Question

我想将Sympy-Lambdify func保存/序列化到一个文件中，然后再用另一个python程序加载/加载它。

案例1 ：效果很好

import dill
import sympy as sp
from sympy.utilities.lambdify import lambdify

dill.settings['recurse'] = True


a,b = sp.symbols('a, b')
expr = a**2 + 2*a + 1 + b

func = lambdify((a,b), expr)

myfunc = dill.loads(dill.dumps(func))

print(myfunc)
print(type(myfunc))
print(myfunc(2,3))

输出：

<function <lambda> at 0x00000210AA0D6598>
<class 'function'>
12

案例2 ：返回错误

import dill
import sympy as sp
from sympy.utilities.lambdify import lambdify

dill.settings['recurse'] = True


a,b = sp.symbols('a, b')
expr = a**2 + 2*a + 1 + b

func = lambdify((a,b), expr)


with open('expr', 'wb') as outf:
    dill.dump(expr, outf)

with open('expr','rb') as inf:
    myfunc= dill.load(inf)

print(myfunc)
print(type(myfunc))
print(myfunc(2,3))

输出：

a**2 + 2*a + b + 1
<class 'sympy.core.add.Add'>
Traceback (most recent call last):
  File "test.py", line 25, in <module>
    print(myfunc(2,3))
TypeError: 'Add' object is not callable

有人可以帮我修理吗？

提前谢谢大家！

Answer 1

而不是expr将func放入dill.dump()：

with open('expr', 'wb') as outf:
    dill.dump(func, outf)

<强>输出

<function <lambda> at 0x7fd3015c4510>
<class 'function'>
12

python 3：dump / load sympy lambdify-epression时出错

1 个答案: