杂记

1. pre-fork的多进程模型

 在web server中广泛使用,即父进程先创建listen socket,然后fork子进程,所有子进程都在accpet同一个socket,这种方法去掉了fork的开销,在早期的linux版本中会存在惊群现象,意思是当有一个连接到来时,所有的进程都会被唤醒,但现在不会

 

2. 如何求一个数是不是平方数:最简单的方法,对1...n 进行binary search

 

3. 求二叉树里两个节点的最低公共祖先: 用一个stack,先根遍历这棵树,当两个节点都已访问后,堆栈里的节点即是

 

4.spell checker:首先对字典里的每一个词进行加工:

def edits1(word):
   s
= [(word[:i], word[i:]) for i in range(len(word) + 1)]
   deletes    
= [a + b[1:] for a, b in s if b]
   transposes
= [a + b[1] + b[0] + b[2:] for a, b in s if len(b)>1]
   replaces  
= [a + c + b[1:] for a, b in s for c in alphabet if b]
   inserts    
= [a + c + b     for a, b in s for c in alphabet]
   
return set(deletes + transposes + replaces + inserts)

这样是计算出这个词的edit1 的距离,然后对输入词查找是否在这个set中,可以继续这样操作得到编辑距离为2的集合
完成的代码如下,参考(http://www.norvig.com/spell-correct.html)
import re, collections

def words(text): return re.findall('[a-z]+', text.lower())

def train(features):
    model
= collections.defaultdict(lambda: 1)
   
for f in features:
        model
[f] += 1
   
return model

NWORDS
= train(words(file('big.txt').read()))

alphabet
= 'abcdefghijklmnopqrstuvwxyz'

def edits1(word):
   s
= [(word[:i], word[i:]) for i in range(len(word) + 1)]
   deletes    
= [a + b[1:] for a, b in s if b]
   transposes
= [a + b[1] + b[0] + b[2:] for a, b in s if len(b)>1]
   replaces  
= [a + c + b[1:] for a, b in s for c in alphabet if b]
   inserts    
= [a + c + b     for a, b in s for c in alphabet]
   
return set(deletes + transposes + replaces + inserts)

def known_edits2(word):
   
return set(e2 for e1 in edits1(word) for e2 in edits1(e1) if e2 in NWORDS)

def known(words): return set(w for w in words if w in NWORDS)

def correct(word):
    candidates
= known([word]) or known(edits1(word)) or known_edits2(word) or [word]
   
return max(candidates, key=NWORDS.get)

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