Question

我有一个字典列表，其值为一些元组：

dic1 = {
    'persuaded ': [[('teacher', '6'), ('group', '5'), ('man', '5'), ('girl', '5')]], 
    'removed ': [[('apple', '5'), ('makeup', '4'), ('trash', '4'), ('stain', '4')]]
}

我需要做的是将嵌套元组转换为字典，以便我可以使用密钥将其与其他类似列表进行比较。理想的结果将是：

dic2 = {
    'persuaded ': [{'teacher': '6'}, {'group': '5'}, {'man': '5'}, {'girl': '5'}], 
    'removed ': [{'apple': '5'}, {'makeup': '4'}, {'trash': '4'}, {'stain': '4'}]
}

我试过了：

dic2 = {}
for x, y in dic1_zipped:
    d.setdefault(x, []).append(y)

和

from collections import defaultdict
dic2= defaultdict( list )
for n,v in dic1_zipped:
    fq[n].append(v)

但是在字典里都没有深入。我真的很感激有关如何解决这个问题的任何建议！谢谢！

Answer 1

以下嵌套字典和列表推导将为您完成：

dic2 = {key: [{k: v} for sublist in value for k, v in sublist] for key, value in dic1.items()}

演示：

>>> {key: [{k: v} for sublist in value for k, v in sublist] for key, value in dic1.items()}
{'removed ': [{'apple': '5'}, {'makeup': '4'}, {'trash': '4'}, {'stain': '4'}], 'persuaded ': [{'teacher': '6'}, {'group': '5'}, {'man': '5'}, {'girl': '5'}]}
>>> from pprint import pprint
>>> pprint(_)
{'persuaded ': [{'teacher': '6'}, {'group': '5'}, {'man': '5'}, {'girl': '5'}],
 'removed ': [{'apple': '5'}, {'makeup': '4'}, {'trash': '4'}, {'stain': '4'}]}

我真的很惊讶你不希望每个值都有一个字典：

dic2 = {key: {k: v for sublist in value for k, v in sublist} for key, value in dic1.items()}

产生：

>>> {key: {k: v for sublist in value for k, v in sublist} for key, value in dic1.items()}
{'removed ': {'stain': '4', 'trash': '4', 'apple': '5', 'makeup': '4'}, 'persuaded ': {'group': '5', 'teacher': '6', 'man': '5', 'girl': '5'}}
>>> pprint(_)
{'persuaded ': {'girl': '5', 'group': '5', 'man': '5', 'teacher': '6'},
 'removed ': {'apple': '5', 'makeup': '4', 'stain': '4', 'trash': '4'}}

Answer 2

如果你摆脱了不必要的列表包装，即

dic1 = {'persuaded ': [('teacher', '6'), ('group', '5'), ('man', '5'), ('girl', '5')], 'removed ': [('apple', '5'), ('makeup', '4'), ('trash', '4'), ('stain', '4')]}

你可以很简单地做到这一点：

dic2 = dict((k1, dict(v1)) for k1, v1 in dic1.items())

这给出了

{'persuaded ': {'girl': '5', 'man': '5', 'group': '5', 'teacher': '6'}, 'removed ': {'apple': '5', 'makeup': '4', 'stain': '4', 'trash': '4'}}

这似乎比单项词典列表更有用。

（如果你无法摆脱额外的列表级别，只需使用dict(v1[0])。）

Answer 3

dict((k, v) for k,v in c1.items())

将嵌套元组转换为嵌套字典

3 个答案: