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python - Python - 将列表从一个函数传递到另一个函数

问题描述

尝试将列表从一个函数传递到另一个函数（teamurls），然后我将在另一个程序中使用它。我有我的程序在使用产量有一个价值输出的地方工作。（yield full_team_urls）如何将列表从第一个函数传递到第二个函数（def - team_urls）还有可能返回列表并继续使用yield吗？每个函数只能返回或输出一个对象吗？

编辑：尝试将 teamurls 传递给第二个函数，如下所示，我收到错误 - TypeError: team_urls() missing 1 required positional argument: 'teamurls'>

def table():
    url = 'https://www.skysports.com/premier-league-table'

    base_url = 'https://www.skysports.com'

    today = str(date.today())

    premier_r = requests.get(url)

    print(premier_r.status_code)

    premier_soup = BeautifulSoup(premier_r.text, 'html.parser')

    headers = "Position, Team, Pl, W, D, L, F, A, GD, Pts\n"

    premier_soup_tr = premier_soup.find_all('tr', {'class': 'standing-table__row'})

    premier_soup_th = premier_soup.find_all('thead')

    f = open('premier_league_table.csv', 'w')
    f.write("Table as of {}\n".format (today))
    f.write(headers)
    premier_soup_tr = premier_soup.find_all('tr', {'class': 'standing-table__row'})
    result = [[r.text.strip() for r in td.find_all('td', {'class': 'standing-table__cell'})][:-1] for td in premier_soup_tr[1:]]
    teamurls = ([a.find("a",href=True)["href"] for a in premier_soup_tr[1:]])
    return teamurls
    for item in result:

        f.write(",".join(item))
        f.write("\n")
    f.close()


    print('\n Premier league teams full urls:\n')
    for item in teamurls:

        entire_team = []

#        full_team_urls.append(base_url+ item)
        full_team_urls = (base_url + item + '-squad')
        yield full_team_urls

table()

def team_urls(teamurls):
    teams = [i.strip('/') for i in teamurls]
    print (teams)
team_urls()

标签： pythonlistfunction

解决方案

要将值传递给方法，请使用参数：

team_urls(teamurls)

您需要在定义中指定该参数team_urls：

def team_urls(teamurls):

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