根据条件将元组合并为字典

Question

输入：

x = ( (1,"A", 10),(1,"B", 10),(1,"B", 10),(1,"C", 10),(1,"C", 10),(1,"B", 10),(1,"A", 10),(1,"A", 10),(1,"C", 10),(1,"B", 10))

预期产出：

{'A': [(1, 'A', 10), (7, 'A', 70), (8, 'A', 80)], 'B': [(2, 'B', 20), (3, 'B', 30), (6, 'B', 60), (10, 'B', 100)], 'C': [(4, 'C', 40), (5, 'C', 50), (9, 'C', 90)]}

基本上按元组中的元素之一对我的元组进行分组

我试过这个，但不觉得pythonic

def consolidate(values):
    res = {}
    for a in values:
        if a[1] in res:
            p = res[a[1]]
            p.append(a)
            res[a[1]] = p
        else:
            res[a[1]] = [a]
    return res

Answer 1

您可以使用

dict.setdefault

：

out = {}
for a, b, c in x:
    out.setdefault(b, []).append((a, b, c))

print(out)

印花：

{
    "A": [(1, "A", 10), (1, "A", 10), (1, "A", 10)],
    "B": [(1, "B", 10), (1, "B", 10), (1, "B", 10), (1, "B", 10)],
    "C": [(1, "C", 10), (1, "C", 10), (1, "C", 10)],
}