Python 为列表中的每个单词搜索webtext

我必须编写代码，从用户选择的网站收集文本，并将搜索这三个选定的单词文本。然后，它必须输出每个单词及其在网站上出现的次数

My attempt at writing this program leaves me with an output that tells me that 0 of the words listed are present on the webtext even when I know they do appear. Does anyone have an idea as to how to make it work?
import requests

def main():

    Asentence="This,is,a,sentence,of,some,kind!"
    print(type(Asentence))
    print(Asentence)
    ListOfWords=Asentence.split(",")
    print(type(ListOfWords))
    print(ListOfWords)
    print(ListOfWords[0])
    print(ListOfWords[-1])
    print(ListOfWords[3])

    SomeOtherList=["Sally", "Fred"]
    print(type(SomeOtherList))
    print(SomeOtherList)
    print(SomeOtherList[0])

    for thing in SomeOtherList:
        print(thing)

    n= eval(input("How many websites would you like to enter? :"))
    while n > 0:
        Word()
        n=n-1  



#------------------------------------------   
def Word():   
    answer=input("please enter the websites to examine in the http format ")

    response=requests.get(answer)
    txt = response.text
    print(txt)
    mywords=Firstpart(list)
    num=FindAWord(txt,mywords)
    print("There are", num, "words called",mywords)

#----------------------------------------    
def FindAWord(TheWebText,word):

    print(TheWebText)
    print(type(TheWebText))
    MyList=TheWebText.split(sep=" ")
    print(MyList[0:100])
    count=0
    for item in MyList:
        if(item==word in Firstpart(list)):
            print(item)
            count=count+1

    return count

#----------------------------------

def Firstpart(list):
 wordchoice=[]
 firstword=input("Please enter the first word you would like to look for")
 wordchoice.append(firstword)  
 secondword=input("Please enter the second word you would like to look for")
 wordchoice.append(secondword) 
 thirdword=input("Please enter the third word you would like to look for")
 wordchoice.append(thirdword)
 return wordchoice


main()

Thank you so much in advance. 

---

您可以使用collections模块中的Counter来帮助您

import requests
from collections import Counter
def main():
    url = input('Please enter the url to the website you want to search: ')
    if not 'http' in url:
        url = 'http://' + url

    words = []
    for i in range(1,4):
        words.append(input('Please enter word number {}: '.format(i)))

    resp = requests.get(url)
    counter = Counter(resp.text.split())
    for word in words:
        print(word, 'found', counter[word], 'times')


if __name__ == '__main__':
    main()

---

您可以使用collections模块中的Counter来帮助您

import requests
from collections import Counter
def main():
    url = input('Please enter the url to the website you want to search: ')
    if not 'http' in url:
        url = 'http://' + url

    words = []
    for i in range(1,4):
        words.append(input('Please enter word number {}: '.format(i)))

    resp = requests.get(url)
    counter = Counter(resp.text.split())
    for word in words:
        print(word, 'found', counter[word], 'times')


if __name__ == '__main__':
    main()

Joakim给出了这一点，这有助于使您的代码更易于阅读和理解，但我将首先为您提供一个它不起作用的原因

在

Word（）

函数中，变量

mywords

是用户输入的单词列表。当您将其传递给

FindAWord

函数时，您给出的是一个列表，而不是一个单词。然后，当您比较if（item==word）（该行的第一部分（list）中确实不应该有

）时，您正在检查单个单词是否等于一个列表

您可以通过执行以下操作来修复该部分：

def Word():   
    answer=input("please enter the websites to examine in the http format ")

    response=requests.get(answer)
    txt = response.text
    print(txt)
    mywords=Firstpart(list)
    for word in mywords:
        num=FindAWord(txt,word)
        print("There are", num, "words called",word)

def FindAWord(TheWebText,word):
    print(TheWebText)
    print(type(TheWebText))
    MyList=TheWebText.split(sep=" ")
    print(MyList[0:100])
    count=0
    for item in MyList:
        if(item==word):
            print(item)
            count=count+1
    return count

您应该真正关注使变量名更具描述性，以帮助您（和其他人）更容易地阅读代码。如您所见，您在
FindAWord
a
word
中命名了参数，该参数是单数，给人的印象是它是一个单词。相反，这是一个单词列表。如果是
users\u words
或其他什么东西，你会立即发现
If（item==users\u words）
Joakim给出了一个让你的代码更容易阅读和理解的方法，但首先我会给你一个它不起作用的原因

在
Word（）
函数中，变量
mywords
是用户输入的单词列表。当您将其传递给
FindAWord
函数时，您给出的是一个列表，而不是一个单词。然后，当您比较if（item==word）

（该行的第一部分（list）中确实不应该有

）时，您正在检查单个单词是否等于一个列表

您可以通过执行以下操作来修复该部分：

def Word():   
    answer=input("please enter the websites to examine in the http format ")

    response=requests.get(answer)
    txt = response.text
    print(txt)
    mywords=Firstpart(list)
    for word in mywords:
        num=FindAWord(txt,word)
        print("There are", num, "words called",word)

def FindAWord(TheWebText,word):
    print(TheWebText)
    print(type(TheWebText))
    MyList=TheWebText.split(sep=" ")
    print(MyList[0:100])
    count=0
    for item in MyList:
        if(item==word):
            print(item)
            count=count+1
    return count

您应该真正关注使变量名更具描述性，以帮助您（和其他人）更容易地阅读代码。如您所见，您在
FindAWord
a
word
中命名了参数，该参数是单数，给人的印象是它是一个单词。相反，这是一个单词列表。如果是
users\u words
或其他什么东西，您会立即看到
If（item==users\u words）
中有错误