More efficient way to look up dictionary values wh

Question

I have a dictionary whose keys come in sets that share the same prefix, like this:

d = { "key1":"valA", "key123":"valB", "key1XY":"valC",
      "key2":"valD", "key2-22":"valE" }

Given a query string, I need to look up all the values associated with keys that start with that prefix, e.g. for query="key1" I need to get ["valA", "valB", "valC"]

My implementation below works but is too slow for a large number of queries since the dictionary d has about 30,000 keys and most of the keys are more than 20 characters long:

result = [d[s] for s in d.keys() if s.startswith(query)]

Is there a faster/more efficient way to implement this?

Answer 1

The sortedContainers lib has a SortedDict implementation, once you have sorted dict you can bisect_left to find where to start, bisect_right to find the last position then use irange to get the keys in the range:

from sortedcontainers import SortedDict
from operator import itemgetter
from itertools import takewhile


d = { "key1":"valA", "key123":"valB", "key1XY":"valC",
  "key2":"valD", "key2-22":"valE","key3":"foo" }

key = "key2"
d = SortedDict(sorted(d.items(), key=itemgetter(0)))
start = d.bisect_left(key)
print([d[key] for key in takewhile(lambda x: x.startswith("key2"), d.irange(d.iloc[start]]))
['valD', 'valE']

Once you maintain a sorteddict using the sorteddict is a lot more efficint:

In [68]: l = ["key{}".format(randint(1,1000000)) for _ in range(100000)] 
In [69]: l.sort()    
In [70]: d = SortedDict(zip(l,range(100000)))

In [71]: timeit [d[s] for s in d.keys() if s.startswith("key2")]
10 loops, best of 3: 124 ms per loop

In [72]: timeit [d[s] for s in d if s.startswith("key2")]
10 loops, best of 3: 24.6 ms per loop

In [73]: %%timeit
key = "key2"
start = d.bisect_left(key)
l2 =[d[k] for k in takewhile(lambda x: x.startswith("key2"),d.irange(d.iloc[start]))]
   ....: 

100 loops, best of 3: 5.57 ms per loop

Answer 2

You could use a suffix tree:

#!/usr/bin/env python2
from SuffixTree import SubstringDict # $ pip install https://github.com/JDonner/SuffixTree/archive/master.zip

d = { "key1":"valA", "key123":"valB", "key1XY":"valC",
      "key2":"valD", "key2-22":"valE" }

a = '\n' # anchor
prefixes = SubstringDict()
for key, value in d.items(): # populated the tree *once*
    prefixes[a + key] = value # assume there is no '\n' in key

for query in ["key1", "key2"]: # perform queries
    print query, prefixes[a + query]

Output

key1 ['valC', 'valA', 'valB']
key2 ['valE', 'valD']

Answer 3

You can avoid producing the intermediate list generated by dict.keys() (in python 2.x):

result = [d[key] for key in d if key.startswith(query)]

But you most likely want to use a trie instead of a dictionary, so you can find all the values associated with a key with a common prefix (a trie is similar to a tree based on prefixes).

Here you can find some different implementation of tries.

A trie for keys "A", "to", "tea", "ted", "ten", "i", "in", and "inn". (source wikipedia)

---

Let's compare the timings for the different solutions:

# create a dictionary with 30k entries
d = {str(x):str(x) for x in xrange(1, 30001)}
query = '108'

# dict with keys()
%timeit [d[s] for s in d.keys() if s.startswith(query)]

    100 loops, best of 3: 8.87 ms per loop

---

# dict without keys()
%timeit [d[s] for s in d if s.startswith(query)]

    100 loops, best of 3: 7.83 ms per loop

# 11.72% improvement

---

# PyTrie (https://pypi.python.org/pypi/PyTrie/0.2)
import pytrie
pt = pytrie.Trie(d)

%timeit [pt[s] for s in pt.iterkeys(query)]

    1000 loops, best of 3: 320 µs per loop

# 96.36% improvement

---

# datrie (https://pypi.python.org/pypi/datrie/0.7)
import datrie
dt = datrie.Trie('0123456789')
for key, val in d.iteritems():
    dt[unicode(key)] = val

%timeit [dt[s] for s in dt.keys(unicode(query))]

    10000 loops, best of 3: 162 µs per loop

# 98.17% improvement