In a dict of dicts, how do you emulate Perl's auto-vivification behavior?

Question

Both Google and the online docs are not delivering much insight on my query, so I thought I would ask the community here.

In Perl, you can easily setup a hash-of-a-hash-of-a-hash and test the final key like so:

my $hash = {};
$hash{"element1"}{"sub1"}{"subsub1"} = "value1";
if (exists($hash{"element1"}{"sub1"}{"subsub1"})) {
   print "found value\n";
}

What's the 'best-practice' equivalent in Python?

score 16 · Accepted Answer · answered Jun 26 '10 at 03:13

16

The closest equivalent is probably something like the following:

import collections

def hasher():
  return collections.defaultdict(hasher)

hash = hasher()
hash['element1']['sub1']['subsub1'] = 'value1'
if 'subsub1' in hash['element1']['sub1']:
  print 'found value'

answered Jun 26 '10 at 03:13

Alex Martelli

762,786
156
1,160
1,345

3

Note that `if sub1 in hash['phony']` creates a key `'phony'`. This might be problematic, especially since the OP wants to test if keys exist... – unutbu Jun 26 '10 at 03:24
@~unutbu: the OP asks: "test the *final* key", therefore `hash['element1']['sub1'] = dict(subsub1='value1')` will do. – jfs Jun 26 '10 at 03:56
Thanks, this is quite helpful. – jbb Jun 26 '10 at 08:32

score 5 · Answer 2 · answered Jun 26 '10 at 03:20

5

As to whether this is a best practice in Python is up to debate:

hash = {}
hash['element1', 'sub1', 'subsub1'] = 'value'
if ('element1', 'sub1', 'subsub1') in hash:
    print "found value"

But, it certainly works and is very elegant, if it works for you.

The major drawback is that you don't have intermediate access. You can't do:

if ('element1', 'sub1') in hash:
   print "found value"

answered Jun 26 '10 at 03:20

carl

47,134
17
71
81

2

You should explain what's happening. This is a single hash/dict with a 3-element [tuple](http://docs.python.org/library/functions.html#tuple) as a key. Really not the same thing, though you did somewhat indicate that with your second example. – Matthew Flaschen Jun 26 '10 at 03:32
Interesting way to handle the "drawback" (if you consider it to be such) in Alex Martelli's answer, pointed out by ~unutbu. – David Z Jun 26 '10 at 05:35
Indeed, this wasn't what I was looking for - but interesting nonetheless. Thanks for the tip. – jbb Jun 26 '10 at 08:33

score 2 · Answer 3 · answered Aug 16 '13 at 10:39

2

from collections import defaultdict

tree = lambda: defaultdict(tree)

t = tree()

t[1][2][3] = 4
t[1][3][3] = 5
t[1][2]['test'] = 6

from wikipedia Autovivification

answered Aug 16 '13 at 10:39

papalagi

640
2
6
19

1

Besides using `lambda`, how is this different from the accepted answer 3 years ago? – MattH Aug 16 '13 at 10:42

score 0 · Answer 4 · answered Jun 26 '10 at 04:09

I don't know if I'll get any agreement, but this is how I typically declare dictionaries of dictionaries:

someObj = {
  'element1': {
    'sub1': {
      'subsub1': 'value1'
    }
  }
}

As for checking for the presence of an element, I agree with this approach:

try:
  someObj['element1']['sub1']['subsub1']
except KeyError:
  print('no value found')
else:
  print('found value')

In a dict of dicts, how do you emulate Perl's auto-vivification behavior?

4 Answers4

Linked

Related