本文整理汇总了Python中numpy.polynomial.hermite_e.herme2poly方法的典型用法代码示例。如果您正苦于以下问题：Python hermite_e.herme2poly方法的具体用法？Python hermite_e.herme2poly怎么用？Python hermite_e.herme2poly使用的例子？那么, 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numpy.polynomial.hermite_e的用法示例。
在下文中一共展示了hermite_e.herme2poly方法的4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_hermefromroots
# 需要导入模块: from numpy.polynomial import hermite_e [as 别名]
# 或者: from numpy.polynomial.hermite_e import herme2poly [as 别名]
def test_hermefromroots(self):
res = herme.hermefromroots([])
assert_almost_equal(trim(res), [1])
for i in range(1, 5):
roots = np.cos(np.linspace(-np.pi, 0, 2*i + 1)[1::2])
pol = herme.hermefromroots(roots)
res = herme.hermeval(roots, pol)
tgt = 0
assert_(len(pol) == i + 1)
assert_almost_equal(herme.herme2poly(pol)[-1], 1)
assert_almost_equal(res, tgt) 示例2: test_herme2poly
# 需要导入模块: from numpy.polynomial import hermite_e [as 别名]
# 或者: from numpy.polynomial.hermite_e import herme2poly [as 别名]
def test_herme2poly(self):
for i in range(10):
assert_almost_equal(herme.herme2poly([0]*i + [1]), Helist[i]) 示例3: test_hermefromroots
# 需要导入模块: from numpy.polynomial import hermite_e [as 别名]
# 或者: from numpy.polynomial.hermite_e import herme2poly [as 别名]
def test_hermefromroots(self) :
res = herme.hermefromroots([])
assert_almost_equal(trim(res), [1])
for i in range(1, 5) :
roots = np.cos(np.linspace(-np.pi, 0, 2*i + 1)[1::2])
pol = herme.hermefromroots(roots)
res = herme.hermeval(roots, pol)
tgt = 0
assert_(len(pol) == i + 1)
assert_almost_equal(herme.herme2poly(pol)[-1], 1)
assert_almost_equal(res, tgt) 示例4: test_herme2poly
# 需要导入模块: from numpy.polynomial import hermite_e [as 别名]
# 或者: from numpy.polynomial.hermite_e import herme2poly [as 别名]
def test_herme2poly(self) :
for i in range(10) :
assert_almost_equal(herme.herme2poly([0]*i + [1]), Helist[i])