原文:http://norvig.com/spell-correct.html
2007年的一個星期,兩位朋友(迪恩和比爾)獨立告訴我,他們對谷歌的拼寫糾正感到驚訝。輸入類似[speling]的搜索,Google會立即顯示結果: spelling。我認爲Dean和Bill是高度成熟的工程師和數學家,他們對這個過程的運作方式有很好的直覺。但他們沒有,並且想到它,爲什麼他們應該知道迄今爲止他們的專長?
我認爲他們和其他人可以從解釋中受益。工業強度的糾正器的全部細節非常複雜(你可以在這裏或這裏閱讀一些關於它的內容)。但我認爲,在橫貫大陸的飛機旅行過程中,我可以編寫和解釋一個玩具拼寫校正器,在大約半頁代碼中以每秒至少10個字的處理速度達到80%或90%的準確度。
這裏是(或參見spell.py):
import re
from collections import Counter
def words(text): return re.findall(r'\w+', text.lower())
WORDS = Counter(words(open('big.txt').read()))
def P(word, N=sum(WORDS.values())):
"Probability of `word`."
return WORDS[word] / N
def correction(word):
"Most probable spelling correction for word."
return max(candidates(word), key=P)
def candidates(word):
"Generate possible spelling corrections for word."
return (known([word]) or known(edits1(word)) or known(edits2(word)) or [word])
def known(words):
"The subset of `words` that appear in the dictionary of WORDS."
return set(w for w in words if w in WORDS)
def edits1(word):
"All edits that are one edit away from `word`."
letters = 'abcdefghijklmnopqrstuvwxyz'
splits = [(word[:i], word[i:]) for i in range(len(word) + 1)]
deletes = [L + R[1:] for L, R in splits if R]
transposes = [L + R[1] + R[0] + R[2:] for L, R in splits if len(R)>1]
replaces = [L + c + R[1:] for L, R in splits if R for c in letters]
inserts = [L + c + R for L, R in splits for c in letters]
return set(deletes + transposes + replaces + inserts)
def edits2(word):
"All edits that are two edits away from `word`."
return (e2 for e1 in edits1(word) for e2 in edits1(e1))
注:edits1() 函數寫的太簡潔了。後邊還有很多進一步分析,我不想翻譯了。