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Python: Create a class and display the namespace of the said class

Python Class ( Basic ): Exercise-2 with Solution

Write a Python program to create a class and display the namespace of the said class.

Sample Solution:

Python Code:

class py_solution:
    def sub_sets(self, sset):
        return self.subsetsRecur([], sorted(sset))    
    def subsetsRecur(self, current, sset):
        if sset:
            return self.subsetsRecur(current, sset[1:]) + self.subsetsRecur(current + [sset[0]], sset[1:])
        return [current]
for name in py_solution.__dict__:
    print(name)

Sample Output:

__module__
sub_sets
subsetsRecur
__dict__
__weakref__
__doc__

Flowchart:

Flowchart: Create a class and display the namespace of the said class

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Next: Write a Python program to create an instance of a specified class and display the namespace of the said instance.

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Python: Tips of the Day

Find current directory and file's directory:

To get the full path to the directory a Python file is contained in, write this in that file:

import os 
dir_path = os.path.dirname(os.path.realpath(__file__))

(Note that the incantation above won't work if you've already used os.chdir() to change your current working directory, since the value of the __file__ constant is relative to the current working directory and is not changed by an os.chdir() call.)

To get the current working directory use

import os
cwd = os.getcwd()

Documentation references for the modules, constants and functions used above:

  • The os and os.path modules.
  • The __file__ constant
  • os.path.realpath(path) (returns "the canonical path of the specified filename, eliminating any symbolic links encountered in the path")
  • os.path.dirname(path) (returns "the directory name of pathname path")
  • os.getcwd() (returns "a string representing the current working directory")
  • os.chdir(path) ("change the current working directory to path")

Ref: https://bit.ly/3fy0R6m