 0.000 0.000 0.000 {method 'astype' of 'numpy.ndarray' objects}
1 0.000 0.000 0.000 0.000 {method 'disable' of '_lsprof.Profiler' objects}
1 0.000 0.000 0.000 0.000 {numpy.core.multiarray.arange}
7995 0.010 0.000 0.010 0.000 {numpy.core.multiarray.array}
3997 0.006 0.000 0.006 0.000 {numpy.core.multiarray.concatenate}
1 0.000 0.000 0.000 0.000 {numpy.core.multiarray.result_type}
来加速吗？
答案 0 :(得分：2)
你不是在比较同样的事情。要查看odeint实际评估ODE函数的点，请放入print t语句（当然不计时）。 odeint通常具有自适应时间步长的方法会产生稀疏的积分样本列表，并从中插入所需的输出。
您必须为RK4方法使用错误估算器，并基于该复制此自适应方案。
当然，使用向量对象解释的python代码永远不会与在执行期间使用简单数组从odeint调用的lsoda编译的FORTRAN代码竞争。
在具有插值的自适应步长方案中使用RK4的示例：
def RK4Step(f, x, y, h, k1):
k2=f(x+0.5*h, y+0.5*h*k1)
k3=f(x+0.5*h, y+0.5*h*k2)
k4=f(x+ h, y+ h*k3)
return (k1+2*(k2+k3)+k4)/6.0
def RK4TwoStep(f, x, y, h, k1):
step1 = RK4Step(f, x , y , 0.5*h, k1 )
x1, y1 = x+0.5*h, y+0.5*h*step1;
step2 = RK4Step(f, x1, y1, 0.5*h, f(x1, y1) )
return (step1+step2)/2
def RK4odeint(fin,times,y0, tol):
# numpy-ify the inputs
f = lambda t,y : np.array(fin(t,y))
y0 = np.array(y0)
# allocate output structure
yout = np.array([y0]*len(times));
# in consequence, yout[0] = y0;
# initialize integrator variables
h = times[1]-times[0];
hmax = abs(times[-1]-times[0]);
# last and current point of the numerical integration
ycurr = ylast = qcurr = qlast = y0;
tcurr = tlast = times[0];
fcurr = flast = f(tcurr, ycurr);
totalerr = 0.0
totalvar = 0.0
for i, t in enumerate(times[1:]):
# remember that t == t[i+1], result goes to yout[i+1]
while (t-tcurr)*h>0:
# advance the integration
k1, k2 = RK4Step(f,tcurr,ycurr,h, fcurr), RK4TwoStep(f,tcurr,ycurr,h, fcurr);
# RK4 is of fourth order, that is,
# k1 = (y(x+h)-y(x))/h + C*h^4
# k2 = (y(x+h)-y(x))/h + C*h^4/16
# Using the double step k2 gives
# C*h^4/16 = (k2-k1)/15 as local error density
# change h to match the global relative error density tol
# use |k2| as scale for the absolute error
# |k1-k2|/15*hfac^4 = tol*|k2|, h <- h*hfac
scale = max(abs(k2))
steperr = max(abs(k1-k2))/2
# compute the ideal step size factor and sanitize the result to prevent ridiculous changes
hfac = ( tol*scale / ( 1e-16+steperr) )**0.25
hfac = min(10, max(0.01, hfac) )
# repeat the step if there is a significant step size correction
if ( abs(h*hfac)<hmax and (0.6 > hfac or hfac > 3 )):
# recompute with new step size
h *= hfac;
k2 = RK4TwoStep(f, tcurr, ycurr, h, fcurr) ;
# update and cycle the integration points
ylast = ycurr; ycurr = ycurr + h*k2;
tlast = tcurr; tcurr += h;
flast = fcurr; fcurr = f(tcurr, ycurr);
# cubic Bezier control points
qlast = ylast + (tcurr-tlast)/3*flast;
qcurr = ycurr - (tcurr-tlast)/3*fcurr;
totalvar += h*scale;
totalerr = (1+h*scale)*totalerr + h*steperr;
reportstr = "internal step to t=%12.8f \t" % tcurr;
#now tlast <= t <= tcurr, can interpolate the value for yout[i+1] using the cubic Bezier formula
s = (t - tlast)/(tcurr - tlast);
yout[i+1] = (1-s)**2*((1-s)*ylast + 3*s*qlast) + s**2*(3*(1-s)*qcurr + s*ycurr)
return np.array(yout)