Question

这就是我在ipython中所做的（我正在使用Python 3.6）

from PyDictionary import PyDictionary
dictionary = PyDictionary()
list = dictionary.synonym("life")

我得到了错误：

/Library/Frameworks/Python.framework/Versions/3.6/lib/python3.6/site-packages/PyDictionary/utils.py:5: UserWarning: No parser was explicitly specified, so I'm using the best available HTML parser for this system ("html5lib"). This usually isn't a problem, but if you run this code on another system, or in a different virtual environment, it may use a different parser and behave differently.

The code that caused this warning is on line 5 of the file /Library/Frameworks/Python.framework/Versions/3.6/lib/python3.6/site-packages/PyDictionary/utils.py. To get rid of this warning, pass the additional argument 'features="html5lib"' to the BeautifulSoup constructor.

  return BeautifulSoup(requests.get(url).text)
life has no Synonyms in the API

这种情况发生在我尝试过的每个单词上，我做错了吗？是我需要添加参数'features =“ html5lib”'的问题，如果是，那么BeautifulSoup构造函数在哪里，我该怎么做？

Answer 1

它是Saran Roy答案的更新版本：

import requests
from bs4 import BeautifulSoup

def synonyms(term):
    response = requests.get('https://www.thesaurus.com/browse/{}'.format(term))
    soup = BeautifulSoup(response.text, 'lxml')
    soup.find('section', {'class': 'css-191l5o0-ClassicContentCard e1qo4u830'})
    return [span.text for span in soup.findAll('a', {'class': 'css-r5sw71-ItemAnchor etbu2a31'})] # 'css-1k3kgmb-ItemAnchor etbu2a31' for less relevant synonyms

word = "Input Your Word Here!"
print(synonyms(word))

Answer 2

PyDictionary.synonym函数尝试在thesaurus.com上查找同义词，但是代码已过期。它正在寻找不再存在的html结构。以下代码将基本完成相同的操作：

import requests
from bs4 import BeautifulSoup

def synonyms(term):
    response = requests.get('http://www.thesaurus.com/browse/{}'.format(term))
    soup = BeautifulSoup(response.text, 'html')
    section = soup.find('section', {'class': 'synonyms-container'})
    return [span.text for span in section.findAll('span')]

您可能要添加一些错误处理。

Answer 3

尝试一下：

import requests
from bs4 import BeautifulSoup

def synonyms(term):
    response = requests.get('https://www.thesaurus.com/browse/{}'.format(term))
    soup = BeautifulSoup(response.text, 'lxml')
    soup.find('section', {'class': 'synonyms-container'})
    return [span.text for span in soup.findAll('a', {'class': 'css-18rr30y'})] # class = .css-7854fb for less relevant

print(synonyms("reticulum"))

它只是NathanVērzemnieks答案的修改版本。

PyDictionary单词“ API中没有同义词”

3 个答案: