Sage 中的奇怪列表输出

Question

考虑以下两行代码：

对于t 字典，t = {1: (1, 0, 0, 0, 0, 0, 0, 0, 0), 2: (1, 1, 1, 1, 1, 1, 1, 1, 0)}，当我尝试这样做时：list(t[1]) 将tuple 转换为list，它给了我输出[(0,1)]。但是当我做list(1,0,0,0) 时，它（应该）给了我[1,0,0,0]。这里出了什么问题？

整个成绩单

# given a prime p, return all A_n representations of dimension = p^2
def rankrep(p):
    bound = p*p
    s = SymmetricFunctions(QQ).schur()
    Sym_p = s[p]
    A = lambda i: WeylCharacterRing("A{0}".format(i))
    deg = []
    index = []
    L = []
    for i in xrange(bound):
        deg.append([])
        fw = A(i+1).fundamental_weights()
        temp = A(i+1)
        for j in fw.keys():
            deg[i].append(temp(fw[j]).degree())
            if temp(fw[j]).degree() == bound:
                index.append('A'+str(i+1)+'(fw['+str(j)+'])')
                L.append(fw[j])
    return index, deg, L
def make_vars2(L):
    return dict(enumerate(L, start=1))

[index, deg, L] = rankrep(3)
t = make_vars2(L)
print(t[1])
print t
list(t[1])

给我

(1, 0, 0, 0, 0, 0, 0, 0, 0)
{1: (1, 0, 0, 0, 0, 0, 0, 0, 0), 2: (1, 1, 1, 1, 1, 1, 1, 1, 0)}
[(0, 1)]

Answer 1

尽管您的 t 看起来像是一个带有整数键和整数值元组的字典，但它不是这样的：

sage: t
{1: (1, 0, 0, 0, 0, 0, 0, 0, 0), 2: (1, 1, 1, 1, 1, 1, 1, 1, 0)}
sage: map(type, t)
[int, int]
sage: map(type, t.values())
[sage.combinat.root_system.ambient_space.AmbientSpace_with_category.element_class,
 sage.combinat.root_system.ambient_space.AmbientSpace_with_category.element_class]
sage: parent(t[1])
Ambient space of the Root system of type ['A', 8]

如果你想得到系数向量，你可以使用.to_vector()。例如，我们有

sage: t[1]
(1, 0, 0, 0, 0, 0, 0, 0, 0)
sage: type(t[1])
<class 'sage.combinat.root_system.ambient_space.AmbientSpace_with_category.element_class'>
sage: list(t[1])
[(0, 1)]

但是

sage: t[1].to_vector()
(1, 0, 0, 0, 0, 0, 0, 0, 0)
sage: type(t[1].to_vector())
<type 'sage.modules.vector_rational_dense.Vector_rational_dense'>
sage: list(t[1].to_vector())
[1, 0, 0, 0, 0, 0, 0, 0, 0]