如何计算字符串中的某些单词（不仅仅是一个单词），然后如果单词数量不同则输出不同的代码？

Question

抱歉，这个问题很长，我不知道如何在不将其标记为重复的情况下具体措辞。无论如何，我正在尝试计算用户输入的一组关键字，这些关键字被写入文本文件。这是代码：

import os
def keyword_searcher():
user_query = input("Enter your problem:\n")
with open("user_query.txt", "a+") as query:
    query.write(user_query.lower())
    query.seek(0)
    for line in query:
        for word in line.split():
            if word in ("backlight", "display", "cracked", "touchscreen"):
                with open("Solutions1.txt", "r") as f:
                    solution_one  = f.read()
                    print ("Keyword for screen problem found:\n")
                    print (solution_one)
            elif word in ("battery", "charger", "usb", "charge"):
                with open("Solutions2.txt", "r") as f_two:
                    solution_two = f_two.read()
                    print ("Keyword for battery problem found:\n")
                    print(solution_two)
            elif word in ("virus", "hacked", "infected", "antivirus"):
                with open("Solutions3.txt", "r") as f_three:
                    solution_three = f_three.read()
                    print ("Keyword for virus problem found:\n")
                    print (solution_three)
            else:
                pass
keyword_searcher()
os.remove("user_query.txt")

我的问题是，当用户输入多个关键字时，它会输出那么多次代码，而我只需要一次。我该如何解决这个问题？

Answer 1

for word in line.split():
    if word in ("backlight", "display", "cracked", "touchscreen"):
       ...

意味着您将在循环的每次迭代中检查一个词，因此如果有多个词匹配，您将得到重复项。要仅在列表中存在至少一个单词时执行此操作，请使用 sets:

for line in query:
    words = set(line.split())
    if words & {"backlight", "display", "cracked", "touchscreen"}:
        ...

或者，您可以执行类似

的操作

for line in query:
    words = line.split()
    if any(word in ("backlight", "display", "cracked", "touchscreen") for word in words):
        ...

这可能更容易阅读，但会更慢。

Answer 2

你真的可以精简这个程序。不需要将用户输入写入文件并在最后删除它，只需将其存储为变量即可。您应该使用 if 语句检查每个语句，以确保它检查每个语句。此外， else: pass 是多余的。希望这对你有所帮助，祝你的程序好运：

list_one = ["backlight", "display", "cracked", "touchscreen"]
list_two = ["battery", "charger", "usb", "charge"]
list_three = ["virus", "hacked", "infected", "antivirus"]

user_query = input("Enter your problem:\n").split()

if any(word in list_one for word in user_query):
    with open("Solutions1.txt") as f:
        solution_one  = f.read()
        print("Keyword for screen problem found:\n")
        print(solution_one)

if any(word in list_two for word in user_query):
    with open("Solutions2.txt") as f_two:
        solution_two = f_two.read()
        print("Keyword for battery problem found:\n")
        print(solution_two)

if any(word in list_three for word in user_query):
    with open("Solutions3.txt") as f_three:
        solution_three = f_three.read()
        print("Keyword for virus problem found:\n")
        print(solution_three)

如何计算字符串中的某些单词（不仅仅是一个单词），然后如果单词数量不同则输出不同的代码？

How do I count up certain words (not just one word) in a string and then output different code if there are different amounts of words?

python

counter

if-statement