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我需要有关我正在开发的文本云程序的帮助。我意识到这是家庭作业,但我自己已经走了很远,只是现在被难住了几个小时。我被困在网络爬虫部分。该程序应该打开一个页面,收集该页面中的所有单词,并按频率对它们进行排序。然后它应该打开该页面上的任何链接并获取该页面上的单词等。深度由全局变量 DEPTH 控制。最后,它应该把所有页面中的所有单词放在一起,形成一个文本云。

我正在尝试使用递归来调用一个函数来保持打开链接,直到达到深度。顶部的 import 语句只是使用了一个名为 getHTML(URL) 的函数,它返回页面上单词列表的元组,以及页面上的任何链接。

到目前为止,这是我的代码。除了 getRecursiveURLs(url, DEPTH) 和 makeWords(i) 之外,每个函数都可以正常工作。我也不是 100% 确定底部的 counter(List) 函数。

from hmc_urllib import getHTML

MAXWORDS = 50
DEPTH = 2

all_links = []

def getURL():
    """Asks the user for a URL"""

    URL = input('Please enter a URL: ')

    #all_links.append(URL)

    return makeListOfWords(URL), getRecursiveURLs(URL, DEPTH)


def getRecursiveURLs(url, DEPTH):
    """Opens up all links and adds them to global all_links list,
    if they're not in all_links already"""

    s = getHTML(url)
    links = s[1]
    if DEPTH > 0:
        for i in links:
            getRecursiveURLs(i, DEPTH - 1)
            if i not in all_links:
                all_links.append(i)
                #print('This is all_links in the IF', all_links)
                makeWords(i)#getRecursiveURLs(i, DEPTH - 1)
            #elif i in all_links:

             #   print('This is all_links in the ELIF', all_links)
              #  makeWords(i) #getRecursiveURLs(i, DEPTH - 1)
    #print('All_links at the end', all_links)
    return all_links





def makeWords(i):
    """Take all_links and create a dictionary for each page.
    Then, create a final dictionary of all the words on all pages."""

    for i in all_links:
        FinalDict = makeListOfWords(i)
        #print(all_links)
        #makeListOfWords(i))
    return FinalDict


def makeListOfWords(URL):
    """Gets the text from a webpage and puts the words into a list"""

    text = getHTML(str(URL))
    L = text[0].split()
    return cleaner(L)


def cleaner(L):

    """Cleans the text of punctuation and removes words if they are in the stop list."""

    stopList = ['', 'a', 'i', 'the', 'and', 'an', 'in', 'with', 'for',
                'it', 'am', 'at', 'on', 'of', 'to', 'is', 'so', 'too',
                'my', 'but', 'are', 'very', 'here', 'even', 'from',
                'them', 'then', 'than', 'this', 'that', 'though']

    x = [dePunc(c) for c in L]

    for c in x:
        if c in stopList:
            x.remove(c)

    a = [stemmer(c) for c in x]

    return counter(a)


def dePunc( rawword ):
    """ de-punctuationifies the input string """

    L = [ c for c in rawword if 'A' <= c <= 'Z' or 'a' <= c <= 'z' ]
    word = ''.join(L)
    return word


def stemmer(word):

    """Stems the words"""

    # List of endings
    endings = ['ed', 'es', 's', 'ly', 'ing', 'er', 'ers']

    # This first case handles 3 letter suffixes WITH a doubled consonant. I.E. spammers -> spam
    if word[len(word)-3:len(word)] in endings and word[-4] == word[-5]:
        return word[0:len(word)-4]

    # This case handles 3 letter suffixes WITHOUT a doubled consonant. I.E. players -> play
    elif word[len(word)-3:len(word)] in endings and word[-4] != word[-5]:
        return word[0:len(word)-3]

    # This case handles 2 letter suffixes WITH a doubled consonant. I.E. spammed -> spam
    elif word[len(word)-2:len(word)] in endings and word[-3] == word[-4]:
        return word[0:len(word)-3]

    # This case handles 2 letter suffixes WITHOUT a doubled consonant. I.E. played -> played
    elif word[len(word)-2:len(word)] in endings and word[-3] != word[-4]:
        return word[0:len(word)-3]

    # If word not inflected, return as-is.
    else:
        return word

def counter(List):
    """Creates dictionary of words and their frequencies, 'sorts' them,
    and prints them from most least frequent"""

    freq = {}
    result = {}
 # Assign frequency to each word
    for item in List:
        freq[item] = freq.get(item,0) + 1

    # 'Sort' the dictionary by frequency
    for i in sorted(freq, key=freq.get, reverse=True):
        if len(result) < MAXWORDS:
            print(i, '(', freq[i], ')', sep='')
            result[i] = freq[i]
    return result
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1 回答 1

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目前尚不完全清楚该作业的确切要求,但据我所知,您希望一次且仅一次访问 DEPTH 的所有页面。此外,您希望从所有页面中获取所有单词并使用聚合结果。下面的代码片段是您正在寻找的,但它未经测试(我没有 hmc_urllib)。all_linksmakeWords并且makeListOfWords已被删除,但其余代码将是相同的。

visited_links = []

def getURL():
    url = input('Please enter a URL: ')
    word_list = getRecursiveURLs(url, DEPTH)
    return cleaner(word_list) # this prints the word count for all pages

def getRecursiveURLs(url, DEPTH):
    text, links  = getHTML(url)
    visited_links.append(url)
    returned_word_list = text.split()
    #cleaner(text.split()) # this prints the word count for the current page

    if DEPTH > 0:
        for link in links:
            if link not in visited_links:
                returned_word_list += getRecursiveURLs(link, DEPTH - 1)
    return returned_word_list

一旦你有一个清理和词干的列表,你可以使用以下函数来生成单词统计字典并分别打印单词统计字典:

def counter(words):
    """
    Example Input: ['spam', 'egg', 'egg', 'egg', 'spam', 'spam', 'egg', 'egg']
    Example Output: {'spam': 3, 'egg', 5}
    """
    return dict((word, x.count(word)) for word in set(words))

def print_count(word_count, word_max):
    """
    Example Input: {'spam': 3, 'egg', 5}
    Prints the word list up to the word_max sorted by frequency
    """
    for word in sorted(word_count, key=word_count.get, reverse=True)[:word_max]:
        print(word,'(', word_count[word], ')', sep= '')
于 2012-05-31T06:58:32.350 回答