Question

我遇到了以下问题，首先我将简要解释发生了什么：

f(x)
g(x, y) = f(x) - y

从那里我们期待

g(x, f(x)) = f(x) - f(x) = 0

lambdified g(x,y)返回非常接近零而非零的东西。这是一个重现问题的代码。仅当我在log

中进行了足够数量的f(x)评估时，它才会到达

要点：https://gist.github.com/marekyggdrasil/39a24213ebaba6293464d116821cc334

源：

from sympy import Symbol, pprint, log, lambdify

# setting symbols
g1 = Symbol("gamma1")
g2 = Symbol("gamma2")
g3 = Symbol("gamma3")
g4 = Symbol("gamma4")
rt = Symbol("rt")

# setting expressions
criteria  = (g1 * log(g1, 2.0))/2.0
criteria += (g2 * log(g2, 2.0))/2.0
criteria += (g3 * log(g3, 2.0))/2.0
criteria += (g4 * log(g4, 2.0))/2.0
rooteq = criteria - rt

print "\ncriteria function: "
pprint(criteria)

print "\ncriteria function - rt: "
pprint(rooteq)

# lambdifying expressions to callable functions
tsymbols = [g1, g2, g3, g4, rt]
lambfun_criteria = lambdify(tsymbols, criteria)
lambfun_rooteq = lambdify(tsymbols, rooteq)

# example point x
x = [0.25006462253641376, 2.2501938662000542, 2.2501938662000542, 2.2501938662000542, 0.0]

# evaluating of criteria on x
rootval = lambfun_criteria(*x)

# setting rt to this evaluation
x[4] = rootval

print "\nactual evaluation of rooteq: " + str(lambfun_rooteq(*x))
print "\nexpected evaluation of rooteq: " + str(- x[4] + lambfun_criteria(*x))

输出

$ python lambdifytest.py 

criteria function: 
0.721347520444482⋅γ₁⋅log(γ₁) + 0.721347520444482⋅γ₂⋅log(γ₂) + 0.721347520444482⋅γ₃⋅log(γ₃) + 0.721347520444482⋅γ₄⋅log(γ₄)

criteria function - rt: 
0.721347520444482⋅γ₁⋅log(γ₁) + 0.721347520444482⋅γ₂⋅log(γ₂) + 0.721347520444482⋅γ₃⋅log(γ₃) + 0.721347520444482⋅γ₄⋅log(γ₄) - rt

actual evaluation of rooteq: 4.4408920985e-16

expected evaluation of rooteq: 0.0

Answer 1

我发现当SymPy表达式被lambdified时，可以指定自定义数学模块。就我而言，mpmath帮助了。

导入mpmath并将精度设置为您认为足够的

from mpmath import mp
mp.dps = 15

告诉SymPy你希望你的lambda函数使用mpmath来处理浮点运算

lambfun_criteria = lambdify(tsymbols, criteria, "mpmath")
lambfun_rooteq = lambdify(tsymbols, rooteq, "mpmath")

然后我终于得到了输出

actual evaluation of rooteq: 0.0
expected evaluation of rooteq: 0.0

问题的原因如下，创建的lambda函数中使用的默认浮点算法与比较结果的脚本中使用的算法不同。

如果有人需要复制粘贴

，我用完整的解决方案文件更新了要点

https://gist.github.com/marekyggdrasil/39a24213ebaba6293464d116821cc334#file-lambdify_test_solution-py

编辑：忘记提供文档参考

http://docs.sympy.org/dev/modules/utilities/lambdify.html

lambdified sympy表达式返回错误的结果

1 个答案: