将字典条目转换为变量 - python
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/18090672/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
convert dictionary entries into variables - python
提问by HappyPy
Is there a pythonic way to assign the values of a dictionary to its keys, in order to convert the dictionary entries into variables? I tried this out:
是否有一种 Pythonic 方法可以将字典的值分配给它的键，以便将字典条目转换为变量？我试过了：
>>> d = {'a':1, 'b':2}
>>> for key,val in d.items():
exec('exec(key)=val')
exec(key)=val
^
SyntaxError: invalid syntax
UPDATE: Perhaps I should have been more specific: I am actually certain that the key-value pairs are correct because they were previously defined as variables by me before. I then stored these variables in a dictionary (as key-value pairs) and would like to reuse them in a different function. I could just define them all over again in the new function, but because I may have a dictionary with about 20 entries, I thought there may be a more efficient way of doing this.
更新：也许我应该更具体：我实际上确定键值对是正确的，因为它们之前被我定义为变量。然后我将这些变量存储在字典中（作为键值对），并希望在不同的函数中重用它们。我可以在新函数中重新定义它们，但是因为我可能有一个包含大约 20 个条目的字典，我认为可能有一种更有效的方法来做到这一点。
采纳答案by HappyPy
This was what I was looking for:
这就是我要找的：
>>> d = {'a':1, 'b':2}
>>> for key,val in d.items():
exec(key + '=val')
回答by user2357112 supports Monica
You already have a perfectly good dictionary. Just use that. If you know what the keys are going to be, and you're absolutely sure this is a reasonable idea, you can do something like
你已经有一本非常好的字典了。就用那个。如果您知道密钥将是什么，并且您绝对确定这是一个合理的想法，您可以执行以下操作
a, b = d['a'], d['b']
but most of the time, you should just use the dictionary. (If using the dictionary is awkward, you are probably not organizing your data well; ask for help reorganizing it.)
但大多数时候，你应该只使用字典。（如果使用字典很尴尬，你可能没有很好地组织你的数据；寻求帮助重新组织它。）
回答by tdelaney
Consider the "Bunch" solution in Python: load variables in a dict into namespace. Your variables end up as part of a new object, not locals, but you can treat them as variables instead of dict entries.
考虑Python 中的“Bunch”解决方案：将 dict 中的变量加载到 namespace。您的变量最终会成为新对象的一部分，而不是本地对象，但您可以将它们视为变量而不是 dict 条目。
class Bunch(object):
def __init__(self, adict):
self.__dict__.update(adict)
d = {'a':1, 'b':2}
vars = Bunch(d)
print vars.a, vars.b
回答by restrepo
Use pandas:
使用熊猫：
import pandas as pd
var=pd.Series({'a':1, 'b':2})
#update both keys and variables
var.a=3
print(var.a,var['a'])
回答by Peque
You can do it in a single line with:
您可以在一行中使用：
>>> d = {'a': 1, 'b': 2}
>>> locals().update(d)
>>> a
1
However, you should be careful with how Python may optimize locals/globals access when using this trick.
但是，在使用此技巧时，您应该注意Python 如何优化本地/全局访问。
Note
笔记
I think editing locals()like that is generally a bad idea. If you think globals()is a better alternative, think it twice! :-D
我认为这样的编辑locals()通常是一个坏主意。如果您认为globals()是更好的选择，请三思！:-D
Instead, I would rather always use a namespace.
相反，我宁愿始终使用命名空间。
With Python 3 you can:
使用 Python 3，您可以：
>>> from types import SimpleNamespace
>>> d = {'a': 1, 'b': 2}
>>> n = SimpleNamespace(**d)
>>> n.a
1
If you are stuck with Python 2 or if you need to use some features missing in types.SimpleNamespace, you can also:
如果您坚持使用 Python 2 或者如果您需要使用 中缺少的某些功能types.SimpleNamespace，您还可以：
>>> from argparse import Namespace
>>> d = {'a': 1, 'b': 2}
>>> n = Namespace(**d)
>>> n.a
1
If you are not expecting to modify your data, you may as well consider using collections.namedtuple, also available in Python 3.
如果您不希望修改您的数据，您不妨考虑使用collections.namedtuple，也可在 Python 3 中使用。
回答by suhailvs
you can use operator.itemgetter
你可以使用operator.itemgetter
>>> from operator import itemgetter
>>> d = {'a':1, 'b':2}
>>> a, b = itemgetter('a', 'b')(d)
>>> a
1
>>> b
2
