python - 在python中将列表转换为矩阵

Question

我正在尝试构建一个拼写校正器使用图。

第1步:我用一些书籍作为语料库和Python networkx包构建了一个直接图节点是单词，我在这个图中为每个节点添加一个属性叫做'DISTANCE'，代表两个单词之间的距离。

 example:
 graph[‘love’][‘you’][‘DISTANCE’] = 37,means, in my corpus，‘love you’ appeals 37 times.
 graph[‘you’][‘love’][‘DISTANCE’] = 39,means, in my corpus，‘you love’ appeals 39 times.

显然，graph[‘love’][‘you’]和graph[‘you’][‘love’]是不同的。我的问题是当我完成一些操作时，我得到一个包含列表的列表。像这样，(长度是可变的):

[

[who,whom,whose],
[are,all,],
[that,than,this]

]

每个子列表都包含可能是正确的单词，我的问题是我想将这个列表转换成这个。

 [
 [who,are,that],
 [who,are,than],
 [who,are,this],
 [who,all,that],
 [who,all,than],
 [who,all,this],
 [whom,are,that],
 [whom,are,than],
 [whom,are,this],
 [whom,all,that],
 [whom,all,than],
 [whom,all,this],
 [whose,are,that],
 [whose,are,than],
 [whose,are,this],
 [whose,all,that],    
 [whose,all,than],
 [whose,all,this],
 ]

所以我可以计算 DISTANCE，确定最佳距离。

我是算法新手，你知道哪种算法可以满足这个要求吗？如果您有什么建议可以帮助我使这个拼写更正器更有效，请告诉我。

谢谢!

Answer 1

您可以使用 itertools.product进行转换:

from itertools import product

d = [
    ['who','whom','whose'],
    ['are','all'],
    ['that','than','this']
]

print list(product(*d))

格式化输出:

[
    ('who', 'are', 'that'), 
    ('who', 'are', 'than'), 
    ('who', 'are', 'this'), 
    ('who', 'all', 'that'), 
    ('who', 'all', 'than'), 
    ('who', 'all', 'this'), 
    ('whom', 'are', 'that'), 
    ('whom', 'are', 'than'), 
    ('whom', 'are', 'this'), 
    ('whom', 'all', 'that'), 
    ('whom', 'all', 'than'), 
    ('whom', 'all', 'this'), 
    ('whose', 'are', 'that'),
    ('whose', 'are', 'than'), 
    ('whose', 'are', 'this'), 
    ('whose', 'all', 'that'), 
    ('whose', 'all', 'than'), 
    ('whose', 'all', 'this')
]