基于控制台的英语词典应用程序

Question

我对编程很陌生。我用Python编写了以下基于控制台的应用程序。这是一款英语词典应用程序。如果用户输入他/她想要搜索的单词，那么这个应用程序应该能够显示搜索单词的定义、示例(用于句子)、同义词、词源、语音和词类。它有点像PyDictionary，它是一个包；另一方面，它是一个功能齐全的(到目前为止.)应用程序和它使用不同的在线资源。这个应用程序依赖于BeautifulSoup和请求包。我希望有人能指点我，使它更好，更短，如果在代码中有任何错误。

import requests as r
from bs4 import BeautifulSoup as BS


def server_one(url):  # Oxford online dictionary
    """ Server_one scrapping algorithm to get essential information regarding the searched word """

    soup = get_soup(url)

    try:
        definition = soup.select_one('.ind').get_text(strip=True)  # Scraps first definition
    except AttributeError:
        definition = 'NOT FOUNT'

    try:
        example = soup.select_one('.exg').get_text(strip=True)  # Scraps first example
    except AttributeError:
        example = 'NOT FOUNT'

    try:
        parts_of_speech = soup.select_one('.pos').get_text(strip=True).capitalize()  # Gets the POS
    except AttributeError:
        parts_of_speech = 'NOT FOUNT'

    try:
        synonyms = soup.select_one('.exs').get_text(strip=True).split(', ')  # Collects all the synonyms
    except AttributeError:
        synonyms = 'NOT FOUNT'

    if 'NOT FOUNT' in synonyms:
        synonyms_tobe_sent = ''
    else:
        synonyms_tobe_sent = [cap.capitalize() for cap in synonyms]  # Capitalizes all the elements in the list

    try:
        origins = []
        for ori in soup.select('.senseInnerWrapper'):  # Scraps all the origins in a list
            if len(ori.text) < 400:
                origins.append(ori.text)
    except AttributeError:
        origins = 'NOT FOUNT'

    try:
        phonetics = soup.select_one('.phoneticspelling').get_text(strip=True)  # Gets the phonetics
    except AttributeError:
        phonetics = 'NOT FOUNT'

    try:
        next_definitions = []
        for tag in soup.select('.ind'):  # Gathers all possible definitions of the searched word
            next_definitions.append(tag.text)
        try:
            next_definitions.pop(0)  # Removes the first definition since it's already used before
        except IndexError:
            pass
    except AttributeError:
        next_definitions = 'NOT FOUNT'

    if example != 'NOT FOUNT':
        example_tobe_sent = example[1:-1].capitalize() + '.'  # Removes colons(') from the string and adds a (.)
    else:
        example_tobe_sent = 'NOT FOUNT'

    return definition, example_tobe_sent, next_definitions, parts_of_speech, synonyms_tobe_sent, origins, phonetics


def search_word():
    """ This function returns the searched word in lower format """

    word = input('Word that you wish to search\n>>> ').lower()
    return word


def url_server_one():
    """ A function to get correct URL for the server_one """

    w = search_word()
    new_w = w.replace(' ', '_')  # If a word has space then it gets _ instead, e.g. 'look up' turns into 'look_up'
    if ' ' in w:
        url = 'https://en.oxforddictionaries.com/definition/%s' % new_w  # URL with underscore(_)
    else:
        url = 'https://en.oxforddictionaries.com/definition/%s' % w  # URL without underscore
    data = (url, w)  # To make two variables as a tuple
    return data


def get_soup(url):
    """ This returns the bs4 soup object that will be used for scrapping """

    source_code = r.get(url)
    plain_text = source_code.text
    soup_data = BS(plain_text, 'html.parser')
    return soup_data


def display_dict():
    """ A function to display all the collected information in a desired format, currently supports only server_one """

    url, title = url_server_one()
    def_, exam, next_defs, part_sp, synos, ori, phonet = server_one(url)

    print('\nDefinition of %s:\n%s\n' % (title.capitalize(), def_))
    print('Parts of Speech:\n%s\n' % part_sp)
    print('Phonetics of %s:\n%s\n' % (title.capitalize(), phonet))
    print('Example of %s as follows:\n%s\n' % (title.capitalize(), exam))

    print('Synonyms:')
    if synos != '' or synos != []:
        print(', '.join(synos))
        print()
    if len(synos) < 1:
        print('NOT FOUND\n')

    print('Origins:')
    if ori != '' or ori != []:
        for i in ori:
            if i != '' or i != 'NOT FOUNT':
                print(i, sep=' ')
            print()
    if len(ori) < 1:
        print('NOT FOUND\n')

    print('Some other definition(s) of %s:' % title.capitalize())
    if next_defs != '' or next_defs != []:
        for i in next_defs:
            if i != '' or i != 'NOT FOUNT':
                print('* ' + i.capitalize() + '\n')
    if len(next_defs) < 1:
        print('NOT FOUND\n')


def try_():
    """ Part of repeat() function that returns integer value """

    while True:
        try:
            x = int(input('Press 1 to search again\n'
                      'Press 2 to quit'))
            if x in {1, 2}:
                return x
            print('Either press 1 or 2')
        except ValueError:
            print('Enter an integer')


def repeat():
    """ A function to repeat the process of searching over again and quit from the app """

    _try = try_()
    if _try == 1:
        display_dict()
    else:
        quit()


def main():
    print('Welcome to PyEngDict V1.0 by AJ\n')
    display_dict()  # Initially launches the app's search option
    while True:
        repeat()

if __name__ == '__main__':
    main()

Answer 1

下面是我要做的一些事情：

• 变量命名--尝试以描述性的方式命名变量--变量名称，如i (注意，它不是用作索引变量，而是作为下一个来源或定义)或ori，part_sp不够清楚--理解它们用于什么需要时间。当变量影响可读性时，缩短变量并不是正确的动机。
• if ori != '' or ori != []:可以用if ori:代替，if next_defs != '' or next_defs != []:可以用if next_defs:代替
• 我不确定您是否真的需要w和new_w在url_server_one()函数中--对于每个搜索字符串: def url_server_one()：“”一个函数来获得server_one的正确URL，您可以用一个undescore替换一个空格。“word = search_word().replace(‘'，'_') url = 'https://en.oxforddictionaries.com/definition/’+单词返回(url，word)
• 在这个部分中处理AttributeError是没有意义的--它不会被.select()或.text抛出，因为如果没有找到匹配的标记，.select()将返回一个空列表，以及一个所有都具有text属性的Tag实例列表。换句话说，AttributeError不会被抛出:soup.select(‘.senseInnerWrapper’)中ori的: try: ori.text= [] (‘.senseInnerWrapper’)：#在len(ori.text) < 400: origins.append(ori.text)列表中删除所有的起源，除了AttributeError: origins.append= 'NOT‘(收集next_definitions的块也是如此)。
• 您可以使用列表理解来定义origins列表:源= ori.text for ori in soup.select('.senseInnerWrapper') if len(ori.text) < 400
• 您可以使用多线串代替多个具有新行字符的print调用。

还有这些try和except AttributeError:重复块。有多种方法可以解决这一问题。一种是切换到基于字典的方法，首先定义字段名和CSS选择器之间的映射，如下所示：

selectors = {
    'definition': '.ind',
    'example': '.exg',
    'parts_of_speech': '.pos',
    'phoneticspelling': '.phoneticspelling',
    # ...
}

field_values = {}
for field, selector in selectors.items():
    tag = soup.select_one(selector)
    field_values[field] = tag.get_text() if tag is not None else 'NOT FOUND'