如果我有一些选择:
In [1]: from model_utils import Choices In [2]: CHOICES = Choices( ...: (1, 'somekey', 'The long title'), ...: ) In [3]: CHOICES[1] Out[3]: 'The long title'
somekey 钥匙? This answer 不适合我。
somekey
In [10]: {v: k for k, v in dict(CHOICES).iteritems()} Out[10]: {'The long title': 1}
解决方案是:
In [11]: {v: k for k, v in CHOICES._identifier_map.items()} Out[11]: {1: 'somekey'}