The new Web 2.0-esqe (or dare i say Web 3.0-esqe) nomenclature is dominated by "double-vowels" (Wii, xumii, mobee) or "vowel-combos" (cuil) not often found in normal english. So this approach will not succeed in giving you a really innovative name. Probably you need to perturb the Markov-chain to achieve that or waste a lot of free time digging your cognitive faculty!
Yeah, but you have to remember that those names will affect the perception of similar names. The Wii, for example, being a monster hit is going to affect the way people perceive names.
don't bother with whois; query the root/gtld nameservers for a valid NS record for the domain to see if it's registered. dns is much quicker and nobody's going to shut you down for using it.
Yep. Thanks for the suggestion. After I logged in I realized that Apache was basically in an infinite loop where it kept trying to create new child processes but those where instantly being killed because of a lack of memory. After a hard reset it seems to be working now.
Funny I did a similar thing recently, specifically for domain names. The code to train a Markov chain (without using any libraries outside standard Python) is this:
import re, collections
from itertools import *
def words(text):
return (w.group() for w in re.finditer(r"\w+", text.lower()) if w.group().isalpha())
F=collections.defaultdict(lambda:collections.defaultdict(lambda:1))
M = 5
def add_spaces(word):
return ' ' * M + word.strip() + ' ' * M
N = 0
from operator import add
def train(words):
for word in words:
w = add_spaces(word)
for i in range(len(w)-M):
F[w[i:i+M]][w[i+M]] += 1
global N
N = reduce(add,(n for d in F.itervalues() for n in d.itervalues()))
train(words(file('big.txt').read()))
Then, I tried to measure how "good" a particular word is (experimenting with different measures):
from math import log
def goodness(word):
w = add_spaces(word)
res = 0
for i in range(len(w)-M):
res += F[w[i:i+M]][w[i+M]]
# res += log(F[w[i:i+M]][w[i+M]])
global N
return float(res) / ((len(w)-M)*N)
print 'test score:', reduce(add, (goodness(w) for w in words(file('test.txt').read())))
Then, tried to find the top N words at less than 2 edits distance:
alphabet = 'abcdefghijklmnopqrstuvwxyz'
import types
def edits1(word):
if type(word)==types.StringType:
n = len(word)
return set(chain((word[0:i]+word[i+1:] for i in range(n)), # deletion
(word[0:i]+word[i+1]+word[i]+word[i+2:] for i in range(n-1)), # transposition
(word[0:i]+c+word[i+1:] for i in range(n) for c in alphabet), # alteration
(word[0:i]+c+word[i:] for i in range(n+1) for c in alphabet))) # insertion
else: #Assume word is actualy an iterable of words
return set(reduce(chain,(edits1(e) for e in word)))
def edits2(word):
return edits1(edits1(word))
def topn(n, word):
words = [(x, goodness(x)) for x in edits2(word)]
words.sort(cmp = lambda x,y: -cmp(x[1],y[1]))
return words[:n]
I have borrowed some code from Peter Norvig's spelling checker, and the style is very much influenced by it.
Try closer to something like 800 lines in the library. It comes with demos, a test suite, etc. And I think you're skipping the base class in your naive count.
I'm not really sure how the line count of a library is relevant to anything?
