CODEWITHSUNDEEP
pythondictionary
user2678074
user2678074

Reputation: 825

Dictionary of directories

I have some paths like (simplified): /rootFolder/subfolder_1/subsubfolder_1/

paths = {
    'rootFolder':'rootFolder/'
    'subFolder_1':paths[rootFolder]+'subFolder/'
    'subsubFolder_1:paths[subfolder]+'subsubFolder/'
}

Basically, I want a static dictionary of paths, but I also want to keep dependencies so I can make just one change in the future. I am unsure if that's doable, and if, then how? Do you have some better ideas than mine? Thanks in advance for help.

Upvotes: 2

Views: 105

Answers (3)

Nizam Mohamed
Nizam Mohamed

Reputation: 9240

You can use functions as values for the keys. Because functions are not called before the dict is constructed, there'll be no NameError. 

In [320]: paths = { 'root': lambda: '/', 'usr': lambda: os.path.join(paths['root'](), 'usr')}

In [321]: paths['root']()
Out[321]: '/'

In [322]: paths['usr']()
Out[322]: '/usr'

In [323]: paths['root'] = lambda: '//'

In [324]: paths['usr']()
Out[324]: '//usr'

If you don't want do paths['root'](), simply subclass dict. 

from types import FunctionType
import os

class DynDict(dict):
    def __getitem__(self, key):
        val = dict.__getitem__(self, key)
        return val() if type(val) is FunctionType else val

paths = DynDict(root='/', tmp=lambda: os.path.join(paths['root'], 'tmp'))
print(paths['root'], paths['tmp'])
paths['root'] = '/var/'
print(paths['root'], paths['tmp'])

Output;

/ /tmp
/var/ /var/tmp

Upvotes: 2

frank
frank

Reputation: 103

paths = {
  'rootFolder'    :  "'rootFolder/'",
  'subFolder_1'   :  "eval(paths['rootFolder']) + 'subFolder/'",
  'subsubFolder_1':  "eval(paths['subFolder_1'])+ 'subsubFolder/'"
}

access via: 

eval(paths["subsubFolder_1"])
# 'rootFolder/subFolder/subsubFolder/'

you can encapsulate the eval-usage by a function getPath("keyInDict") which do the user-unfriendly part of accessing the dict

Upvotes: 0

arewm
arewm

Reputation: 649

You can just create a function to return the dictionary that you want.

def path_dict(root):
    return {'rootFolder':root,
            'subFolder_1':root+'subFolder/'
            'subsubFolder_1:root+'subfolder/subsubFolder/'
           }

Then,

>>> path_dict('rootFolder/')
{'subsubfolder_1': 'rootFolder/subfolder/subsubfolder/', 'rootFolder': 'rootFolder/', 'subfolder_1': 'rootFolder/subfolder/'}

If you want to use keys in the dictionary to build off of, you can do it by initializing the dictionary in several steps.

paths = {'rootFolder':'rootFolder/'}
paths['subFolder_1'] = paths['rootFolder']+'subFolder/'

It is not a one-line instantiation of the directory, but you can combine the two to have a one-line instantiation with less code reuse.

def path_dict(root):
    paths = {'rootFolder':root}
    paths['subFolder_1'] = paths['rootFolder']+'subFolder/'
    paths['subsubFolder_1'] = paths['subFolder_1']+'subsubFolder/'

    return paths

Upvotes: 0

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