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Python: recursively create dictionary from paths

Asked 2010-01-06T23:08:32.883
12

I have several hundred thousand endpoint URLs that I want to generate stats for. For example I have:

/a/b/c
/a/b/d
/a/c/d
/b/c/d
/b/d/e
/a/b/c
/b/c/d

I want to create a dictionary that looks like this

{
    'a': {
        'b': {
            'c': {
                '_count': 2
            },
            'd': {
                '_count': 1
            }
        },
        'c': {
            'd': {
                '_count': 1
            }
        }
    },
    'b': {
        'c': {
            'd': {
                '_count': 2
            }
        },
        'd': {
            'e': {
                '_count': 1
            }
        }
    }
}

Any clever ways to do this?

EDIT

I should mention that the paths are not always 3 parts. There might be /a/b/c/d/e/f/g/h... etc, etc.

python recursion
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1 Answer

13

If the paths all look like in your example, this would work:

counts = {}
for p in paths:
   parts = p.split('/')
   branch = counts
   for part in parts[1:-1]:
      branch = branch.setdefault(part, {})
   branch[parts[-1]] = 1 + branch.get(parts[-1], 0)

This uses dictionary methods like setdefault() and get() to avoid having to write a lot of if-statements.

Note that this will not work if a path that has subdirectories can also appear on it's own. Then it's not clear if the corresponding part of counts should contain a number or another dictionary. In this case it would probably be best to store both a count and a dict for each node, using a tuple or a custom class.

The basic algorithm stays the same:

class Stats(object):
   def __init__(self):
      self.count = 0
      self.subdirs = {}

counts = Stats()
for p in paths:
   parts = p.split('/')
   branch = counts
   for part in parts[1:]:
      branch = branch.subdirs.setdefault(part, Stats())
   branch.count += 1

With some pretty printing you get:

def printstats(stats, indent=''):
   print indent + str(stats.count) + ' times'
   for (d, s) in stats.subdirs.items():
       print indent + d + ':'
       printstats(s, indent + '  ')

>>> printstats(counts)
0 times
a:
  0 times
  c:
    0 times
    d:
      1 times
  b:
    0 times
    c:
      2 times
    d:
      1 times
...
answered 2010-01-06T23:18:03.597

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