Elegant way to create dict from local variables in Python

Question

Somethings I create dict from locals() like this:

var1 = 1
var2 = 2
var3 = 3
...
my_dict = dict(
    var1=var1,
    var2=var2,
    var3=var3,
    ...
)

Or:

var1 = 1
var2 = 2
var3 = 3
...
keys = 'var1 var2 var3 ...'.split()
my_dict = {k: v for k, v in locals().items() if k in keys}

Note: keys not only var1 , var2 , var3 , may also be foo , key ,...

Is it possible to create a class like this:

class LocalDict(dict):
    ...

LocalDict('var1 var2 var3') == dict(var1=var1, var2=var2, var3=var3)

Answer 1

It all depends on the way you need to work with the data, really. Your methods above are sound, especially the dict comprehension. So other than the old-fashioned way:

{"key": "value", ... }

I'd say you're doing it right. If you have a specific example, you can edit your question and I will try to work off that.

Hope that helps!

Answer 2

How about this?

var1 = 1
var2 = 2
var3 = 3
my_dict = {k: v for k, v in locals().items() if k.startswith('var')}

or

var1 = 1
var2 = 2
var3 = 3
var4 = "abc"
my_dict = {k: v for k, v in locals().items() if isinstance(v, int) or isinstance(v, str)}

Answer 3

Since all the built-in members of locals() starts and ends with __ , you can filter them out by these prefix and suffix. In addition, you can filter out functions by using callable :

For example:

a = 1
b = 2
output = {k:v for k,v in locals().items() if not (k.startswith("__") and k.endswith("__")) and not callable(v)}
print(output) # output: {'a': 1, 'b': 2}

Answer 4

After trying a lot, the following code is nearly what I want:

class LocalDict(dict):
    def __new__(cls, spaces, attrs):
         return {k: v for k, v in spaces.items() if k in attrs.split()}

def foo():
    a = 1
    b = c = 2
    my_dict = LocalDict(locals(), 'a b c')
    return my_dict

print(foo())  # {'a': 1, 'b': 2, 'c': 2}

And I wonder whether there is some way that I don't need to pass locals to the LocalDict class.