Python 有没有更好的方法来读取许多html url？

我需要使用一堆

url

和

Sett.

值执行一系列命令

如何使下面的代码更干净

import pandas as pd

url_ita_a = ('https://fbref.com/it/comp/11/calendario/Risultati-e-partite-di-Serie-A')
df_ita_a = pd.read_html(url_ita_a)[0]
df2_ita_a = df_ita_a[['Sett.', 'Data', 'Casa', 'Punteggio', 'Ospiti']]
zero_df_ita_a = df2_ita_a[(df2_ita_a['Punteggio'] == '0–0') & (df2_ita_a['Sett.'] > 15)]

url_tur_a = ('https://fbref.com/it/comp/26/calendario/Risultati-e-partite-di-Super-Lig')
df_tur_a = pd.read_html(url_tur_a)[0]
df2_tur_a = df_tur_a[['Sett.', 'Data', 'Casa', 'Punteggio', 'Ospiti']]
zero_df_tur_a = df2_tur_a[(df2_tur_a['Punteggio'] == '0–0') & (df2_tur_a['Sett.'] > 15)]

...
...
...

url_rom_a = ('https://fbref.com/it/comp/47/calendario/Risultati-e-partite-di-Liga-I')
df_rom_a = pd.read_html(url_rom_a)[0]
df2_rom_a = df_rom_a[['Sett.', 'Data', 'Casa', 'Punteggio', 'Ospiti']]
zero_df_rom_a = df2_rom_a[(df2_rom_a['Punteggio'] == '0–0') & (df2_rom_a['Sett.'] > 13)]

url_ind_a = ('https://fbref.com/it/comp/82/calendario/Risultati-e-partite-di-Indian-Super-League')
df_ind_a = pd.read_html(url_ind_a)[0]
df2_ind_a = df_ind_a[['Sett.', 'Data', 'Casa', 'Punteggio', 'Ospiti']]
zero_df_ind_a = df2_ind_a[(df2_ind_a['Punteggio'] == '0–0') & (df2_ind_a['Sett.'] > 10)]

frames = [zero_df_ita_a, zero_df_tur_a, ... zero_df_rom_a,
          zero_df_ind_a]

result = pd.concat(frames)

在元组列表中添加所有

URL

和

Sett.

值

使用

for

循环对它们进行迭代

代码如下：

import pandas as pd

urls = [('https://fbref.com/it/comp/11/calendario/Risultati-e-partite-di-Serie-A', 15),
        ('https://fbref.com/it/comp/26/calendario/Risultati-e-partite-di-Super-Lig', 15),
        ('https://fbref.com/it/comp/12/calendario/Risultati-e-partite-di-La-Liga', 13),
        ('https://fbref.com/it/comp/13/calendario/Risultati-e-partite-di-Ligue-1', 13),
        ('https://fbref.com/it/comp/20/calendario/Risultati-e-partite-di-Bundesliga', 13)
        ] #Add all URLS and 'Sett.' values

frames = []

for url, sett in urls:
    df2 = pd.read_html(url)[0][['Sett.', 'Data', 'Casa', 'Punteggio', 'Ospiti']]
    frames.append(df2[(df2['Punteggio'] == '0–0') & (df2['Sett.'] > sett)]) #Here will come in handy the 'sett' variable

result = pd.concat(frames)

我没有手动将所有URL添加到列表中，因为它们太多了

---

把DF放在字典里，在一个循环中完成。回答得很好。建议将url:中我的元组的

替换为url的
，在url:
中设置。少了一行。哇，我不知道这是一个想法，谢谢！：）请用
绿色箭头给出答案
，以便接受！：）