使用 Python 在字符串中的字符串列表中查找项目的索引

Question

I'm looking for a fast approach to find all the indexes in string which match with items (one or multiple words). Actually I do not need index in list I need index in string.

I have a list of words and an string like these:

words = ['must', 'shall', 'may','should','forbidden','car',...]
string= 'you should wash the car every day'

desired output:
[1,4]# should=1, car=4

The length of list some times can be more than hundreds of items and string more that tens of thousands.

I'm looking for a so fast approach because it is called a thousand times in each iteration.

I know how to implement it with loops and check all the items one-by-one but it's so slow!

Answer 1

One solution is make words set instead of list and then do simple list comprehension:

words = {'must', 'shall', 'may','should','forbidden','car'}
string= 'you should wash the car every day'

out = [i for i, w in enumerate(string.split()) if w in words]

print(out)

Prints:

[1, 4]

Answer 2

You need the Aho Corasick algorithm to this.

Given a set of strings and a text, it finds occurrences of all strings from the set in the given text in O(len+ans) , where len is the length of the text and ans is the size of the answer.

It uses an automaton and can be modified to suit your needs.

Answer 3

You can use dictionaries time complexity for look up dictionary is O(1)

string = 'you should wash the car every day'

wordToIndex = {word: index for index, word in enumerate(string.split())}

words = ['must', 'shall', 'may','should','forbidden','car']

result = [wordToIndex[word] for word in words if word in wordToIndex]

# [1,4]

Answer 4

Use list comprehension,

print([string.split().index(i) for i in string.split() if i in words]) 
#[1,4]