Python变量作为dict的键

for i in ('apple', 'banana', 'carrot'):
    fruitdict[i] = locals()[i]

这个 globals()

>>> apple = 1
>>> banana = 'f'
>>> carrot = 3
>>> globals()
{'carrot': 3, 'apple': 1, '__builtins__': <module '__builtin__' (built-in)>, '__name__': '__main__', '__doc__': None, 'banana': 'f'}

还有一个类似的函数叫做 locals() .

我意识到这可能并不是您想要的,但它可以提供一些关于Python如何提供对变量的访问的见解。

编辑 :听起来你的问题也许可以通过简单地使用字典来更好地解决:

fruitdict = {}
fruitdict['apple'] = 1
fruitdict['banana'] = 'f'
fruitdict['carrot'] = 3

一行是:-

fruitdict = dict(zip(('apple','banana','carrot'), (1,'f', '3'))

这里是一行,无需重新键入任何变量或其值:

fruitdict.update({k:v for k,v in locals().copy().iteritems() if k[:2] != '__' and k != 'fruitdict'})

根据mouad的回答,这里有一种更像蟒蛇的方法来根据前缀选择变量:

# All the vars that I want to get start with fruit_
fruit_apple = 1
fruit_carrot = 'f'
rotten = 666

prefix = 'fruit_'
sourcedict = locals()
fruitdict = { v[len(prefix):] : sourcedict[v]
              for v in sourcedict
              if v.startswith(prefix) }
# fruitdict = {'carrot': 'f', 'apple': 1}

您甚至可以将其放入一个以prefix和sourcedict作为参数的函数中。

如果要绑定变量本身的位置,可以这样做:

>>> apple = 1
>>> banana = 'f'
>>> carrot = 3
>>> fruitdict = {}
>>> fruitdict['apple'] = lambda : apple
>>> fruitdict['banana'] = lambda : banana
>>> fruitdict['carrot'] = lambda : carrot
>>> for k in fruitdict.keys():
...     print k, fruitdict[k]()
... 
carrot 3
apple 1
banana f
>>> apple = 7
>>> for k in fruitdict.keys():
...     print k, fruitdict[k]()
... 
carrot 3
apple 7
banana f

to_dict = lambda **k: k
apple = 1
banana = 'f'
carrot = 3
to_dict(apple=apple, banana=banana, carrot=carrot)
#{'apple': 1, 'banana': 'f', 'carrot': 3}

嗯,这有点,嗯。。。非蟒蛇。。。丑陋的。。。粗俗的。。。

下面是一段代码,假设您要创建一个包含所有本地变量的字典 在执行特定检查点后创建:

checkpoint = [ 'checkpoint' ] + locals().keys()[:]
## Various local assigments here ...
var_keys_since_checkpoint = set(locals().keys()) - set(checkpoint)
new_vars = dict()
for each in var_keys_since_checkpoint:
   new_vars[each] = locals()[each]

注意,我们在捕获 locals().keys() 我还明确地从中分了一杯羹,不过在这种情况下不必这样做,因为要将引用添加到“['checkpoint']列表中,就必须将其展平。但是,如果您正在使用此代码的变体,并试图将 ['checkpoint'] + portion (because that key was already in 局部变量() , for example) ... then, without the [:] slice you could end up with a reference to the locals().keys()`其值将随着您添加变量而更改。

我想不出一种方法来称呼 new_vars.update() 带有要添加/更新的密钥列表。所以 for

为什么你不做相反的事:

fruitdict = { 
      'apple':1,
      'banana':'f',
      'carrot':3,
}

locals().update(fruitdict)

更新:

顺便说一句,你为什么不标记你想得到的变量我不知道 可能是这样的:

# All the vars that i want to get are followed by _fruit
apple_fruit = 1
carrot_fruit = 'f'

for var in locals():
    if var.endswith('fruit'):
       you_dict.update({var:locals()[var])

这一点也不奇怪,比如 将所有局部变量放入 字典。

你想要的是:

apple = 1
banana = 'f'
carrot = 3
fruitdict = {}

# I want to set the key equal to variable name, and value equal to variable value
# is there a more Pythonic way to get {'apple': 1, 'banana': 'f', 'carrot': 3}?

names= 'apple banana carrot'.split() # I'm just being lazy for this post
items = globals()                    # or locals()

for name in names:
    fruitdict[name] = items[name]

老实说,你所做的只是把一本字典里的东西复制到另一本字典里。

(格雷格·休吉尔几乎给出了全部答案,我刚刚完成)

……就像人们建议的那样,你应该首先把这些放进字典里,但我想,出于某种原因,你不能

a = "something"
randround = {}
randround['A'] = "%s" % a

工作。