本文整理汇总了Python中sympy.core.C.acos方法的典型用法代码示例。如果您正苦于以下问题：Python C.acos方法的具体用法？Python C.acos怎么用？Python C.acos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.core.C的用法示例。
在下文中一共展示了C.acos方法的5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: eval
# 需要导入模块: from sympy.core import C [as 别名]
# 或者: from sympy.core.C import acos [as 别名]
def eval(cls, arg):
arg = sympify(arg)
if arg.is_Number:
if arg is S.NaN:
return S.NaN
elif arg is S.Infinity:
return S.Infinity
elif arg is S.NegativeInfinity:
return S.Infinity
elif arg is S.Zero:
return S.Pi*S.ImaginaryUnit / 2
elif arg is S.One:
return S.Zero
elif arg is S.NegativeOne:
return S.Pi*S.ImaginaryUnit
if arg.is_number:
cst_table = {
S.ImaginaryUnit : C.log(S.ImaginaryUnit*(1+sqrt(2))),
-S.ImaginaryUnit : C.log(-S.ImaginaryUnit*(1+sqrt(2))),
S.Half : S.Pi/3,
-S.Half : 2*S.Pi/3,
sqrt(2)/2 : S.Pi/4,
-sqrt(2)/2 : 3*S.Pi/4,
1/sqrt(2) : S.Pi/4,
-1/sqrt(2) : 3*S.Pi/4,
sqrt(3)/2 : S.Pi/6,
-sqrt(3)/2 : 5*S.Pi/6,
(sqrt(3)-1)/sqrt(2**3) : 5*S.Pi/12,
-(sqrt(3)-1)/sqrt(2**3) : 7*S.Pi/12,
sqrt(2+sqrt(2))/2 : S.Pi/8,
-sqrt(2+sqrt(2))/2 : 7*S.Pi/8,
sqrt(2-sqrt(2))/2 : 3*S.Pi/8,
-sqrt(2-sqrt(2))/2 : 5*S.Pi/8,
(1+sqrt(3))/(2*sqrt(2)) : S.Pi/12,
-(1+sqrt(3))/(2*sqrt(2)) : 11*S.Pi/12,
(sqrt(5)+1)/4 : S.Pi/5,
-(sqrt(5)+1)/4 : 4*S.Pi/5
}
if arg in cst_table:
if arg.is_real:
return cst_table[arg]*S.ImaginaryUnit
return cst_table[arg]
if arg is S.ComplexInfinity:
return S.Infinity
i_coeff = arg.as_coefficient(S.ImaginaryUnit)
if i_coeff is not None:
if _coeff_isneg(i_coeff):
return S.ImaginaryUnit * C.acos(i_coeff)
return S.ImaginaryUnit * C.acos(-i_coeff)
else:
if _coeff_isneg(arg):
return -cls(-arg)示例2: angle_between
# 需要导入模块: from sympy.core import C [as 别名]
# 或者: from sympy.core.C import acos [as 别名]
def angle_between(l1, l2):
"""The angle formed between the two linear entities.
Parameters
----------
l1 : LinearEntity
l2 : LinearEntity
Returns
-------
angle : angle in radians
Notes
-----
From the dot product of vectors v1 and v2 it is known that:
dot(v1, v2) = |v1|*|v2|*cos(A)
where A is the angle formed between the two vectors. We can
get the directional vectors of the two lines and readily
find the angle between the two using the above formula.
Examples
--------
>>> from sympy import Point, Line
>>> p1, p2, p3 = Point(0, 0), Point(0, 4), Point(2, 0)
>>> l1, l2 = Line(p1, p2), Line(p1, p3)
>>> l1.angle_between(l2)
pi/2
"""
v1 = l1.p2 - l1.p1
v2 = l2.p2 - l2.p1
return C.acos((v1[0]*v2[0] + v1[1]*v2[1]) / (abs(v1)*abs(v2)))示例3: angle_between
# 需要导入模块: from sympy.core import C [as 别名]
# 或者: from sympy.core.C import acos [as 别名]
def angle_between(l1, l2):
"""
Returns an angle formed between the two linear entities.
Description of Method Used:
===========================
From the dot product of vectors v1 and v2 it is known that:
dot(v1, v2) = |v1|*|v2|*cos(A)
where A is the angle formed between the two vectors. We can
get the directional vectors of the two lines and readily
find the angle between the two using the above formula.
"""
v1 = l1.p2 - l1.p1
v2 = l2.p2 - l2.p1
return C.acos( (v1[0]*v2[0]+v1[1]*v2[1]) / (abs(v1)*abs(v2)) )示例4: angle_between
# 需要导入模块: from sympy.core import C [as 别名]
# 或者: from sympy.core.C import acos [as 别名]
def angle_between(self, o):
"""Angle between the plane and other geometric entity.
Parameters
==========
LinearEntity3D, Plane.
Returns
=======
angle : angle in radians
Notes
=====
This method accepts only 3D entities as it's parameter, but if you want
to calculate the angle between a 2D entity and a plane you should
first convert to a 3D entity by projecting onto a desired plane and
then proceed to calculate the angle.
Examples
========
>>> from sympy import Point3D, Line3D, Plane
>>> a = Plane(Point3D(1, 2, 2), normal_vector=[1, 2, 3])
>>> b = Line3D(Point3D(1, 3, 4), Point3D(2, 2, 2))
>>> a.angle_between(b)
-asin(sqrt(21)/6)
"""
from sympy.geometry.line3d import LinearEntity3D
if isinstance(o, LinearEntity3D):
a = Matrix(self.normal_vector)
b = Matrix(o.direction_ratio)
c = a.dot(b)
d = sqrt(sum([i**2 for i in self.normal_vector]))
e = sqrt(sum([i**2 for i in o.direction_ratio]))
return C.asin(c/(d*e))
if isinstance(o, Plane):
a = Matrix(self.normal_vector)
b = Matrix(o.normal_vector)
c = a.dot(b)
d = sqrt(sum([i**2 for i in self.normal_vector]))
e = sqrt(sum([i**2 for i in o.normal_vector]))
return C.acos(c/(d*e))示例5: angle_between
# 需要导入模块: from sympy.core import C [as 别名]
# 或者: from sympy.core.C import acos [as 别名]
def angle_between(l1, l2):
"""The angle formed between the two linear entities.
Parameters
==========
l1 : LinearEntity
l2 : LinearEntity
Returns
=======
angle : angle in radians
Notes
=====
From the dot product of vectors v1 and v2 it is known that:
``dot(v1, v2) = |v1|*|v2|*cos(A)``
where A is the angle formed between the two vectors. We can
get the directional vectors of the two lines and readily
find the angle between the two using the above formula.
See Also
========
is_perpendicular
Examples
========
>>> from sympy import Point3D, Line3D
>>> p1, p2, p3 = Point3D(0, 0, 0), Point3D(1, 1, 1), Point3D(-1, 2, 0)
>>> l1, l2 = Line3D(p1, p2), Line3D(p2, p3)
>>> l1.angle_between(l2)
acos(-sqrt(2)/3)
"""
v1 = l1.p2 - l1.p1
v2 = l2.p2 - l2.p1
return C.acos(v1.dot(v2)/(abs(v1)*abs(v2)))