我试图从字典中的值创建一个字典,name_num其中列表的长度是新的键,name_num字典键和值是值.所以:
name_num = {"Bill": [1,2,3,4], "Bob":[3,4,2], "Mary": [5, 1], "Jim":[6,17,4], "Kim": [21,54,35]}
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我想创建以下字典:
new_dict = {4:{"Bill": [1,2,3,4]}, 3:{"Bob":[3,4,2], "Jim":[6,17,4], "Kim": [21,54,35]}, 2:{"Mary": [5, 1]}}
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我尝试了很多变化,但这段代码让我最接近:
for mykey in name_num:
new_dict[len(name_num[mykey])] = {mykey: name_num[mykey]}
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输出:
new_dict = {4:{"Bill": [1,2,3,4]}, 3:{"Jim":[6,17,4]}, 2:{"Mary": [5, 1]}}
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我知道我需要以某种方式遍历代码,所以我可以将其他值添加到键3.
Kar*_*rin 22
这是一个很好的用例defaultdict:
from collections import defaultdict
name_num = {
'Bill': [1, 2, 3, 4],
'Bob': [3, 4, 2],
'Mary': [5, 1],
'Jim': [6, 17, 4],
'Kim': [21, 54, 35],
}
new_dict = defaultdict(dict)
for name, nums in name_num.items():
new_dict[len(nums)][name] = nums
print(dict(new_dict))
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输出:
{
2: {'Mary': [5, 1]},
3: {'Bob': [3, 4, 2], 'Jim': [6, 17, 4], 'Kim': [21, 54, 35]},
4: {'Bill': [1, 2, 3, 4]}
}
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字典,关联数组或映射(许多名称,基本上是相同的功能)属性是键是唯一的.
如果长度相同,您希望拥有的键(整数)不是唯一的,这就是您的代码不起作用的原因.为现有密钥设置新值意味着替换旧值.
您必须将键值对添加到现有值字典中.
for mykey in name_num:
length = len(name_num[mykey])
if length in new_dict: # key already present in new dictionary
new_dict[length][mykey] = name_num[mykey]
else:
new_dict[length] = {mykey: name_num[mykey]}
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应该做的伎俩