Question

我正在尝试创建一个功能，我在其中提供了一个经过301跳的URL列表，它为我展平了它。我想将结果列表保存为CSV，这样我就可以将它交给可以实现它并且摆脱301跳的开发人员。

例如，我的抓取工具会生成301跳的列表：

    URL1          | URL2              | URL3              | URL4
example.com/url1  | example.com/url2  |                   | 
example.com/url3  | example.com/url4  | example.com/url5  | 
example.com/url6  | example.com/url7  | example.com/url8  | example.com/10
example.com/url9  | example.com/url7  | example.com/url8  | 
example.com/url23 | example.com/url10 |                   | 
example.com/url24 | example.com/url45 | example.com/url46 | 
example.com/url25 | example.com/url45 | example.com/url46 | 
example.com/url26 | example.com/url45 | example.com/url46 | 
example.com/url27 | example.com/url45 | example.com/url46 | 
example.com/url28 | example.com/url45 | example.com/url46 | 
example.com/url29 | example.com/url45 | example.com/url46 | 
example.com/url30 | example.com/url45 | example.com/url46 |

我想要获得的输出是

URL1              | URL2 
example.com/url1  | example.com/url2
example.com/url3  | example.com/url5
example.com/url4  | example.com/url5
example.com/url6  | example.com/10
example.com/url7  | example.com/10
example.com/url8  | example.com/10
example.com/url23 | example.com/url10
...

我已使用以下代码将Pandas数据框转换为列表列表：

import pandas as pd
import numpy as np

csv1 = pd.read_csv('Example_301_sheet.csv', header=None)
outlist = []

def link_flat(csv):

    for row in csv.iterrows():
        index, data = row
        outlist.append(data.tolist())

    return outlist

这会将每一行作为一个列表返回，它们全部嵌套在一个列表中，如下所示：

[['example.com/url1', 'example.com/url2', nan, nan],
 ['example.com/url3', 'example.com/url4', 'example.com/url5', nan],
 ['example.com/url6',
  'example.com/url7',
  'example.com/url8',
  'example.com/10'],
 ['example.com/url9', 'example.com/url7', 'example.com/url8', nan],
 ['example.com/url23', 'example.com/url10', nan, nan],
 ['example.com/url24', 'example.com/url45', 'example.com/url46', nan],
 ['example.com/url25', 'example.com/url45', 'example.com/url46', nan],
 ['example.com/url26', 'example.com/url45', 'example.com/url46', nan],
 ['example.com/url27', 'example.com/url45', 'example.com/url46', nan],
 ['example.com/url28', 'example.com/url45', 'example.com/url46', nan],
 ['example.com/url29', 'example.com/url45', 'example.com/url46', nan],
 ['example.com/url30', 'example.com/url45', 'example.com/url46', nan]]

如何将每个嵌套列表中的每个网址与同一列表中的最后一个网址相匹配，以生成上述列表？

Answer 1

您需要使用groupby + last确定每行的最后一个有效项目，然后重塑您的dataFrame并使用melt构建两列映射。< / p>

df.columns = range(len(df.columns))
df = (
    df.assign(URL2=df.stack().groupby(level=0).last())
      .melt('URL2', value_name='URL1')  
      .drop('variable', 1)
      .dropna()
      .drop_duplicates()
      .query('URL1 != URL2')
      .sort_index(axis=1)
      .reset_index(drop=True)
)

df
                 URL1               URL2
0    example.com/url1   example.com/url2
1    example.com/url3   example.com/url5
2    example.com/url6     example.com/10
3    example.com/url9   example.com/url8
4   example.com/url23  example.com/url10
5   example.com/url24  example.com/url46
6   example.com/url25  example.com/url46
7   example.com/url26  example.com/url46
8   example.com/url27  example.com/url46
9   example.com/url28  example.com/url46
10  example.com/url29  example.com/url46
11  example.com/url30  example.com/url46
12   example.com/url4   example.com/url5
13   example.com/url7     example.com/10
14   example.com/url7   example.com/url8
15  example.com/url45  example.com/url46
16   example.com/url8     example.com/10

将每个项目与行中的最后一个非空值配对

1 个答案: