how to get the current scope in python

Question

is there any reference to point to the current scope , i look up many articles and couldn't find the answer ,for example i want to print every var's content in current scope

for x in list(locals()):
    print(x)

but only give me this,the var's name

__builtins__
__file__
__package__
__cached__
__name__
__doc__

i dont want code like this

print(__builtins__)
print(__file__)
print(__package__)
print(__cached__)
print(__name__)
print(__doc__)
....

The underscore variables ARE part of your current scope. If you don't want them, just skip them. — gecco, Aug 31 '12 at 09:05
@gecco he wants to print the values of the variables, not to skip them. The for loop in his post prints the name of the variables, not there vaues — miracle173, May 27 '17 at 10:47

Simon Bergot · Answer 1 · 2012-08-31T09:28:25.370

4

To also get the values, you can use:

for symbol, value in locals().items():
    print symbol, value

locals() gives you a dict. Iterating over a dict gives you its keys. To get the list of (key, value) pairs, use the items method.

edited Aug 31 '12 at 09:28

answered Aug 31 '12 at 09:11

Simon Bergot

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if u run your code,will show you these,RuntimeError: dictionary changed size during iteration – Max Aug 31 '12 at 09:20
what version of python are you using? Ihave changed the code to something working with 2.7 – Simon Bergot Aug 31 '12 at 09:29
1

This fails in Python 3.6 with the same RuntimeError. – miguelmorin Dec 10 '18 at 15:12
to avoid the RuntimeError: dictionary changed size during iteration, use this code: d = { k : v for k,v in locals().items() if v} to copy the dictionary and then print the symbol and value like above. – aksinghdce Mar 24 '19 at 19:43

score 4 · Accepted Answer · answered Aug 31 '12 at 09:56

Massive overkill... Wrap the filtering and printing of local namespace in a function.

I don't recommend this. I'm posting it predominantly to show it can be done and to get comments.

import inspect

def print_local_namespace():
    ns = inspect.stack()[1][0].f_locals
    for k, v in ns.iteritems():
        if not k.startswith('__'):
            print '{0} : {1}'.format(k, v)


def test(a, b):
    c = 'Freely'
    print_local_namespace()
    return a + b + c


test('I', 'P')

This is iterating over call stack frames, not definition scopes. You want to got to parent scopes, not parent stack frames. — Mitar, Apr 30 '18 at 18:57

score 2 · Answer 3 · answered Aug 31 '12 at 09:01

2

To print only those variable names in locals() which do not start with '__':

for local_var in list(locals()):
    if not local_var.startswith('__'): print local_var

answered Aug 31 '12 at 09:01

Andy Hayden

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Great hack for getting Python2 "f-strings", `"i wish I had f-strings".format(**locals())` – ignorance Jan 15 '18 at 20:15
This worked for me in Python 3 with the tweak `print(local_var)`. – miguelmorin Dec 10 '18 at 15:15

score 2 · Answer 4 · edited May 23 '17 at 12:15

2

What do you mean by "current scope"? If you mean only the local variables, then locals() is the correct answer. If you mean all the identifiers that you can use[locals + globals + nonlocals] than things get messier. Probably the simpler solution is this one.

If you don't want the __.*__ variables, just filter them out.

edited May 23 '17 at 12:15

Community

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answered Aug 31 '12 at 09:44

Bakuriu

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how to get the current scope in python

4 Answers4