本文整理汇总了Python中sympy.assumptions.Q.antihermitian方法的典型用法代码示例。如果您正苦于以下问题：Python Q.antihermitian方法的具体用法？Python Q.antihermitian怎么用？Python Q.antihermitian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.assumptions.Q的用法示例。
在下文中一共展示了Q.antihermitian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: Pow
# 需要导入模块: from sympy.assumptions import Q [as 别名]
# 或者: from sympy.assumptions.Q import antihermitian [as 别名]
def Pow(expr, assumptions):
"""
Hermitian**Integer -> !Antihermitian
Antihermitian**Even -> !Antihermitian
Antihermitian**Odd -> Antihermitian
"""
if expr.is_number:
return AskImaginaryHandler._number(expr, assumptions)
if ask(Q.hermitian(expr.base), assumptions):
if ask(Q.integer(expr.exp), assumptions):
return False
elif ask(Q.antihermitian(expr.base), assumptions):
if ask(Q.even(expr.exp), assumptions):
return False
elif ask(Q.odd(expr.exp), assumptions):
return True示例2: Mul
# 需要导入模块: from sympy.assumptions import Q [as 别名]
# 或者: from sympy.assumptions.Q import antihermitian [as 别名]
def Mul(expr, assumptions):
"""
As long as there is at most only one noncommutative term:
Hermitian*Hermitian -> !Antihermitian
Hermitian*Antihermitian -> Antihermitian
Antihermitian*Antihermitian -> !Antihermitian
"""
if expr.is_number:
return AskImaginaryHandler._number(expr, assumptions)
nccount = 0
result = False
for arg in expr.args:
if ask(Q.antihermitian(arg), assumptions):
result = result ^ True
elif not ask(Q.hermitian(arg), assumptions):
break
if ask(~Q.commutative(arg), assumptions):
nccount += 1
if nccount > 1:
break
else:
return result