[英]Removing nested dict items with dict comprehension
我有兩個決定:
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]}'}
我想首先檢查.keys()
比較( ==
)是否返回True
,在這里它確實是-5.00
),然后我要從all_odds
中刪除所有具有在blocked.values()
列出的鍵的項。
對於上述情況,應導致:
all_odds_final = {'-5.00': '{"16":[2.18,1.61],"18": [2.12,1.79]}'}
我嘗試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]
您知道這很丑陋,而且無法正常工作。
首先,使用ast.literal_eval()
使字符串成為字典。 不要使用eval()
:
>>> import ast
>>> all_odds['-5.00'] = ast.literal_eval(all_odds['-5.00'])
然后,您可以使用字典理解:
>>> 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]}}
但是如果您希望將-5.00
的值作為字符串...
>>> {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]}"}
這是大約2行的操作方法。 在這里我不會使用ast或eval,但是如果您想使用它,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]}}
這似乎可行:
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,[])
如果您不想復制字典,則可以從原始all_odds字典中彈出項目:
for key, blocks in blocked.iteritems():
map(all_odds[key].pop,blocks,[])
map函數中的空列表是如此,因此pop被調用None作為第二個參數。 如果沒有它,則pop僅獲得一個參數,如果不存在該鍵,則將返回錯誤。
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.