如何从函数结果创建新列

How to create a new column from function result

目前运行正在使用下面的脚本检查一长串 url 中的错误。此代码首先在 df['Final_URL'] 中查找唯一的 url,测试每个单独的 url 和 returns link url 的状态.当我 运行 下面的代码时,我在我的笔记本上得到了当前输出,这很好。现在我想将状态代码(例如 200、404、BAD 等)推送到我的 df 中一个名为 "Status" 的新列,用于所有 url,它等于我唯一的 urls在代码的开头得到。

创建新列 df['Status'] 的最佳方法是什么,因为我想将其导出到 google 工作表,您知道在使用 pygsheets 更新单元格时是否保留了文本颜色吗?

Input code:
#get unique urls and check for errors
URLS = []

for unique_link in df['Final_URL'].unique():
    URLS.append(unique_link)

try:

    GREEN = '3[92m'
    YELLOW = '3[93m'
    RED = '3[91m'
    ENDC = '3[0m'

    def main():
        while True:
            print ("\nTesting URLs.", time.ctime())
            checkUrls()
            time.sleep(10) #Sleep 10 seconds
            break

    def checkUrls():     
        for url in URLS:
            status = "N/A"
            try:
                #check if regex contains bet3.com
                if re.search(".*bet3\.com.*", url):
                    status = checkUrl(url)
                else:
                    status = "BAD"

            except requests.exceptions.ConnectionError:
                status = "DOWN"

            printStatus(url, status)

            #for x in df['Final_URL']:
            #    if x == url:
            #        df['Status'] = printStatus(status)



    def checkUrl(url):
        r = requests.get(url, timeout=5)
        #print r.status_code
        return str(r.status_code)

    def printStatus(url, status):
        color = GREEN

        if status != "200":
            color=RED

        print (color+status+ENDC+' '+ url)



    #
    # Main app
    #
    if __name__ == '__main__':
        main()

except:

    print('Something went wrong!')



Current output:

200 https://www.bet3.com/dl/~offer
404 http://extra.bet3.com/promotions/en/soccer/soccer-accumulator-bonus
BAD https://extra.betting3.com/features/en/bet-builder
200 https://www.bet3.com/dl/6

你可以这样重写你的函数

def checkUrl(url):
    if re.search(".*bet3\.com.*", url):
        try:
            r = requests.get(url, timeout=5)
        except requests.exceptions.ConnectionError:
            return 'DOWN'
        return str(r.status_code)
    return 'BAD'

然后像这样应用它

df['Status'] = df['Final_URL'].apply(checkUrl)

不过,正如 user32185 所注意到的,如果有重复的 URL,这将调用它们两次。

为了避免这种情况,您可以按照 user32185 的建议并像这样编写您的函数:

def checkUrls(urls):
    results = []
    for url in urls:
        if re.search(".*bet3\.com.*", url):
            try:
                r = requests.get(url, timeout=5)
            except requests.exceptions.ConnectionError:
                results.append([url, 'DOWN'])
            results.append([url, str(r.status_code)])
        else:
            results.append([url, 'BAD'])
    return pd.DataFrame(data=results, columns=['Final_URL', 'Status'])

然后像这样使用它:

status_df = checkUrls(df['Final_URL'].unique())
df = df.merge(status_df, how='left', on='Final_URL')