将每个项目与一行中的最后一个非空值配对

Question

我正在尝试创建一个函数，我向它提供一个 URL 的列表，这些列表经过 301 跳，它会为我展平它。我想将结果列表保存为 CSV，这样我就可以将它交给能够实现它并摆脱 301 跳的开发人员。

例如，我的爬虫将生成这个 301 跃点列表：

    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 |

我想要得到的输出是

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returns每一行作为一个列表，它们都嵌套在一个列表中，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 URL 与同一列表中的最后一个 URL 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，然后使用 melt 重塑您的 dataFrame 并构建一个两列映射。

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

将每个项目与一行中的最后一个非空值配对

Pair each item with the last non-null value in a row

python

dataframe

melt

pandas