与Python中提供的引用列表匹配的近似字符串

Question

我想在一个长字符串中查找最频繁出现的近似匹配，条件是该词也来自所提供的列表。在

示例：

# provided list 
>> jobskill = ["scrum", "customer experience improvement", "python"]

# long string 
>> jobtext = ["We are looking for Graduates in our Customer Experience department in Swindon, you will be responsible for improving customer experience and will also be working with the digital team. Send in your application by 31st December 2018", 
"If you are ScrumMaster at the top of your game with ability to communicate inspire and take people with you then there could not be a better time, we are the pioneer in digital relationship banking, and we are currently lacking talent in our Scrum team, if you are passionate about Scrum, apply to our Scrum team, knowledge with python is a plus!"]

# write a function that returns most frequent approximate match
>> mostfrequent(input = jobtext, lookup = jobskill)
# desired_output: {"customer experience improvement, "scrum"}

感谢任何形式的帮助，谢谢！在

Answer 1

模糊使用

from collections import defaultdict
from fuzzywuzzy import fuzz

# provided list
jobskill = ["scrum", "customer experience improvement", "python"]

# long string
jobtext = [
    "We are looking for Graduates in our Customer Experience department in Swindon, you will be responsible for improving customer experience and will also be working with the digital team. Send in your application by 31st December 2018",
    "If you are ScrumMaster at the top of your game with ability to communicate inspire and take people with you then there could not be a better time, we are the pioneer in digital relationship banking, and we are currently lacking talent in our Scrum team, if you are passionate about Scrum, apply to our Scrum team, knowledge with python is a plus!",
]


def k_most_frequent(k, text, queries, threshold=70):
    """Return k most frequent queries using fuzzywuzzy to match."""

    frequency = defaultdict(int)
    text = " ".join(text).split()
    for query in queries:
        for window in range(len(query.split()) + 1):
            frequency[query] += sum(
                [
                    fuzz.ratio(query, " ".join(text[i : i + window])) > threshold
                    for i in range(len(text))
                ]
            )

    return sorted(frequency.keys(), key=frequency.get, reverse=True)[:k]


print(k_most_frequent(2, jobtext, jobskill))

# output: ["customer experience improvement, "scrum"]