返回计数在不同列表上的单词列表

Question

大家下午好

今天，我被要求编写以下函数：

def compareurl(url1,url2,enc,n)

此函数比较两个URL，并返回包含以下内容的列表：

[word,occ_in_url1,occ_in_u2]

哪里：

字--->长度为n的字

occ_in_url1 --->在url1中输入单词

occ_in_url2 ---> url2中的单词次数

所以我开始写函数，这是我到目前为止写的：

def compare_url(url1,url2,enc,n):
    from urllib.request import urlopen
    with urlopen('url1') as f1:
        readpage1 = f1.read()
        decodepage1 = readpage1.decode('enc')
    with urlopen('url2') as f2:
        readpage2 = f2.read()
        decodepage2 = readpage2.decode('enc')
    all_lower1 = decodepage1.lower()
    all_lower2 = decodepage2.lower()
    import string
    all_lower1nopunctuation = "".join(l for l in all_lower1 if l not in string.punctuation)
    all_lower2nopunctuation = "".join(l for l in all_lower2 if l not in string.punctuation)
    for word1 in all_lower1nopunctuation:
        if len(word1) == k:
            all_lower1nopunctuation.count(word1)
    for word2 in all_lower2nopunctuation:
        if len(word2) == k:
            all_lower2opunctuation.count(word2)
    return(word1,all_lower1nopunctuation.count(word1),all_lower2opunctuation.count(word2))
    return(word2,all_lower1nopunctuation.count(word1),all_lower2opunctuation.count(word2))

但是这段代码并不像我想的那样起作用，实际上它根本不起作用。

我也想要：

1. 对返回列表进行递减排序（从返回次数最多的单词开始）

2. 如果2个单词出现相同的次数，则必须按字母顺序返回

Answer 1

您的代码中有一些错别字（将来要小心），但是也有一些python问题（或可以改进的地方）。

---

首先，您的导入内容应位于文档的顶部

from urllib.request import urlopen
import string

---

您应该使用一个string调用urlopen ，这就是您正在做的，但是这个字符串是'url1'而不是'http://...' 。 您不要在引号内使用变量：

with urlopen(url1) as f1: #remove quotes
    readpage1 = f1.read()
    decodepage1 = readpage1.decode(enc) #remove quotes
with urlopen(url2) as f2: #remove quotes
    readpage2 = f2.read()
    decodepage2 = readpage2.decode(enc) #remove quotes

---

您需要改善all_lower1nopunctuation初始化。 要更换stackoverflow.com与stackoverflowcom ，这实际上应该stackoverflow com 。

#all_lower1nopunctuation = "".join(l for l in all_lower1 if l not in string.punctuation)
#the if statement should be after 'l' and before 'for'
#you should include 'else' to replace the punctuation with a space
all_lower1nopunctuation = ''.join(l if l not in string.punctuation
else ' ' for l in all_lower1)
all_lower2nopunctuation = ''.join(l if l not in string.punctuation
else ' ' for l in all_lower2)

---

将两者合并for一个。 还将找到的单词添加到集合中（唯一元素列表）。

all_lower1nopunctuation.count(word1)返回word1在all_lower1nopunctuation中出现的次数 。 它不会增加计数器。

因为all_lower1nopunctuation是字符串 （而不是list ），所以all_lower1nopunctuation中的for word1 in all_lower1nopunctuation 。 使用.split(' ')将其转换为列表 。

.replace('\\n', '')删除所有换行符，否则它们也将被视为单词 。

#for word1 in all_lower1nopunctuation:
#    if len(word1) == k: #also, this should be == n, not == k
#        all_lower1nopunctuation.count(word1)
#for word2 in all_lower2nopunctuation:
#    if len(word2) == k:
#        all_lower2opunctuation.count(word2)

word_set = set([])
for word in all_lower1nopunctuation.replace('\n', '').split(' '):
    if len(word) == n and word in all_lower2nopunctuation:
        word_set.add(word) #set uses .add() instead of .append()

---

现在您已经在两个URL上出现了一组单词，您需要存储每个URL中有多少个单词 。 以下代码将确保您具有所要求 的元组列表

count_list = []
for final_word in word_set:
    count_list.append((final_word,
    all_lower1nopunctuation.count(final_word),
    all_lower2nopunctuation.count(final_word)))

---

返回意味着函数已完成，并且解释器将在调用该函数之前的任何位置继续运行，因此返回之后的内容无关紧要。

正如RemcoGerlich所说 。

您的代码将始终仅返回第一个 return ，因此您需要将两个return合并为一个。

#return(word1,all_lower1nopunctuation.count(word1),all_lower2opunctuation.count(word2))
#return(word2,all_lower1nopunctuation.count(word1),all_lower2opunctuation.count(word2))
return(count_list) # which contains a list of tuples with all words and its counts

---

TL; DR

from urllib.request import urlopen
import string

def compare_url(url1,url2,enc,n):
    with urlopen(url1) as f1:
        readpage1 = f1.read()
        decodepage1 = readpage1.decode(enc)
    with urlopen(url2) as f2:
        readpage2 = f2.read()
        decodepage2 = readpage2.decode(enc)

    all_lower1 = decodepage1.lower()
    all_lower2 = decodepage2.lower()

    all_lower1nopunctuation = ''.join(l if l not in string.punctuation
    else ' ' for l in all_lower1)
    all_lower2nopunctuation = ''.join(l if l not in string.punctuation
    else ' ' for l in all_lower2)

    word_set = set([])
    for word in all_lower1nopunctuation.replace('\n', '').split(' '):
        if len(word) == n and word in all_lower2nopunctuation:
            word_set.add(word)

    count_list = []
    for final_word in word_set:
        count_list.append((final_word,
        all_lower1nopunctuation.count(final_word),
        all_lower2nopunctuation.count(final_word)))

    return(count_list)

url1 = 'https://www.tutorialspoint.com/python/list_count.htm'
url2 = 'https://stackoverflow.com/a/128577/7067541'

for word_count in compare_url(url1,url2, 'utf-8', 5):
    print (word_count)