Python 3 在文本文件中每行不止一次搜索关键字

Question

处理一个在特定文件中搜索关键字的简单脚本。一旦找到关键字，它就会计算它被使用的次数，然后将每次找到关键字的行号记录到一个列表中。然后打印出它被找到的次数和在哪几行。 该程序对关键字进行计数，但在一行中找到一次后，它会转到下一行。它不计算多个关键字。我知道它这样做是因为 if 语句：if keyword in line: Number_Of_Key_Words = Number_Of_Key_Words + 1 found_at.append(num) continue

如何让它考虑文件每行可能存在的多个关键字？

完整代码：

def search():
Number_Of_Key_Words = 0
found_at = [];
keyword = input("Enter a key word to search for: ")
with open("WordList.txt") as file:
    for num, line in enumerate(file, 1):
        if keyword in line:
            Number_Of_Key_Words = Number_Of_Key_Words + 1
            found_at.append(num)
            continue
    print(Number_Of_Key_Words)
    print("Found on lines: ", found_at)
search()

示例 单词表

单词单词单词

搜索“单词” 输出：1
在第 [1] 行找到

想要： 输出：3
在第 [1] 行找到

Answer 1

一种方法是使用re 模块findall，它会在字符串中找到所有出现的地方。

类似这样的东西（我也提出了一些改变）：

import re

keyword = input("...")
found_at = []
counter = 0  # Number_Of_Key_Words is not a good python name
# file is a PY2 built in, so I use 'f' instead
# also you should be explicit for the open mode, 'r' == read mode
with open("...", "r") as f:
    for num, line in enumerate(f, 1):
        # re.findall() will return a list of all keyword occurrence
        # len() will then measure the occurrence effectively
        count = len(re.findall(keyword, line))
        if count > 0:  # ie. keyword in line at least once
            found_at.append(num)
            counter += count

    print(found_at)
    print(counter)

希望这会有所帮助。

Answer 2

或者，您可以使用“str.count (keyword, start, end)”

样本数据“wordtext.txt”：

red, blue, red, green.
blue, yellow, white.
green, orange.
red, blue, green, red, black, yellow, red.

输出：

>>> with open ('wordtext.txt') as f:
    for i, line in enumerate (f):
        found = line.count ('red')
        if found:
            print ('Line: %d  Red: %d' % (i, found))


Line: 0  Red: 2
Line: 3  Red: 3
>>>