Question

我试图计算政治家在某些演讲中使用的收缩次数。我有很多演讲，但这里有一些网址作为样本：

every_link_test = ['http://www.millercenter.org/president/obama/speeches/speech-4427',
 'http://www.millercenter.org/president/obama/speeches/speech-4424',
 'http://www.millercenter.org/president/obama/speeches/speech-4453',
 'http://www.millercenter.org/president/obama/speeches/speech-4612',
 'http://www.millercenter.org/president/obama/speeches/speech-5502']

我现在有一个非常粗略的计数器 - 它只计算所有这些链接中使用的总收缩数。例如，以下代码返回上面五个链接的79,101,101,182,224。但是，我想链接filename，我在下面创建的变量，所以我会有类似(speech_1, 79),(speech_2, 22),(speech_3,0),(speech_4,81),(speech_5,42)的内容。这样，我就可以跟踪每个单独演讲中使用的收缩次数。我的代码出现以下错误：AttributeError: 'tuple' object has no attribute 'split'

这是我的代码：

import urllib2,sys,os
from bs4 import BeautifulSoup,NavigableString
from string import punctuation as p
from multiprocessing import Pool
import re, nltk
import requests
reload(sys)

url = 'http://www.millercenter.org/president/speeches'
url2 = 'http://www.millercenter.org'

conn = urllib2.urlopen(url)
html = conn.read()

miller_center_soup = BeautifulSoup(html)
links = miller_center_soup.find_all('a')

linklist = [tag.get('href') for tag in links if tag.get('href') is not None]

# remove all items in list that don't contain 'speeches'
linkslist = [_ for _ in linklist if re.search('speeches',_)]
del linkslist[0:2]

# concatenate 'http://www.millercenter.org' with each speech's URL ending
every_link_dups = [url2 + end_link for end_link in linkslist]

# remove duplicates
seen = set()
every_link = [] # no duplicates array
for l in every_link_dups:
    if l not in seen:
        every_link.append(l)
        seen.add(l)

def processURL_short_2(l):
    open_url = urllib2.urlopen(l).read()
    item_soup = BeautifulSoup(open_url)
    item_div = item_soup.find('div',{'id':'transcript'},{'class':'displaytext'})
    item_str = item_div.text.lower()

    splitlink = l.split("/")
    president = splitlink[4]
    speech_num = splitlink[-1]
    filename = "{0}_{1}".format(president, speech_num)
    return item_str, filename

every_link_test = every_link[0:5]
print every_link_test
count = 0
for l in every_link_test:
    content_1 = processURL_short_2(l)
    for word in content_1.split():
        word = word.strip(p)
        if word in contractions:
            count = count + 1        
    print count, filename

Answer 1

而不是print count, filename，您应该将这些数据保存到数据结构中，例如字典。由于processURL_short_2已被修改为返回元组，因此您需要将其解压缩。

data = {} # initialize a dictionary
for l in every_link_test:
    content_1, filename = processURL_short_2(l) # unpack the content and filename
    for word in content_1.split():
        word = word.strip(p)
        if word in contractions:
            count = count + 1        
    data[filename] = count # add this to the dictionary as filename:count

这会为您提供{'obama_4424':79, 'obama_4453':101,...}之类的字典，让您轻松存储和访问已解析的数据。

Answer 2

正如错误消息所述，您无法按照使用方式使用拆分。 split是用于字符串的。

所以你需要改变这个：

for word in content_1.split():

到此：

for word in content_1[0]:

我通过运行您的代码选择[0]，我认为这会为您提供您要搜索的文本块。

@ TigerhawkT3有一个很好的建议你也应该遵循他们的答案：

https://stackoverflow.com/a/32981533/1832539

尝试计算列表中的元素时出现问题？

2 个答案: