python – 在字符串列表中搜索的高效且最快的方法

2019年8月4日 359次阅读

以下函数返回列表中包含与输入单词完全相同的字符的单词数.单词中字符的顺序并不重要.但是,有一个包含数百万字的列表.执行此搜索的最有效和最快的方法是什么？

例：

words_list = ['yek','lion','eky','ekky','kkey','opt'];

如果我们将单词“key”与列表中的单词匹配,则该函数仅返回“yek”和“eky”,因为它们与“key”共享相同的完全字符,而不管顺序如何.

以下是我写的功能

def find_a4(words_list, word):
    # all possible permutations of the word that we are looking for
    # it's a set of words 
    word_permutations = set([''.join(p) for p in permutations(word)])
    word_size = len(word)
    count = 0

    for word in word_list:
        # in the case of word "key", 
        # we only accept words that have 3 characters 
        # and they are in the word_permutations 
        if len(word) == word_size and word in word_permutations:
            count += 1

    return count

最佳答案 一个字典,其键是单词的排序版本：

word_list = ['yek','lion','eky','ekky','kkey','opt']

from collections import defaultdict
word_index = defaultdict(set)

for word in word_list:
    idx = tuple(sorted(word))
    word_index[idx].add(word)

# word_index = {
#    ('e', 'k', 'y'): {'yek', 'eky'},
#    ('i', 'l', 'n', 'o'): {'lion'},
#    ('e', 'k', 'k', 'y'): {'kkey', 'ekky'},
#    ('o', 'p', 't'): {'opt'}
# }

然后查询你会做：

def find_a4(word_index, word):
    idx = tuple(sorted(word))
    return len(word_index[idx])

或者,如果您需要返回实际单词,请将其更改为返回word_index [idx].

效率：查询运行in average in O(1) time.

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