Is there any way to print all the local variables without printing them expilictly ?
def some_function(a,b):
name='mike'
city='new york'
#here print all the local variables inside this function?
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Does this answer your question? Viewing all defined variablesCountour-Integral– Countour-Integral2020-12-18 21:51:24 +00:00Commented Dec 18, 2020 at 21:51
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@Countour-Integral No, my question specific to the function variables. The below answer satisfies my question.Ravi– Ravi2020-12-18 21:54:53 +00:00Commented Dec 18, 2020 at 21:54
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The very first and at least half of the answers there, mention how to get local variables despite the more generilized title. It does indeed answer your question.Countour-Integral– Countour-Integral2020-12-18 21:57:56 +00:00Commented Dec 18, 2020 at 21:57
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2 Answers
That would be the locals() built-in function
Python 3.9.0 (tags/v3.9.0:9cf6752, Oct 5 2020, 15:34:40) [MSC v.1927 64 bit (AMD64)] on win32
Type "help", "copyright", "credits" or "license" for more information.
>>> locals()
{'__name__': '__main__', '__doc__': None, '__package__': None, '__loader__': <class '_frozen_importlib.BuiltinImporter'>, '__spec__': None, '__annotations__': {}, '__builtins__': <module 'builtins' (built-in)>}
>>> x = 5
>>> locals()
{'__name__': '__main__', '__doc__': None, '__package__': None, '__loader__': <class '_frozen_importlib.BuiltinImporter'>, '__spec__': None, '__annotations__': {}, '__builtins__': <module 'builtins' (built-in)>, 'x': 5}
You can filter out the builtins with a list comprehension:
>>> [_ for _ in locals() if not (_.startswith('__') and _.endswith('__'))]
['x']
Comments
If you just want variable names you can use dir() as well:
def some_function(a, b):
name='Mike'
city='New York'
# print all the local variables inside this function?
print(dir())
some_function('a', 'b')
