在 python 中编写重复代码的更简洁方法？

Question

我正在使用 selenium 从网页下载一些文件。在星期一，我需要下载星期五、星期六和星期日的信息。每隔一天我只需要昨天。我写了一个 if/else 语句来完成这个，然后将代码复制并粘贴到 else 语句中。一定有更 pythonic 的方式来写这个，但我还是新手。

today = datetime.date.today()
yesterday = str(today - timedelta(days=1))
if today.weekday() == 0:
    fri = str(today - timedelta(days=3))
    sat = str(today - timedelta(days=2))
    weekend = [fri, sat, yesterday]
    for day in weekend:
        # Needs to go first otherwise page won't load
        date_field = driver.find_element_by_xpath(
            """//*[@id="id blah blah"]""")
        date_field.send_keys(day)
        org_list = driver.find_element_by_xpath(
            """//*[@id="id blah blah"]/option[text()=\"string\"]""").click()

        delay = 5
        try:
            table_chk = WebDriverWait(driver, delay).until(
                EC.presence_of_element_located((By.XPATH, """//*[@id="id blah blah"]""")))
            export_btn = driver.find_element_by_xpath(
                """//*[@id="id blah blah"]""")
            export_btn.click()
            date_field = driver.find_element_by_xpath(
                """//*[@id="id blah blah"]""")
            date_field.clear()
            org_list = driver.find_element_by_xpath(
                """//*[@id="id blah blah"]/option[1]""").click()
        except TimeoutException:
            print("Loading took too much time!")
        time.sleep(2)
else:
    # Needs to go first otherwise it doesn't work
    date_field = driver.find_element_by_xpath(
        """//*[@id="id blah blah"]""")
    date_field.send_keys(yesterday)
    org_list = driver.find_element_by_xpath(
        """//*[@id="id blah blah"]/option[text()=\"string\"]""").click()

    delay = 5
    try:
        table_chk = WebDriverWait(driver, delay).until(
            EC.presence_of_element_located((By.XPATH, """//*[@id="id blah blah"]""")))
        export_btn = driver.find_element_by_xpath(
            """//*[@id="id blah blah"]""")
        export_btn.click()
    except TimeoutException:
        print("Loading took too much time!")

如何有效地重复代码，但在周一周五、周六、周日重复运行多次，而前一天仅重复一次，一周中每隔一天重复一次？

Answer 1

使其始终循环，但以编程方式将集合定义为大部分时间作为单个元素循环，需要时多天：

today = datetime.date.today()
# No need to define yesterday; we'll make it as needed next
if today.weekday() == 0:
    # Today is Monday, quickly get the days for Friday-Sunday
    days = [today - timedelta(days=i) for i in (3, 2, 1)]
else:
    # Today is not Monday, just check yesterday
    days = [today - timedelta(days=1)]
# days is now either one element list of just yesterday, or the whole weekend
# loop runs once or three times, as needed, with the same code
for day in days:
    # Complete body of original for day in weekend: loop goes here

如果你真的想把代码重复减到最少，你可以将循环之前的代码减少到：

today = datetime.date.today()
num_days_to_check = 3 if today.weekday() == 0 else 1
days = [today - timedelta(days=i) for i in range(num_days_to_check, 0, -1)]

因为真的，所有不同的是你需要检查多少天，1 或 3，所以条件可以简化为两者之间的单线选择，剩下的只是基于最初的决定点.

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