Reference a dictionary within itself

Question

Is it possible to get a value from a dictionary entry, within that same dictionary? I'd like to build up a list of directories whilst referencing previously added directories..

common_dirs = {
    'root': '/var/tmp',
    'java_path': os.path.join(dict.get('root'), 'java'),
    'application_path': os.path.join(dict.get('java_path'), 'my_app')
}

Well, for one thing, you're shadowing the built in name here. And two, no, you can't because the name doesn't exist until it's created. — Morgan Thrapp, Jul 07 '16 at 20:47
I understand that it's not created, hence me looking for a solution and pointing that out within the question.. the question is, can it be referenced using some internal command? — AK47, Jul 07 '16 at 20:48
I think you'll want to first create the dictionary by saying `dict = {}`, and then add values to it. — Tanner Swett, Jul 07 '16 at 20:48

Moses Koledoye · Accepted Answer · 2016-07-08T09:51:19.193

8

Why not update the dictionary:

my_dict = {'root': '/var/tmp'}
my_dict.update({'file': os.path.join(my_dict.get('root'), 'file')})

Don't use dict as a name. You may need the real dict builtin later on.

edited Jul 08 '16 at 09:51

answered Jul 07 '16 at 20:48

Moses Koledoye

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Why even use update? Just do `my_dict['file'] = os.path.join(my_dict.get('root'), 'file')` – Morgan Thrapp Jul 07 '16 at 20:49
2

@MorganThrapp It's the same. Just needed my approach to match OP's question by using as many `{}` as possible :) – Moses Koledoye Jul 07 '16 at 20:50

score 2 · Answer 2 · answered Jul 08 '16 at 09:26

No. At the point at which you are filling in the dictionary initializer, the dictionary does not exist, so you cannot reference it!

But in principle there is no reason why a dictionary cannot contain itself as one of the values, as the other answers explain, but you just can't refer to it in its own initializer {}.

Alec · Answer 3 · 2016-07-07T20:55:43.387

0

You could also try using a class instead of a dictionary:

import os

class Paths:
    def __init__(self, root, filename):
        self.root = root
        self.file = os.path.join(self.root, filename)

root = '/var/tmp'
filename = 'file'

p = Paths(root, filename)
print(p.file)
# /var/tmp/file

edited Jul 07 '16 at 20:55

answered Jul 07 '16 at 20:50

Alec

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That's some crazy overkill, and could potentially break if the rest of OP's code is expecting a dictionary. – Morgan Thrapp Jul 07 '16 at 20:51
@MorganThrapp Probably, but that doesn't mean they couldn't be passed in dynamically, assuming this is a minimal verifiable example. – Alec Jul 07 '16 at 20:53
1

@MorganThrapp Fair enough, I edited the response to show that. I do agree that a class is overkill in all likelihood though! – Alec Jul 07 '16 at 20:56

Reference a dictionary within itself

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