Question

如果我的帖子使用了错误的线程，我事先表示歉意。我似乎对尝试使用的这个标记“系统”感到困惑，或者由于我的conditions可以有多个键/值而使逻辑迷失了字典的使用。

目前，只有在我坚持使用and或or方法的情况下，我试图实现的结果才是可能的，但如果可能，我会尝试同时实现两者

#conditions = {'motions': ['walk']}
#conditions = {'motions': ['run']}
conditions = {'motions': ['run', 'walk']}

# 1. if only 'run', item02 and item04 should be shown
# 2. if only 'walk' + 'run', item04 should be shown


my_items = [
     {'item01' : {'motions': ['walk']}},
     {'item02' : {'motions': ['run']}},
     {'item03' : {'motions': ['crawl']}},
     {'item04' : {'motions': ['run', 'walk']}},
]

result = []

for m in my_items:
    for mk, mv in m.items():
        res1 = all(any(x in mv.get(kc, []) for x in vc) for kc, vc in conditions.items())

        all_conditions_met = len(conditions) == len(mv) #False
        for cond in conditions:
            if cond in mv:
                res2 = all_conditions_met and (set(mv[cond]) == set(conditions[cond]))
                all_conditions_met = bool(res1 and res2)

                # # if I use bool(res1 and res2)
                # returns me only item02, incorrect for #1
                # returns me only item04, correct for #2

                # if I use bool(res1 or res2)
                # returns me item02 and item04, incorrect for #1
                # returns me item01, item02 and item04, incorrect for #2

            else:
                all_conditions_met = False
                break
        if all_conditions_met:
            result.append(mk)

有人可以分享一些见解吗？

Answer 1

如果要查找完全匹配的内容，则应检查all，而不是any。实际上，您正在尝试检查一个集合是否是另一个集合的子集-您可以从Python的built-in set type中受益：

names = []
for row in my_items:
  for name, subconditions in row.items():
    full_match = all(
      set(conditions[key]) <= set(subconditions.get(key, []))
      for key in conditions.keys()
    )

    if full_match:
      names.append(name)

给定键/值对，从字典中导出项目

1 个答案: