本文整理匯總了Python中numpy.polynomial.hermite_e.herme2poly方法的典型用法代碼示例。如果您正苦於以下問題：Python hermite_e.herme2poly方法的具體用法？Python hermite_e.herme2poly怎麽用？Python hermite_e.herme2poly使用的例子？那麽, 這裏精選的方法代碼示例或許可以為您提供幫助。您也可以進一步了解該方法所在類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個代碼示例，這些例子默認根據受歡迎程度排序。您可以為喜歡或者感覺有用的代碼點讚，您的評價將有助於係統推薦出更棒的Python代碼示例。
示例1: test_hermefromroots
# 需要導入模塊: from numpy.polynomial import hermite_e [as 別名]
# 或者: from numpy.polynomial.hermite_e import herme2poly [as 別名]
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
# 需要導入模塊: from numpy.polynomial import hermite_e [as 別名]
# 或者: from numpy.polynomial.hermite_e import herme2poly [as 別名]
def test_herme2poly(self):
for i in range(10):
assert_almost_equal(herme.herme2poly([0]*i + [1]), Helist[i]) 示例3: test_hermefromroots
# 需要導入模塊: from numpy.polynomial import hermite_e [as 別名]
# 或者: from numpy.polynomial.hermite_e import herme2poly [as 別名]
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
# 需要導入模塊: from numpy.polynomial import hermite_e [as 別名]
# 或者: from numpy.polynomial.hermite_e import herme2poly [as 別名]
def test_herme2poly(self) :
for i in range(10) :
assert_almost_equal(herme.herme2poly([0]*i + [1]), Helist[i])