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如何让这个 for 循环在仍然使用范围的同时不重复这个列表以前的输出。这个 for 循环重复前一个数字的输出。每次它转到下一个数字。而不是从 0 到 20 一次。它去 0-1,0-2,0-3,0-4 .....等等。我希望它从 0 到 20 一次，而不是重复自己。

import time
from selenium import webdriver
import selenium
from selenium.webdriver.chrome import service
from selenium.webdriver.chrome.service import Service
from selenium.webdriver.common.keys import Keys
from selenium.webdriver.common.by import By
from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.support import expected_conditions as EC
import pandas as pd

#class scraperdata():

ser= Service("C:\Program Files (x86)\chromedriver.exe")
options = webdriver.ChromeOptions()
options.add_experimental_option('excludeSwitches', ['enable-logging'])
driver = webdriver.Chrome(options=options,service=ser)
driver.get('https://soundcloud.com/jujubucks')
print(driver.title)

wait = WebDriverWait(driver,30)

wait.until(EC.element_to_be_clickable((By.ID,"onetrust-accept-btn-handler"))).click()

try:  
    song_list = []

    i = 1
    for _ in range(20):
        song_contents = driver.find_element(By.XPATH, "//li[@class='soundList__item'][{}]".format(i))
        driver.execute_script("arguments[0].scrollIntoView(true);",song_contents)
        search = song_contents.find_element(By.XPATH, ".//a[contains(@class,'soundTitle__username')]/span").text
        search_song = song_contents.find_element(By.XPATH, ".//a[contains(@class,'soundTitle__title')]/span").text
        search_date = song_contents.find_element(By.XPATH, ".//time[contains(@class,'relativeTime')]/span").text
        search_plays = song_contents.find_element(By.XPATH, ".//span[contains(@class,'sc-ministats-small')]/span").text
        i+=1
        if _ == Exception:
            break

        option ={
        'Artist': search, 
        'Song_title': search_song, 
        'Date': search_date,
        'Streams': search_plays
        }
        song_list.append(option)

        df = pd.DataFrame(song_list)
        print(df)

except Exception:
    pass        

driver.quit()

输出

Stream Juju Bucks music | Listen to songs, albums, playlists for free on SoundCloud
       Artist                              Song_title               Date   Streams
0  Juju Bucks  Squad Too Deep Ft. Cool Prince (Outro)  Posted 1 year ago  31 plays
       Artist                              Song_title               Date   Streams
0  Juju Bucks  Squad Too Deep Ft. Cool Prince (Outro)  Posted 1 year ago  31 plays
1  Juju Bucks            Tropikana ft. P-Dogg Amazing  Posted 1 year ago  48 plays
       Artist                              Song_title               Date   Streams
0  Juju Bucks  Squad Too Deep Ft. Cool Prince (Outro)  Posted 1 year ago  31 plays
1  Juju Bucks            Tropikana ft. P-Dogg Amazing  Posted 1 year ago  48 plays
2  Juju Bucks              Party Ka Mngani Ft. X-Poll  Posted 1 year ago  72 plays
       Artist                              Song_title               Date    Streams
0  Juju Bucks  Squad Too Deep Ft. Cool Prince (Outro)  Posted 1 year ago   31 plays
1  Juju Bucks            Tropikana ft. P-Dogg Amazing  Posted 1 year ago   48 plays
2  Juju Bucks              Party Ka Mngani Ft. X-Poll  Posted 1 year ago   72 plays
3  Juju Bucks      Joy Ft. Black Sushi & Gavin Bowden  Posted 1 year ago  122 plays
4

2 回答 2

0

您应该将数据帧分配移到 for 循环之外：

for _ in range(20):
    …
    song_list.append(option)  
df = pd.DataFrame(song_list)
print(df)
于 2021-10-28T21:50:59.723 回答
0

for 循环的范围很好。问题是，对于循环的每次迭代，您都将一个新项目附加到song_list，它位于循环范围之外。song_list = []进入循环以使打印语句按您想要的方式工作。

但是，当循环结束时，您将不再跟踪所有歌曲。您可能根本不想在循环内打印。在循环外打印一次。

于 2021-10-28T21:47:18.850 回答