假设您将字典存储在set()， 以便查找是O(1)平均情况下（最坏情况O(n)) https://wiki.python.org/moin/TimeComplexity.
您可以从一个单词生成汉明距离为 1 处的所有有效单词：
>>> def neighbours(word):
... for j in range(len(word)):
... for d in string.ascii_lowercase:
... word1 = ''.join(d if i==j else c for i,c in enumerate(word))
... if word1 != word and word1 in words: yield word1
...
>>> {word: list(neighbours(word)) for word in words}
{'bot': ['lot'], 'lol': ['lot'], 'lot': ['bot', 'lol']}
If M是单词的长度，L字母表的长度（即 26），最坏的情况下使用这种方法查找邻近单词的时间复杂度是O(L*M*N).
“简单方法”方法的时间复杂度是O(N^2).
这种方法什么时候比较好？什么时候L*M < N，即如果仅考虑小写字母，则当M < N/26。 （我这里只考虑了最坏的情况）
Note: 英语单词的平均长度为 5.1 个字母 http://arxiv.org/pdf/1208.6109.pdf。因此，如果您的词典大小超过 132 个单词，您应该考虑这种方法。
也许可以实现比这更好的性能。然而，这实现起来非常简单。
实验基准：
“简单方法”算法（A1):
from itertools import zip_longest
def hammingdist(w1,w2): return sum(1 if c1!=c2 else 0 for c1,c2 in zip_longest(w1,w2))
def graph1(words): return {word: [n for n in words if hammingdist(word,n) == 1] for word in words}
该算法（A2):
def graph2(words): return {word: list(neighbours(word)) for word in words}
基准测试代码：
for dict_size in range(100,6000,100):
words = set([''.join(random.choice(string.ascii_lowercase) for x in range(3)) for _ in range(dict_size)])
t1 = Timer(lambda: graph1()).timeit(10)
t2 = Timer(lambda: graph2()).timeit(10)
print('%d,%f,%f' % (dict_size,t1,t2))
Output:
100,0.119276,0.136940
200,0.459325,0.233766
300,0.958735,0.325848
400,1.706914,0.446965
500,2.744136,0.545569
600,3.748029,0.682245
700,5.443656,0.773449
800,6.773326,0.874296
900,8.535195,0.996929
1000,10.445875,1.126241
1100,12.510936,1.179570
...
我用较小的 N 步长运行了另一个基准测试，以更接近地观察它：
10,0.002243,0.026343
20,0.010982,0.070572
30,0.023949,0.073169
40,0.035697,0.090908
50,0.057658,0.114725
60,0.079863,0.135462
70,0.107428,0.159410
80,0.142211,0.176512
90,0.182526,0.210243
100,0.217721,0.218544
110,0.268710,0.256711
120,0.334201,0.268040
130,0.383052,0.291999
140,0.427078,0.312975
150,0.501833,0.338531
160,0.637434,0.355136
170,0.635296,0.369626
180,0.698631,0.400146
190,0.904568,0.444710
200,1.024610,0.486549
210,1.008412,0.459280
220,1.056356,0.501408
...
您会看到权衡非常低（长度= 3 的单词词典为 100）。对于小字典，O(N^2) 算法执行slightly更好，但是随着 N 的增长，O(LMN) 算法很容易击败它。
对于具有较长单词的词典，O(LMN) 算法在 N 中保持线性，只是具有不同的斜率，因此权衡稍微向右移动（长度 = 5 时为 130）。
