<p>我调整了给定链接中的代码,并使用<a href="https://docs.python.org/2/library/profile.html" rel="nofollow noreferrer">cProfile</a>来比较这两种技术:</p>
<pre><code>import numpy as np
import cProfile as cP
def theory(t):
return (t**2 + 4.)**2 / 16.
def f(x, y):
return x * np.sqrt(y)
def RK4(f):
return lambda t, y, dt: (
lambda dy1: (
lambda dy2: (
lambda dy3: (
lambda dy4: (dy1 + 2*dy2 + 2*dy3 + dy4)/6
)( dt * f( t + dt , y + dy3 ) )
)( dt * f( t + dt/2, y + dy2/2 ) )
)( dt * f( t + dt/2, y + dy1/2 ) )
)( dt * f( t , y ) )
def test_RK4(dy=f, x0=0., y0=1., x1=10, n=10):
vx = np.empty(n+1)
vy = np.empty(n+1)
dy = RK4(f=dy)
dx = (x1 - x0) / float(n)
vx[0] = x = x0
vy[0] = y = y0
i = 1
while i <= n:
vx[i], vy[i] = x + dx, y + dy(x, y, dx)
x, y = vx[i], vy[i]
i += 1
return vx, vy
def rk4_step(dy, x, y, dx):
k1 = dx * dy(x, y)
k2 = dx * dy(x + 0.5 * dx, y + 0.5 * k1)
k3 = dx * dy(x + 0.5 * dx, y + 0.5 * k2)
k4 = dx * dy(x + dx, y + k3)
return x + dx, y + (k1 + k2 + k2 + k3 + k3 + k4) / 6.
def test_rk4(dy=f, x0=0., y0=1., x1=10, n=10):
vx = np.empty(n+1)
vy = np.empty(n+1)
dx = (x1 - x0) / float(n)
vx[0] = x = x0
vy[0] = y = y0
i = 1
while i <= n:
vx[i], vy[i] = rk4_step(dy=dy, x=x, y=y, dx=dx)
x, y = vx[i], vy[i]
i += 1
return vx, vy
cP.run("test_RK4(n=10000)")
cP.run("test_rk4(n=10000)")
</code></pre>
<p>得到了:</p>
^{pr2}$
<p>所以我想说,“<code>lambda</code>”实现中的<a href="https://stackoverflow.com/questions/31779335/why-is-there-overhead-when-calling-functions">function call</a><a href="https://stackoverflow.com/questions/2860234/what-is-overhead">overhead</a>使它变慢了。在</p>
<p>不过,请注意,我似乎在某种程度上失去了一些精确性,因为尽管结果彼此一致,但结果比示例中的结果更差:</p>
<pre><code>>>> vx, vy = test_rk4()
>>> vy
array([ 1. , 1.56110667, 3.99324757, ..., 288.78174798,
451.27952013, 675.64427775])
>>> vx, vy = test_RK4()
>>> vy
array([ 1. , 1.56110667, 3.99324757, ..., 288.78174798,
451.27952013, 675.64427775])
</code></pre>