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我有一个包含数十万个单词的 Python 列表。单词按照它们在文本中的顺序出现。

我正在寻找与包含该单词的字符串相关联的每个单词的字典，其中包含在它之前和之后出现的 2 个（比如说）单词。

例如列表：“This”“is”“an”“example”“sentence”

应该变成字典：

"This" = "This is an"
"is" = "This is an example"
"an" = "This is an example sentence"
"example" = "is an example sentence"
"sentence" = "an example sentence"

就像是：

WordsInContext = Dict()
ContextSize = 2
wIndex = 0
for w in Words:
    WordsInContext.update(w = ' '.join(Words[wIndex-ContextSize:wIndex+ContextSize]))
    wIndex = wIndex + 1

这可能包含一些语法错误，但即使这些错误得到纠正，我相信这将是一种非常低效的方法。

有人可以建议一个更优化的方法吗？

4

2 回答 2

5

我的建议：

words = ["This", "is", "an", "example", "sentence" ]

dict = {}

// insert 2 items at front/back to avoid
// additional conditions in the for loop
words.insert(0, None)
words.insert(0, None)
words.append(None)
words.append(None)

for i in range(len(words)-4):   
    dict[ words[i+2] ] = [w for w in words[i:i+5] if w]
于 2012-04-20T08:02:11.440 回答
0 
>>> from itertools import count
>>> words = ["This", "is", "an", "example", "sentence" ]
>>> context_size = 2
>>> dict((word,words[max(i-context_size,0):j]) for word,i,j in zip(words,count(0),count(context_size+1)))
{'This': ['This', 'is', 'an'], 'is': ['This', 'is', 'an', 'example'], 'sentence': ['an', 'example', 'sentence'], 'example': ['is', 'an', 'example', 'sentence'], 'an': ['This', 'is', 'an', 'example', 'sentence']}

在 python2.7+或3.x

{word:words[max(i-context_size,0):j] for word,i,j in zip(words,count(0),count(context_size+1))}
于 2012-04-20T10:52:53.593 回答