比较python中一个非常大的列表中的项目的最快方法

时间:2018-02-24 11:14:24

标签: python list twitter difflib sequencematcher

我在python列表中存储了很长的推文列表(超过50k)。我正处于比较每个项目与其余项目相比的阶段,通过使用difflib来找到推文之间的相似性(删除那些755相似的人,同时只保留一些相似的推文)。我使用itertools.combinations循环遍历所有项目但是花了很长时间(即几天)。这是我的代码:

import pandas as pd
from difflib import SequenceMatcher
import itertools
import re
import time


def similar(a, b):
    return SequenceMatcher(None, a, b).ratio()

df1=pd.read_csv("50k_TweetSheet.csv")
data = df1['text'].tolist()

orginalData = data
outList = []

data[:] = [re.sub(r"http\S+", "", s) for s in data]
data[:] = [re.sub(r"@\S+", "", s) for s in data]
data[:] = [re.sub(r"RT|rt\S+", "", s) for s in data]
data[:] = [s.replace('\r+', ' ') for s in data]
data[:] = [s.replace('\n+', ' ') for s in data]
data[:] = [s.replace(' +', ' ') for s in data]


numOfRows = len(data)

start_time = time.time()
for a, b in itertools.combinations(range(numOfRows), 2):
    if len(data[a].split()) < 4: continue
    if a in outList: continue
    similarity = similar(data[a],data[b])
    if similarity > 0.75:
        if len(data[a].split()) > len(data[b].split()):
            outList.append(b)
            print(data[a])
        else:
            outList.append(a)
            print(data[b])

有更快的方法吗?

0 个答案:

没有答案