关于django：在python中按字符串名称获取元组列表元素


get tuple list element by string name in python

我有以下tuple列表：它用于django模型中的选项字段。

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ENTITY_TYPE_CHOICES = (
    (0,'choice1'),
    (1,'choice2'),
)

我想按字符串名称获取选项，例如：

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entity_type_index = ENTITY_TYPE_CHOICES['choice1']

号

我得到错误：

tuple indices must be integers, not str

号

相关讨论

你可以通过在迭代的元组。元组对象可以只好accessed模式指数(as，not元素词典)P＞＜／

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ENTITY_TYPE_CHOICES = (
    (0,'choice1'),
    (1,'choice2'),
)

choice = 0

for i in ENTITY_TYPE_CHOICES:
 if i[1] =="choice1":
  choice = i[0]

如果你现在打印：P＞＜／

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print(choice)  # Output: 0

现在你可以使用它与choice。P＞＜／

你可以在字典的建立：P＞＜／

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ENTITY_TYPE_CHOICES = (
(0,'choice1'),
(1,'choice2'),
)

ENTITY_TYPE_CHOICES = dict([i[::-1] for i in ENTITY_TYPE_CHOICES])

entity_type_index = ENTITY_TYPE_CHOICES['choice1']

输出：P＞＜／

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0

如果你不能使用字典：P＞＜／

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new_choice = [i[0] for i in ENTITY_TYPE_CHOICES if i[1] == 'choice1']

print new_choice[0]
相关讨论

tupletype of ENTITY_TYPE_CHOICESis，but as dict你使用它。P＞＜／

method for You can write and find：迭代过关键元组P＞＜／

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get_entity_type_index(k):
    for index, key in ENTITY_TYPE_CHOICES:
         if key == k: return index

but this is not the最佳实践方式。P＞＜／

回答：基本上ajax1234 @P＞＜／

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ENTITY_TYPE_CHOICES = (
   (0,'choice1'),
   (1,'choice2'),
)

ENTITY_TYPE_CHOICES_INDEX = {value: key for key, value in ENTITY_TYPE_CHOICES}

print(ENTITY_TYPE_CHOICES_INDEX['choice2']) # output: 1

你可以选择使用find the index of the ENTITY_TYPE_CHOICES.index('choice1')P＞＜／

然后你可以使用它给the number to接入：恩，by usingP＞＜／

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 ENTITY_TYPE_CHOICES[ENTITY_TYPE_CHOICES.index('choice1')]