我已经尝试了python 3.7.3，你可以使用globals()或vars()

>>> food #Error
>>> milkshake #Error
>>> food="bread"
>>> drink="milkshake"
>>> globals()[food] = "strawberry flavor"
>>> vars()[drink] = "chocolate flavor"
>>> bread
'strawberry flavor'
>>> milkshake
'chocolate flavor'
>>> globals()[drink]
'chocolate flavor'
>>> vars()[food]
'strawberry flavor'

参考: https://www.daniweb.com/programming/software-development/threads/111526/setting-a-string-as-a-variable-name#post548936

Python中的变量

"""
<?php
$a = 'hello';
$e = 'wow'
?>
<?php
$$a = 'world';
?>
<?php
echo "$a ${$a}\n";
echo "$a ${$a[1]}\n";
?>
<?php
echo "$a $hello";
?>
"""

a = 'hello'  #<?php $a = 'hello'; ?>
e = 'wow'   #<?php $e = 'wow'; ?>
vars()[a] = 'world' #<?php $$a = 'world'; ?>
print(a, vars()[a]) #<?php echo "$a ${$a}\n"; ?>
print(a, vars()[vars()['a'][1]]) #<?php echo "$a ${$a[1]}\n"; ?>
print(a, hello) #<?php echo "$a $hello"; ?>

输出:

hello world
hello wow
hello world

使用globals()、locals()或vars()将产生相同的结果

#<?php $a = 'hello'; ?>
#<?php $e = 'wow'; ?>
#<?php $$a = 'world'; ?>
#<?php echo "$a ${$a}\n"; ?>
#<?php echo "$a ${$a[1]}\n"; ?>
#<?php echo "$a $hello"; ?>

print('locals():\n')
a = 'hello'
e = 'wow'
locals()[a] = 'world'
print(a, locals()[a])
print(a, locals()[locals()['a'][1]])
print(a, hello)

print('\n\nglobals():\n')
a = 'hello'
e = 'wow'
globals()[a] = 'world'
print(a, globals()[a])
print(a, globals()[globals()['a'][1]])
print(a, hello)

输出:

locals():

hello world
hello wow
hello world


globals():

hello world
hello wow
hello world

奖励(从字符串创建变量)

# Python 2.7.16 (default, Jul 13 2019, 16:01:51)
# [GCC 8.3.0] on linux2

创建变量并解包元组:

g = globals()
listB = []
for i in range(10):
    g["num%s" % i] = i ** 10
    listB.append("num{0}".format(i))

def printNum():
    print "Printing num0 to num9:"
    for i in range(10):
        print "num%s = " % i, 
        print g["num%s" % i]

printNum()

listA = []
for i in range(10):
    listA.append(i)

listA = tuple(listA)
print listA, '"Tuple to unpack"'

listB = str(str(listB).strip("[]").replace("'", "") + " = listA")

print listB

exec listB

printNum()

输出:

Printing num0 to num9:
num0 =  0
num1 =  1
num2 =  1024
num3 =  59049
num4 =  1048576
num5 =  9765625
num6 =  60466176
num7 =  282475249
num8 =  1073741824
num9 =  3486784401
(0, 1, 2, 3, 4, 5, 6, 7, 8, 9) "Tuple to unpack"
num0, num1, num2, num3, num4, num5, num6, num7, num8, num9 = listA
Printing num0 to num9:
num0 =  0
num1 =  1
num2 =  2
num3 =  3
num4 =  4
num5 =  5
num6 =  6
num7 =  7
num8 =  8
num9 =  9

使用globals()(免责声明:这是一个糟糕的做法，但这是对您的问题最直接的回答，请使用其他数据结构作为接受的答案)。

实际上，您可以动态地将变量分配给全局作用域，例如，如果您想要在全局作用域中访问10个变量i_1, i_2…i_10:

for i in range(10):
    globals()['i_{}'.format(i)] = 'a'

这将给所有这10个变量赋值“a”，当然你也可以动态地改变这个值。所有这些变量现在都可以像其他全局声明的变量一样访问:

>>> i_5
'a'

我已经尝试了python 3.7.3，你可以使用globals()或vars()

>>> food #Error
>>> milkshake #Error
>>> food="bread"
>>> drink="milkshake"
>>> globals()[food] = "strawberry flavor"
>>> vars()[drink] = "chocolate flavor"
>>> bread
'strawberry flavor'
>>> milkshake
'chocolate flavor'
>>> globals()[drink]
'chocolate flavor'
>>> vars()[food]
'strawberry flavor'

参考: https://www.daniweb.com/programming/software-development/threads/111526/setting-a-string-as-a-variable-name#post548936

您可以使用字典来实现这一点。字典是键和值的存储。

>>> dct = {'x': 1, 'y': 2, 'z': 3}
>>> dct
{'y': 2, 'x': 1, 'z': 3}
>>> dct["y"]
2

可以使用可变键名实现可变变量的效果，而不存在安全风险。

>>> x = "spam"
>>> z = {x: "eggs"}
>>> z["spam"]
'eggs'

如果你想做些什么

var1 = 'foo'
var2 = 'bar'
var3 = 'baz'
...

清单可能比字典更合适。list表示对象的有序序列，索引为整数:

lst = ['foo', 'bar', 'baz']
print(lst[1])           # prints bar, because indices start at 0
lst.append('potatoes')  # lst is now ['foo', 'bar', 'baz', 'potatoes']

对于有序序列，列表比使用整数键的字典更方便，因为列表支持索引顺序迭代、切片、追加和其他需要使用字典进行笨拙的键管理的操作。

当你想使用变量变量时，最好使用字典。所以与其写

$foo = "bar"
$$foo = "baz"

你写

mydict = {}
foo = "bar"
mydict[foo] = "baz"

这样就不会意外地覆盖先前存在的变量(这是安全方面的问题)，并且可以使用不同的“名称空间”。