比较单词列表和句子列表并打印匹配行的方法

下面是使用线程的示例解决方案。此代码将数据分成相等的块,并将它们用作 compare() 根据我们声明的线程数量。

strings = ("string1", "string2", "string3")
lines = ['some random', 'lines with string3', 'and without it',\
         '1234', 'string2', 'string1',\
         "string1", 'abcd', 'xyz']

def compare(x, thread_idx):
    print('Thread-{} started'.format(thread_idx))
    for line in x:
        if any(s in line for s in strings):
            print("We got one of strings in line: {}".format(line))
    print('Thread-{} finished'.format(thread_idx))

穿线部分:

from threading import Thread

threads = []
threads_amount = 3
chunk_size = len(lines) // threads_amount

for chunk in range(len(lines) // chunk_size):
    threads.append(Thread(target=compare, args=(lines[chunk*chunk_size: (chunk+1)*chunk_size], chunk+1)))
    threads[-1].start()

for i in range(threads_amount):
    threads[i].join()

输出:

Thread-1 started
Thread-2 started
Thread-3 started
We got one of strings in line: string2
We got one of strings in line: string1
We got one of strings in line: string1
We got one of strings in line: lines with string3
Thread-2 finished
Thread-3 finished
Thread-1 finished

一种可能是使用 set 存储电子邮件。这就是账单 if word in emails O(1) . 因此,所做的工作与文件中的单词总数成比例:

emails = {"string1", "string2", "string3"} # this is a set

for line in f:
    if any(word in emails for word in line.split()):
        print("yay!")

你原来的解决方案是 O(nm) (用于 n 单词和 米 电子邮件)与 o(n) 与 设置 .