根据元组索引排序行

huangapple
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英文:

Sort Rows Based on Tuple Index

问题

Lambda函数如下，它可以根据第3个元组索引元素的相应'Num'值进行排序：

  1. dataTable = dataTable.reindex(sorted(dataTable.index, key=lambda x: (not isinstance(x, tuple), x, -dataTable.loc[x, 'Num'])))
英文:

Overview:

Pandas dataframe with a tuple index and corresponding 'Num' column:

  1. Index                               Num
  3. ('Total', 'A')                      23
  5. ('Total', 'A', 'Pandas')            3
  7. ('Total', 'A', 'Row')               7
  9. ('Total', 'A', 'Tuple')             13
  11. ('Total', 'B')                      35
  13. ('Total', 'B', 'Rows')              12
  15. ('Total', 'B', 'Two')               23
  17. ('Total', 'C')                      54
  19. ('Total', 'C', 'Row')               54
  21. Total                               112

The index and 'Num' column are already sorted with a lambda function by Alphabetical Order and based on the length of tuple elements:

  1. dataTable = dataTable.reindex(sorted(dataTable.index, key=lambda x: (not isinstance(x, tuple), x)))

Problem:

Now, I want to sort only the 3rd tuple index element based on it's corresponding 'Num' value. Here would be an updated example of the dataframe:

  1. Index                               Num
  3. ('Total', 'A')                      23
  5. ('Total', 'A', 'Tuple')             13
  7. ('Total', 'A', 'Row')               7
  9. ('Total', 'A', 'Pandas')            3
  11. ('Total', 'B')                      35
  13. ('Total', 'B', 'Two')               23
  15. ('Total', 'B', 'Rows')              12
  17. ('Total', 'C')                      54
  19. ('Total', 'C', 'Row')               54
  21. Total                               112

Question:

What Lambda function can achieve this?

答案1

得分: 1

  1. 您可以尝试：
  3. ```py
  4. def fn(x):
  5.     vals = x.sort_values(by='Num', ascending=False)
  6.     df.loc[x.index] = vals.values
  8. m = df['Index'].apply(len).eq(3)
  9. df[m].groupby(df.loc[m, 'Index'].str[1], group_keys=False).apply(fn)
  11. print(df)

打印：

  1.                 Index  Num
  2. 0          (Total, A)   23
  3. 1   (Total, A, Tuple)   13
  4. 2     (Total, A, Row)    7
  5. 3  (Total, A, Pandas)    3
  6. 4          (Total, B)   35
  7. 5     (Total, B, Two)   23
  8. 6    (Total, B, Rows)   12
  9. 7          (Total, C)   54
  10. 8     (Total, C, Row)   54
  11. 9               Total  112

初始 df：

  1.                 Index  Num
  2. 0          (Total, A)   23
  3. 1  (Total, A, Pandas)    3
  4. 2     (Total, A, Row)    7
  5. 3   (Total, A, Tuple)   13
  6. 4          (Total, B)   35
  7. 5    (Total, B, Rows)   12
  8. 6     (Total, B, Two)   23
  9. 7          (Total, C)   54
  10. 8     (Total, C, Row)   54
  11. 9               Total  112
  2. <details>
  3. <summary>英文:</summary>
  5. You can try:
  7. ```py
  8. def fn(x):
  9.     vals = x.sort_values(by=&#39;Num&#39;, ascending=False)
  10.     df.loc[x.index] = vals.values
  12. m = df[&#39;Index&#39;].apply(len).eq(3)
  13. df[m].groupby(df.loc[m, &#39;Index&#39;].str[1], group_keys=False).apply(fn)
  15. print(df)

Prints:

  1.                 Index  Num
  2. 0          (Total, A)   23
  3. 1   (Total, A, Tuple)   13
  4. 2     (Total, A, Row)    7
  5. 3  (Total, A, Pandas)    3
  6. 4          (Total, B)   35
  7. 5     (Total, B, Two)   23
  8. 6    (Total, B, Rows)   12
  9. 7          (Total, C)   54
  10. 8     (Total, C, Row)   54
  11. 9               Total  112

Initial df:

  1.                 Index  Num
  2. 0          (Total, A)   23
  3. 1  (Total, A, Pandas)    3
  4. 2     (Total, A, Row)    7
  5. 3   (Total, A, Tuple)   13
  6. 4          (Total, B)   35
  7. 5    (Total, B, Rows)   12
  8. 6     (Total, B, Two)   23
  9. 7          (Total, C)   54
  10. 8     (Total, C, Row)   54
  11. 9               Total  112

huangapple
  • 本文由 发表于 2023年6月2日 01:05:51
  • 转载请务必保留本文链接：https://go.coder-hub.com/76384189.html
  • dataframe
  • lambda
  • pandas
  • python
  • tuples
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