像计算机科学家一样 思考Python 12.元祖

若要新建只包含一个元素的元祖，需要在后面添加一个逗号
也可以使用tuple函数，不能修改元祖的元素
内置函数divmod接收两个参数，并返回两个值的元祖，即商和余数，可以将结果存为一个元祖
import random
def make_histogram(word):
 hist = {}
 for i in word:
  hist[i] = hist.get(i,0) + 1
 return hist
 
def most_frequent(word):
 s = make_histogram(word)
 t = []
 for x,fqre in s.iteritems():
  t.append((fqre,x))
 
 t.sort(reverse = True)
 res = []
 for fqre,x in t:
  res.append(x)
 return res
 
def read_file(filename):
 return open(filename).read()
 
if __name__ == '__main__':
 t = read_file('words.txt')
 s = most_frequent(t)
 for x in s:
  print x
def signature(s):
 t = list(s):
 t.sort()
 t = ''.join(t)
 return t
 
def all_aragrams(filname):
 d ={}
 for line in open(filname):
  word = line.strip().lower()
  t = signature(word)
  
  if t not in d:
   d[t] = [word]
  else:
   d[t].append(word)
   
 return d
 
 
def print_anagram_set_in_order(d):
 t = []
 for v in d.values():
  if len(v) > 1:
   t.append((len(v),v))
   
 t.sort()
 
 for x in t :
  print x
  
def filter_length(d,n):
 res = {}
 for word,anagrams in d.iteritems():
  if len(word) == n:
   res[word] = anagrams
 return res
 
 
if __name__ == '__main__':
 d = all_aragrams('words.txt')
 print_anagram_set_in_order(d)
 
 eight_letters = filter_length(d, 8)
    print_anagram_sets_in_order(eight_letters)
from anagram_sets import *
def metathesis_pair(d):
 for anagrams in d.itervalues():
  for word1 in anagrams:
   for word2 in anagrams:
    if word1 < word2 and word_distance(word1,word2) == 2:
     print word1,word2
     
     
def word_distance(word1,word2):
 assert len(word1) == len(word2)
 
 count = 0
 for c1,c2 in zip(word1,word2):
  if c1 != c2:
   count += 1
   
 return count
 
 
if __name__ == '__main__':
 d = all_anagrams('words.txt')
 metathesis_pair(d)
def make_word_dict():
    """Reads the words in words.txt and returns a dictionary
    that contains the words as keys."""
    d = dict()
    fin = open('words.txt')
    for line in fin:
        word = line.strip().lower()
        d[word] = word
    # have to add single letter words to the word list;
    # also, the empty string is considered a word.
    for letter in ['a', 'i', '']:
        d[letter] = letter
    return d
 
 
memo = {}
memo[''] = ['']
def is_reducible(word,word_dict):
 if word in memo:
  return memo[word]
  
 res = []
 
 for child in children(word,word_dict):
  t = is_reducible(child,word_dict)
  if t:
   res.append(child)
   
 memo[word] = res
 return res
 
 
def children(word,word_dict):
 res = []
 for i in range(len(word)):
  child = word[:i] +word[i+1:]
  if child in word_dict:
   res.append(child)
 return res
 
 
def all_reducible(word_dict):
 res = []
 for word in word_dict:
  t = is_reducible(word,word_dict)
  if t != []:
   res.append(word)
 return res
 
 
def print_trail(word):
 if len(word) == 0 :
  return
 print word,
 t = is_reducible(word,word_dict)
 print_trail(t[0])
 
 
def print_longest_words(word_dict):
 words = all_reducible(word_dict)
 
 t = []
 for word in words:
  t.append((len(word),word))
 t.sort(reverse = True)
 
 for length,word in t[0:5]:
  print_trail(word)
  print '\n'
  
if __name__ == '__main__':
 word_dict = make_word_dict()
 print_longest_words(word_dict)
 
还是不太会！