Question

我有一个字典

 tag_list = [{'Key': 'backup', 'Value': 'true'},
 {'Key': 'backup_daily', 'Value': '7'}]

我想将其转换为变量，例如：

tag_backup = 'true'
tag_backup_daily = '7'

有什么简单的方法吗？

如果不需要将其转换为变量，是否还有其他更好的方法直接引用这些键/值？

我最好要做的是按以下方式引用。

tag_list['backup'] = "true"
tag_list['backup_daily'] = "7"

更新＃1

由于@Lev Zakharov的回答，我做到了以下方法。

>>> new_list = {x['Key']: x['Value'] for x in tag_list}
>>> new_list['backup']
'true'
>>> new_list['backup_daily']
'7'

Answer 1

您可以使用globals()：

tag_list = [{'Key': 'backup', 'Value': 'true'},
 {'Key': 'backup_daily', 'Value': '7'}]

for x in tag_list:
    globals()[f"tag_{x['Key']}"] = x['Value']

print(tag_backup)  # true
print(tag_backup_daily)  # 7

Answer 2

使用简单的dict理解：

{x['Key']:x['Value'] for x in tag_list}

结果：

{'backup': 'true', 'backup_daily': '7'}

Answer 3

您只需使用for循环

In [184]: def to_dict(tags):
     ...:     d = {}
     ...:     for t in tags:
     ...:         d[t['Key']] = t['Value']
     ...:     return d
     ...:

In [185]: to_dict(tag_list)
Out[185]: {'backup': 'true', 'backup_daily': '7'}

您还可以使用map和lambda创建一个tuple s的容器，然后使用其构造函数转换为dict < / p>

In [178]: dict(map(lambda d: (d['Key'], d['Value']), tag_list))
Out[178]: {'backup': 'true', 'backup_daily': '7'}

请注意，由于dict构造函数的存在，该过程要慢得多

In [207]: %timeit to_dict(tag_list)
The slowest run took 31.74 times longer than the fastest. This could mean that an intermediate result is being cached.
1000000 loops, best of 3: 431 ns per loop

In [181]: %timeit {d['Key']: d['Value'] for d in tag_list}
The slowest run took 44.42 times longer than the fastest. This could mean that an intermediate result is being cached.
1000000 loops, best of 3: 457 ns per loop

In [182]: %timeit dict(map(lambda d: (d['Key'], d['Value']), tag_list))
The slowest run took 7.72 times longer than the fastest. This could mean that an intermediate result is being cached.
1000000 loops, best of 3: 1.21 µs per loop

将具有dict的数组转换为变量

更新＃1

3 个答案: