Python 如何将值列表迭代为字符串？

我是Python新手，请原谅我的理解力不足

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

股票

列表，并为列表中的不同公司多次运行相同的代码，最后为每个公司生成一个字典

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)

它为

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，以最终获得如下所示的多个词典，这样我就可以通过在

库存

列表

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}

并且输出应该是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。

我不确定是否正确理解了您的问题

您想对

股票=['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，可以使用for循环对列表或dict进行迭代，如下所示：

for stock in stocks:
    # Do your processing here

在您的情况下，您可以执行以下操作：

{
    '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'
        }
    }
}

使用dict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，我使用了

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'
        }
    }
}

---

祝您度过愉快的一天。

将每个响应添加到一个列表中，如下图所示，并从DictionAnry中返回值：

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}}]

---

您必须循环使用股票中的每个值以获得所需的结果。