本文整理汇总了Python中sympy.concrete.expr_with_limits.AddWithLimits类的典型用法代码示例。如果您正苦于以下问题：Python AddWithLimits类的具体用法？Python AddWithLimits怎么用？Python AddWithLimits使用的例子？那么恭喜您, 这里精选的类代码示例或许可以为您提供帮助。
在下文中一共展示了AddWithLimits类的2个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于系统推荐出更棒的Python代码示例。
示例1: __new__
def __new__(cls, function, *symbols, **assumptions):
obj = AddWithLimits.__new__(cls, function, *symbols, **assumptions)
if not hasattr(obj, 'limits'):
return obj
if any(len(l) != 3 or None in l for l in obj.limits):
raise ValueError('Sum requires values for lower and upper bounds.')
return obj示例2: __new__
def __new__(cls, function, *symbols, **assumptions):
"""Create an unevaluated integral.
Arguments are an integrand followed by one or more limits.
If no limits are given and there is only one free symbol in the
expression, that symbol will be used, otherwise an error will be
raised.
>>> from sympy import Integral
>>> from sympy.abc import x, y
>>> Integral(x)
Integral(x, x)
>>> Integral(y)
Integral(y, y)
When limits are provided, they are interpreted as follows (using
``x`` as though it were the variable of integration):
(x,) or x - indefinite integral
(x, a) - "evaluate at" integral is an abstract antiderivative
(x, a, b) - definite integral
The ``as_dummy`` method can be used to see which symbols cannot be
targeted by subs: those with a preppended underscore cannot be
changed with ``subs``. (Also, the integration variables themselves --
the first element of a limit -- can never be changed by subs.)
>>> i = Integral(x, x)
>>> at = Integral(x, (x, x))
>>> i.as_dummy()
Integral(x, x)
>>> at.as_dummy()
Integral(_x, (_x, x))
"""
#This will help other classes define their own definitions
#of behaviour with Integral.
if hasattr(function, '_eval_Integral'):
return function._eval_Integral(*symbols, **assumptions)
obj = AddWithLimits.__new__(cls, function, *symbols, **assumptions)
return obj