CODEWITHSUNDEEP
pythonshelve
Nescio
Nescio

Reputation: 513

Is shelve really slow and taking a lot of memory or am I doing something wrong?

I'm trying to write a program that uses a shelve database of sorted letters as key, and a list of words that can be created from them as values. e.g: 

db['mnoo'] = ['moon', 'mono']

So I wrote a function that takes a filename and loads it into a shelve. The first part, that turns the file into a dictionary with the same layout as the shelve works fine, but the shelve part takes really long.

I'm trying it with a dictionary of ~100k entries, each value being a list. It seems to take 15-20 seconds for each 1000 enteries, and each entry seems to take ~1kb of space. Is this nromal?
the code:

def save_to_db(filename, shelve_in='anagram_db'):
    dct = anagrams_from_list(process_file(filename))

    with shelve.open(shelve_in, 'c') as db:
        for key, wordlist in dct.items():
            if not key in db:
                db[key] = wordlist
            else:
                db[key].extend(wordlist)

edit: just a quick clarification: each list in dict is about 1-3 words long, shouldn't be too large

Upvotes: 3

Views: 1382

Answers (1)

Charles Duffy
Charles Duffy

Reputation: 295500

First -- yes, shelve's default pickle backend is slow and inefficient, and your best choice is to use something different.

Second -- you're making it worse by editing entries once they're there, rather than getting them into their final state in-memory before serializing them only once.

dct = anagrams_from_list(process_file(filename))
for key, wordlist in dct.items():
  content = {}
  for key, wordlist in dct.iteritems():
    if not key in content:
      content[key] = wordlist
    else:
      content[key].extend(wordlist)

for k, v in content.iteritems():
  db[k] = v

If you want an efficient database, I'd look elsewhere. tokyocabinet, kyotocabinet, SQLite, BDB; the options are numerous.

Upvotes: 5

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