[英]Transform assignments to dict
我很難鍵入很多作業:
definition = ['basename', 'dirname', 'supports_unicode_filenames']
condition = ['isabs', 'isdir', 'isfile', 'islink', 'ismount']
.
.
.
打算將它們轉換為字典,避免重復輸入:
{'definition': ['basename', 'dirname', 'supports_unicode_filenames'],
'condition': ['isabs', 'isdir', 'isfile', 'islink', 'ismount'] ...}
我試圖將它們封裝在課堂上。
class OsPath:
definition = ['basename', 'dirname', 'supports_unicode_filenames']
condition = ['isabs', 'isdir', 'isfile', 'islink', 'ismount']
在控制台上工作
In [125]: dt = dict(vars(OsPath))
In [127]: {i:dt[i] for i in dt if not i.startswith('__')}
Out[127]:
{'condition': ['isabs', 'isdir', 'isfile', 'islink', 'ismount'],
'definition': ['basename', 'dirname', 'supports_unicode_filenames']}
如何在快捷方式上做到?
你可以做:
definition = ['basename', 'dirname', 'supports_unicode_filenames']
condition = ['isabs', 'isdir', 'isfile', 'islink', 'ismount']
d = {'definition': definition, 'condition': condition }
請注意,應避免使用變量名稱dict
(它是一個內置對象)。
但老實說,如果您要大規模執行此操作,則不是很好的做法。 您應該重新考慮對象的設計。
這可能是您想要的嗎?
definition = ['basename', 'dirname', 'supports_unicode_filenames']
condition = ['isabs', 'isdir', 'isfile', 'islink', 'ismount']
result = {i: globals()[i] for i in ['definition', 'condition']}
使用SimpleNamespace
from types import SimpleNamespace
sn = SimpleNamespace(
definition = ['basename', 'dirname', 'supports_unicode_filenames'],
condition = ['isabs', 'isdir', 'isfile', 'islink', 'ismount'],
)
它輸出:
In [44]: sn
Out[44]: namespace(condition=['isabs', 'isdir', 'isfile', 'islink', 'ismount'], definition=['basename', 'dirname', 'supports_unicode_filenames'])
In [45]: vars(sn)
Out[45]:
{'condition': ['isabs', 'isdir', 'isfile', 'islink', 'ismount'],
'definition': ['basename', 'dirname', 'supports_unicode_filenames']}
In [17]: dict(definition = ['basename', 'dirname', 'supports_unicode_filenames'],condition = ['isabs', 'isdir'
...: , 'isfile', 'islink', 'ismount'])
Out[17]:
{'definition': ['basename', 'dirname', 'supports_unicode_filenames'],
'condition': ['isabs', 'isdir', 'isfile', 'islink', 'ismount']}
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