- 146 名前：デフォルトの名無しさん [2015/11/23(月) 23:01:25.81 ID:cyyDybXh.net]
- >>135
from itertools import groupby
def get_unique_ary(ary):
grouped_ary = groupby(ary, key=lambda row: row['name'])
for k, g in grouped_ary: yield sorted(g, key=lambda g: g['age'])[0]
result = get_unique_ary(ary)
