[英]How can I use a dictionary to iterate through my kivy properties and make changes to them?
[英]How can I iterate through a dictionary whose length changes in python?
我有一個要迭代並更改其值的字典,但是我認為更改長度可能會導致錯誤,有什么好的方法嗎?
例:
d = {'a':'orange','b':'pineapple','c':'dragonfruit'}
rating = {'orange':3, 'pineapple':4, 'dragonfruit':4.5, 'apple':3, 'pear': 3.5}
for key, value in d.items():
if some conditions met:
del d['a'] or d['e']='apple' # (or is not operator, i mean literally one of the operation add or delete item...)
例如:
for key, value in d.items():
if rating[value]<4:
del d[key]
要么
d = {'a':['orange'],'b':['pineapple','peach'],'c':['dragonfruit']}
rating = {'orange':3, 'pineapple':4, 'dragonfruit':4.5, 'apple':3, 'pear': 3.5, 'peach': 5}
for key, value in d.items():
rating_sum = sum([ rating[v] for v in value])
if rating_sum > 8 and len(value)>1:
d['e'] = d[key].pop(-1)
可能不是最好的例子……意味着字典的長度在循環中改變。
迭代字典時,請勿添加或刪除鍵。 另外,不要遮蓋內置dict
。
您可以對所有條件使用字典理解:
d = {k: v for k, v in d.items() if \
not (conditions and \
((k == 'a') or ((k, v) == ('e', 'apple'))))}
我假設您的conditions
是k
和v
的函數,否則您應該在字典理解范圍之外計算並應用此測試。
但這可能會有些混亂,因此將邏輯分離為一個函數沒有任何危害:
def keep_pair(k, v):
if conditions:
if (k == 'a') or ((k, v) == ('e', 'apple')):
return False
return True
d = {k, v for k, v in d.items() if keep_pair(k, v)}
您可以將dict復制到某個臨時dict變量,然后遍歷該臨時dict,但對原始dict進行更改。
temp = dict.copy()
for key, value in temp.items():
if some conditions met:
del dict['a'] or dict['e']='apple'
別。 這不是不可能的,但是在大多數情況下,沒有理由不創建新詞典並重新分配名稱,這樣更易於讀寫。
d = <SOME-DICT>
# filter d with dict comprehension, reassign the name d
d = {k:v for k,v in d.items() if <FILTERING-CONDITION>}
一些演示:
>>> d = {'a':'orange','b':'pineapple','c':'dragonfruit'}
>>> {k:v for k,v in d.items() if v != 'pineapple'}
{'a': 'orange', 'c': 'dragonfruit'}
>>> {k:v for k,v in d.items() if not (k == 'd' and v == 'dragonfruit')}
{'a': 'orange', 'b': 'pineapple', 'c': 'dragonfruit'}
>>> {k:v for k,v in d.items() if not (k == 'c' and v == 'dragonfruit')}
{'a': 'orange', 'b': 'pineapple'}
>>> {k:v for k,v in d.items() if not (k == 'a' or (k == 'e' and v == 'pineapple'))}
{'b': 'pineapple', 'c': 'dragonfruit'}
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