我如何从_ast.Dict变成实际的dict？

Question

i have an ast object that returns type '_ast.Dict'

how do i convert that into a regular dict?

my failed attempts -

type(item.value) -> <type '_ast.Dict'>

eval(item.value) -> *** TypeError: eval: argument 1 must be string or code object

ast.parse(item.value,mode="eval") -> *** TypeError: expected a readable buffer object

dict(item.value) -> *** TypeError: '_ast.astlist' object is not callable

len(item.value) -> *** TypeError: object of type '_ast.Dict' has no len()

Answer 1

There are two decent ways to go about getting a regular Python dictionary from an _ast.Dict . Which is better may depend on what the contents of the dictionary are.

If your ast.Dict instance contains regular Python objects (like lists and strings), you can directly convert it to a regular dict with dict(zip(d.keys, d.values)) .

However, if your dictionary contains other ast nodes rather than the corresponding regular objects (eg ast.Str for strings and ast.Num for numbers), you probably want to recursively evaluate everything. The best way I see to do this is to wrap your dictionary in an ast.Expression , then pass it to the built in compile function to turn it into a code object. The code object can then be passed to eval .

In your example, I suspect that item is an ast.Expression already, so you can skip the step of packing the ast.Dict into an Expression . You can get a regular dict with:

dct = eval(compile(item, "<ast expression>", "eval"))

Answer 2

Try this:

my_dict = ast.Dict(keys=('first','second'), values=(1,2))

print type(my_dict)                   #<class '_ast.Dict'>

answer = dict(zip(t.keys,t.values))
print answer                          #{'second': 2, 'first': 1}