如何在python中更改长度的字典中进行迭代?

时间:2018-12-06 16:45:36

标签: python dictionary iteration

我有一个要迭代并更改其值的字典,但是我认为更改长度可能会导致错误,有什么好的方法吗?

示例:

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]

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...意味着字典的长度在循环中改变。

3 个答案:

答案 0 :(得分:5)

在迭代字典时不要添加或删除键。另外,请勿遮挡内置dict

您可以在所有条件下使用字典理解:

d = {k: v for k, v in d.items() if \
     not (conditions and \
     ((k == 'a') or ((k, v) == ('e', 'apple'))))}

我假设您的conditionskv的函数,否则您应该在字典理解之外计算和应用此测试。

但是这可能会有点混乱,因此将逻辑分离为一个函数没有什么害处:

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)}

答案 1 :(得分:0)

您可以将dict复制到某个临时dict变量,然后遍历该临时dict,但对原始dict进行更改。

temp = dict.copy()

for key, value in temp.items():
    if some conditions met:
        del dict['a'] or dict['e']='apple'

答案 2 :(得分:0)

不要。这不是不可能的,但是在大多数情况下,没有理由不创建新字典并重新分配名称,这样更易​​于读写。

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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