本文整理汇总了Python中sympy.Mul.flatten方法的典型用法代码示例。如果您正苦于以下问题：Python Mul.flatten方法的具体用法？Python Mul.flatten怎么用？Python Mul.flatten使用的例子？那么恭喜您, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类sympy.Mul的用法示例。
在下文中一共展示了Mul.flatten方法的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: test_zoo
# 需要导入模块: from sympy import Mul [as 别名]
# 或者: from sympy.Mul import flatten [as 别名]
def test_zoo():
b = Symbol('b', bounded=True)
nz = Symbol('nz', nonzero=True)
p = Symbol('p', positive=True)
n = Symbol('n', negative=True)
im = Symbol('i', imaginary=True)
c = Symbol('c', complex=True)
pb = Symbol('pb', positive=True, bounded=True)
nb = Symbol('nb', negative=True, bounded=True)
imb = Symbol('ib', imaginary=True, bounded=True)
for i in [I, S.Infinity, S.NegativeInfinity, S.Zero, S.One, S.Pi, S.Half, S(3), log(3),
b, nz, p, n, im, pb, nb, imb, c]:
if i.is_bounded and (i.is_real or i.is_imaginary):
assert i + zoo is zoo
assert i - zoo is zoo
assert zoo + i is zoo
assert zoo - i is zoo
elif i.is_bounded is not False:
assert (i + zoo).is_Add
assert (i - zoo).is_Add
assert (zoo + i).is_Add
assert (zoo - i).is_Add
else:
assert (i + zoo) is S.NaN
assert (i - zoo) is S.NaN
assert (zoo + i) is S.NaN
assert (zoo - i) is S.NaN
if i.is_nonzero and (i.is_real or i.is_imaginary):
assert i*zoo is zoo
assert zoo*i is zoo
elif i.is_zero:
assert i*zoo is S.NaN
assert zoo*i is S.NaN
else:
assert (i*zoo).is_Mul
assert (zoo*i).is_Mul
if (1/i).is_nonzero and (i.is_real or i.is_imaginary):
assert zoo/i is zoo
elif (1/i).is_zero:
assert zoo/i is S.NaN
elif i.is_zero:
assert zoo/i is zoo
else:
assert (zoo/i).is_Mul
assert (I*oo).is_Mul # allow directed infinity
assert zoo + zoo is S.NaN
assert zoo * zoo is zoo
assert zoo - zoo is S.NaN
assert zoo/zoo is S.NaN
assert zoo**zoo is S.NaN
assert zoo**0 is S.One
assert zoo**2 is zoo
assert 1/zoo is S.Zero
assert Mul.flatten([S(-1), oo, S(0)]) == ([S.NaN], [], None)示例2: test_issue_6611a
# 需要导入模块: from sympy import Mul [as 别名]
# 或者: from sympy.Mul import flatten [as 别名]
def test_issue_6611a():
assert Mul.flatten([3**Rational(1, 3),
Pow(-Rational(1, 9), Rational(2, 3), evaluate=False)]) == \
([Rational(1, 3), (-1)**Rational(2, 3)], [], None)