python - 分配简单变量命名后的KeyError

我已经很长时间没有使用python了，我正试图跟随一个网站从ESPN的梦幻足球网站上抓取一些信息。我正在使用的wesbite是https://stmorse.github.io/journal/espn-fantasy-v3.html。当我尝试下面列出的代码时，出现KeyError：“ schedule”。

我尝试完全按照网站上的代码进行操作，但给出了相同的错误。我在https://fantasy.espn.com/apis/v3/games/ffl/seasons/2019/segments/0/leagues/14380834?view=mMatchup上打开了api，并看到有安排的地方，所以不确定我还需要做什么。

league_id = 14380834
year = 2019
url = "https://fantasy.espn.com/apis/v3/games/ffl/seasons/2019/segments/0/leagues/14380834/"

r = requests.get(url)
d = r.json()
r = requests.get(url, params={"view": "mMatchup"})

df = [[
        game['matchupPeriodId'],
        game['home']['teamId'], game['home']['totalPoints'],
        game['away']['teamId'], game['away']['totalPoints']
    ] for game in d['schedule']]
df = pd.DataFrame(df, columns=['Week', 'Team1', 'Score1', 'Team2', 'Score2'])
df['Type'] = ['Regular' if w<=14 else 'Playoff' for w in df['Week']]
df.head()

我的预期结果将是我本周要求通过espn网站填充得分2的5列。

追溯为：

File "C:/Users/jacob/PycharmProjects/FF_stuff/app.py", line 15, in <module>
    ] for game in d['schedule']]
KeyError: 'schedule'

Process finished with exit code 1

最佳答案

您需要从包含参数而不是第一个请求的请求中获取json d。

这有效

r = requests.get(url, params={"view": "mMatchup"})
d = r.json()

df = [[
        game['matchupPeriodId'],
        game['home']['teamId'], game['home']['totalPoints'],
        game['away']['teamId'], game['away']['totalPoints']
    ] for game in d['schedule']]
df = pd.DataFrame(df, columns=['Week', 'Team1', 'Score1', 'Team2', 'Score2'])
df['Type'] = ['Regular' if w<=14 else 'Playoff' for w in df['Week']]
df.head()

KeyError即将到来，因为没有参数的请求中没有'schedule'信息。您可以执行d.keys()来查看dict具有哪些键

关于python - 分配简单变量命名后的KeyError，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/58438633/