Question

我有两个词：

blocked = {'-5.00': ['121', '381']}
all_odds = {'-5.00': '{"121":[1.85,1.85],"381":[2.18,1.73],"16":[2.18,1.61],"18":\
            [2.12,1.79]}'}

我想首先检查.keys()比较（==）是否返回True，这里是否（-5.00）然后我要删除all_odds中的所有项目已blocked.values()中列出密钥的{1}}。

对于上述情况，应该导致：

all_odds_final = {'-5.00': '{"16":[2.18,1.61],"18": [2.12,1.79]}'}

我尝试了for loop：

if blocked.keys() == all_odds.keys():
    for value in blocked.values():
        for v in value:
            for val in all_odds.values():
                val = eval(val)
                if val.has_key(v):
                    del val[v]

你知道它非常丑陋加上它还没有正常工作。

Answer 1

首先，将字符串设为ast.literal_eval()的字典。 Don't use eval()：

>>> import ast
>>> all_odds['-5.00'] = ast.literal_eval(all_odds['-5.00'])

然后你可以使用词典理解：

>>> if blocked.keys() == all_odds.keys():
...     print {blocked.keys()[0] : {k:v for k, v in all_odds.values()[0].iteritems() if k not in blocked.values()[0]}}
... 
{'-5.00': {'18': [2.12, 1.79], '16': [2.18, 1.61]}}

但是如果你想要-5.00的值作为字符串......

>>> {blocked.keys()[0]:str({k: v for k, v in all_odds.values()[0].iteritems() if k not in blocked.values()[0]})}
{'-5.00': "{'18': [2.12, 1.79], '16': [2.18, 1.61]}"}

Answer 2

以下是约2行中你可以做同样的事情。我不会在这里使用ast或eval，但如果你想使用它，你可以添加它。

>>> blocked = {'-5.00': ['121', '381']}
>>> all_odds = {'-5.00': {'121':[1.85,1.85],'381':[2.18,1.73],'16':[2.18,1.61],'18':\
...      [2.12,1.79]}}
>>> bkeys = [k for k in all_odds.keys() if k in blocked.keys()]
>>> all_odds_final = {pk: {k:v for k,v in all_odds.get(pk).items() if k not in blocked.get(pk)} for pk in bkeys}
>>> all_odds_final
{'-5.00': {'18': [2.12, 1.79], '16': [2.18, 1.61]}}

Answer 3

这似乎有效：

blocked = {'-5.00': ['121', '381']}
all_odds = {'-5.00': {"121":[1.85,1.85],"381":[2.18,1.73],"16":[2.18,1.61],"18":\
        [2.12,1.79]}}
all_odds_final = dict(all_odds)
for key, blocks in blocked.iteritems():
    map(all_odds_final[key].pop,blocks,[])

如果您不想复制字典，只需从原始all_odds字典中弹出项目：

for key, blocks in blocked.iteritems():
    map(all_odds[key].pop,blocks,[])

map函数中的空列表是pop，因为它是第二个参数，所以使用None调用。如果没有它，pop只会获得一个参数，并且如果该键不存在则会返回错误。

使用dict理解删除嵌套的dict项

3 个答案: