如何将值列表迭代到字符串中？

Question

我是 Python 新手，所以请原谅我的理解不足。

我可以使用 BeatifulSoup4 下载 aapl 的数据，将数据分配给特定变量并使用变量创建字典。不过，我想做的是通过网站链接迭代stock 列表，并为列表中的不同公司多次运行相同的代码，最终为每个公司提供一个字典。

import bs4 as bs
import urllib.request


stocks = ['aapl', 'nvda', 'amgn']

sauce = urllib.request.urlopen('https://www.zacks.com/stock/quote/aapl/balance-sheet')
soup = bs.BeautifulSoup(sauce,'lxml')

#Cash & Cash Equivalents
cash_and_equivalents2017 = soup.find_all('td')[33].string
cash_and_equivalents2017 = int(cash_and_equivalents2017.replace(',',''))
cash_and_equivalents2016 = soup.find_all('td')[34].string
cash_and_equivalents2016 = int(cash_and_equivalents2016.replace(',',''))

#Receivables
receivables2017 = soup.find_all('td')[40].string
receivables2017 = int(receivables2017.replace(',',''))
receivables2016 = soup.find_all('td')[41].string
receivables2016 = int(receivables2016.replace(',',''))


aapl = {'Cash & Cash Equivalents':
            {'2017': cash_and_equivalents2017,
             '2016': cash_and_equivalents2016},
        'Receivables':
            {'2017': receivables2017,
             '2016': receivables2016}
        }
print(aapl)

为aapl 输出以下内容：

{'Cash & Cash Equivalents': {'2017': 74181, '2016': 67155}, 'Receivables': {'2017': 29299, '2016': 30343}}

以上是原始代码，我想用sauce中的aapl替换stocks中的其他字符串值，最终得到多个字典，如下所示，以便我能够自动下载各种数据只需在stock 列表中输入公司，即可将它们分成 3 个不同的字典。

sauce = urllib.request.urlopen('https://www.zacks.com/stock/quote/[stocks]/balance-sheet')

[stocks] = {'Cash & Cash Equivalents':
                {'2017': cash_and_equivalents2017,
                 '2016': cash_and_equivalents2016},
            'Receivables':
                {'2017': receivables2017,
                 '2016': receivables2016}

输出应该是 3 个字典，名称为 aapl、nvda、amgn。

aapl = {'Cash & Cash Equivalents':
            {'2017': cash_and_equivalents2017,
             '2016': cash_and_equivalents2016},
        'Receivables':
            {'2017': receivables2017,
             '2016': receivables2016}

nvda = {'Cash & Cash Equivalents':
            {'2017': cash_and_equivalents2017,
             '2016': cash_and_equivalents2016},
        'Receivables':
            {'2017': receivables2017,
             '2016': receivables2016}

amgn = {'Cash & Cash Equivalents':
            {'2017': cash_and_equivalents2017,
             '2016': cash_and_equivalents2016},
        'Receivables':
            {'2017': receivables2017,
             '2016': receivables2016}

感谢您的时间和帮助，非常感谢。

Answer 1

像这样将每个响应附加到一个列表中，并从字典中提取值：

aapl, nvda, amgn = [{'Cash & Cash Equivalents': {'2017': 74181, '2016': 67155}, 'Receivables': {'2017': 29299, '2016': 30343}},{'Cash & Cash Equivalents': {'2017': 74181, '2016': 67155}, 'Receivables': {'2017': 29299, '2016': 30343}},{'Cash & Cash Equivalents': {'2017': 74181, '2016': 67155}, 'Receivables': {'2017': 29299, '2016': 30343}}]

Answer 2

我不确定我是否正确理解了您的问题。

您要对stocks = ['aapl', 'nvda', 'amgn'] 中的每个条目进行一次请求和处理吗？

如果是这样，在 Python 中，您可以使用 for 循环遍历列表或 dict，如下所示：

for stock in stocks:
    # Do your processing here

在你的情况下你可以做的是：

使用字典而不是列表并将处理后的数据存储在其中。

stocks = {'aapl': {}, 'nvda': {}, 'amgn': {}}
for stock in stocks:

    sauce = urllib.request.urlopen(
        f'https://www.zacks.com/stock/quote/{stock}/balance-sheet')
    soup = bs.BeautifulSoup(sauce, 'lxml')

    # Cash & Cash Equivalents
    cash_and_equivalents2017 = soup.find_all('td')[33].string
    cash_and_equivalents2017 = int(cash_and_equivalents2017.replace(',', ''))
    cash_and_equivalents2016 = soup.find_all('td')[34].string
    cash_and_equivalents2016 = int(cash_and_equivalents2016.replace(',', ''))

    # Receivables
    receivables2017 = soup.find_all('td')[40].string
    receivables2017 = int(receivables2017.replace(',', ''))
    receivables2016 = soup.find_all('td')[41].string
    receivables2016 = int(receivables2016.replace(',', ''))

    stocks[stock] = {'Cash & Cash Equivalents':
                     {'2017': cash_and_equivalents2017,
                      '2016': cash_and_equivalents2016},
                     'Receivables':
                     {'2017': receivables2017,
                         '2016': receivables2016}
                     }

print(stocks)

至于url字符串，我用litteral string interpolation method

将输出以下内容：

{
    'aapl': 
    {
        'Cash & Cash Equivalents': 
        {
            '2017': 'some value', 
            '2016': 'some value'
        }, 
        'Receivables': 
        {
            '2017': 'some value', 
            '2016': 'some value'
            }
    }, 
    'nvda': 
    {
        'Cash & Cash Equivalents': 
        {
            '2017': 'some value', 
            '2016': 'some value'
        }, 
        'Receivables':
        {
            '2017': 'some value', 
            '2016': 'some value'
        }
    }, 
    'amgn': 
    {
        'Cash & Cash Equivalents': 
        {
            '2017': 'some value', 
            '2016': 'some value'
        }, 
        'Receivables': 
        {
            '2017': 'some value', 
            '2016': 'some value'
        }
    }
}

祝你有美好的一天。