如何序列化sympy lambdified函数? [英] How to serialize sympy lambdified function?

问题描述

标题说明了一切.有什么方法可以序列化sympy.lambdify生成的函数吗?:

The title says it all. Is there any way to serialize a function generated by sympy.lambdify?:

import sympy as sym
import pickle
import dill
a, b = sym.symbols("a, b")
expr = sym.sin(a) + sym.cos(b)
lambdified_expr = sym.lambdify((a, b), expr, modules="numpy")
pickle.dumps(lambdified_expr) # won't work
dill.dumps(lambdified_expr) # won't work either

...之所以要这样做，是因为我的代码生成了很多lambdified函数，但是我发现每次都花很长时间.

... The reason I want to do this is because my code generates so many lambdified functions but I found it takes too long every time.

推荐答案

您实际上可以使用dill对其进行腌制. dill的最新版本(例如，在github上)具有设置"，可用于在dump上构造泡菜的方式的变体.是的，dill的默认设置在该对象上失败，但是如果您使用递归跟踪全局引用的设置(即recurse = True)，则不会失败.此设置类似于cloudpickle默认为您提供的设置.

You actually can use dill to pickle it. The most recent versions of dill (e.g. on github) has "settings" that allow variants of how the pickle is constructed on dump. Yes, the default settings for dill fail on this object, but not if you use the setting that recursively traces global references (i.e. recurse = True). This setting is similar to what cloudpickle gives you by default.

>>> import sympy as sym
>>> import pickle
>>> import dill
>>> a, b = symbols("a, b")
>>> a, b = sym.symbols("a, b")
>>> expr = sym.sin(a) + sym.cos(b)
>>> lambdified_expr = sym.lambdify((a, b), expr, modules="numpy")
>>> 
>>> dill.settings
{'recurse': False, 'byref': False, 'protocol': 2, 'fmode': 0}
>>> dill.settings['recurse'] = True
>>> dill.dumps(lambdified_expr)
'\x80\x02cdill.dill\n_create_function\nq\x00(cdill.dill\n_unmarshal\nq\x01U\x83c\x02\x00\x00\x00\x02\x00\x00\x00\x03\x00\x00\x00C \x00\x00s\x14\x00\x00\x00t\x00\x00|\x00\x00\x83\x01\x00t\x01\x00|\x01\x00\x83\x01\x00\x17S(\x01\x00\x00\x00N(\x02\x00\x00\x00t\x03\x00\x00\x00sint\x03\x00\x00\x00cos(\x02\x00\x00\x00t\x01\x00\x00\x00at\x01\x00\x00\x00b(\x00\x00\x00\x00(\x00\x00\x00\x00s\x08\x00\x00\x00<string>t\x08\x00\x00\x00<lambda>\x01\x00\x00\x00s\x00\x00\x00\x00q\x02\x85q\x03Rq\x04}q\x05(U\x03cosq\x06cnumpy.core.umath\ncos\nq\x07U\x03sinq\x08cnumpy.core.umath\nsin\nq\tuU\x08<lambda>q\nNN}q\x0btq\x0cRq\r.'

P.S.我是dill作者，所以我知道.

P.S. I'm the dill author, so I'd know.

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