删除具有两个空行的列，每次出现一系列的0时。

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英文:

Pandas: Dropping columns with 2 blank lines every time a series of 0 appear

问题

我有一大块数据，有很多列，但这些列在某些点上都为0。每当在列"two"中出现0时，我希望将该列删除，并在下面留下2个空行。

  2. one  two three 
  3. 1     4     4
  4. 3     5     5
  5. 5     7     5
  6. 666   0     6
  7. 785   0     8 
  8. 455   0     9 
  9. 454   0     9
  10. 12    2     8
  11. 23    5     9
  12. 2     3     7
  13. 1     5     5 
  14. 123   0     7 
  15. 123   0     7
  16. 3     5     5
  17.  
  19. （desired）
  20. output:
  23. one  two three 
  24. 1     4     4
  25. 3     5     5
  26. 5     7     5
  29. 12    2     8
  30. 23    5     9
  31. 2     3     7
  32. 1     5     5 
  35. 3     5     5
英文:

I have a big chunk of data, with a lot of columns, but this columns present at some points 0. I want every time that a 0 appears in the column "two", that column to drop it with 2 blank lines.

  2. one  two three 
  3. 1     4     4
  4. 3     5     5
  5. 5     7     5
  6. 666   0     6
  7. 785   0     8 
  8. 455   0     9 
  9. 454   0     9
  10. 12    2     8
  11. 23    5     9
  12. 2     3     7
  13. 1     5     5 
  14. 123   0     7 
  15. 123   0     7
  16. 3     5     5
  17.  
  19. (desired)
  20. output:
  23. one  two three 
  24. 1     4     4
  25. 3     5     5
  26. 5     7     5
  29. 12    2     8
  30. 23    5     9
  31. 2     3     7
  32. 1     5     5 
  35. 3     5     5

I tried different function: split, groupby, drop with conditions......., but noone didn't meet my request (most probably because I suck at coding)

答案1

得分: 1

以下是您要翻译的代码部分：

使用自定义的 groupby：

  1. m = df['two'].ne(0)
  3. out = (df[m]
  4.         .groupby((m & ~m.shift(fill_value=False)).cumsum(), group_keys=False)
  5.         .apply(lambda g: pd.concat([g, pd.DataFrame('', columns=g.columns, index=[0, 1])]))
  6.         .reset_index(drop=True)
  7.       )
  9. print(out.to_string(index=False))

或者使用一个巧妙的 repeat 方法：

  1. N = 2
  3. m1 = df['two'].ne(0)
  4. m2 = (m1 & ~m1.shift(fill_value=True))
  5. idx = df.index[m1].repeat(m2[m1]*N+1)
  7. out = df.loc[idx]
  8. out[out.index.duplicated()] = ''
  10. print(out.to_string(index=False))

输出：

  1. one two three
  2.   1   4     4
  3.   3   5     5
  4.   5   7     5
  5.              
  7.  12   2     8
  8.  23   5     9
  9.   2   3     7
  10.   1   5     5
  11.              
  13.   3   5     5
  14.              
  16.
英文:

You can use a custom groupby:

  1. m = df['two'].ne(0)
  3. out = (df[m]
  4.         .groupby((m & ~m.shift(fill_value=False)).cumsum(), group_keys=False)
  5.         .apply(lambda g: pd.concat([g, pd.DataFrame('', columns=g.columns, index=[0, 1])]))
  6.         .reset_index(drop=True)
  7.       )
  9. print(out.to_string(index=False))

Or with a hacky repeat:

  1. N = 2
  3. m1 = df['two'].ne(0)
  4. m2 = (m1 & ~m1.shift(fill_value=True))
  5. idx = df.index[m1].repeat(m2[m1]*N+1)
  7. out = df.loc[idx]
  8. out[out.index.duplicated()] = ''
  10. print(out.to_string(index=False))

Output:

  1. one two three
  2.   1   4     4
  3.   3   5     5
  4.   5   7     5
  5.              
  6.              
  7.  12   2     8
  8.  23   5     9
  9.   2   3     7
  10.   1   5     5
  11.              
  12.              
  13.   3   5     5
  14.              
  15.

答案2

得分: 0

pd.Index.union() 方法有一个 sort 选项，所以在重新索引时，它应该按照正确的顺序进行操作。

  1. m = df['two'].ne(0)
  2. df.reindex(df.loc[m].index.union(df.loc[m.diff().ne(0) & ~m].index.repeat(2) + .5)).reset_index(drop=True)

输出：

  1.      one  two  three
  2. 0    1.0  4.0    4.0
  3. 1    3.0  5.0    5.0
  4. 2    5.0  7.0    5.0
  5. 3    NaN  NaN    NaN
  6. 4    NaN  NaN    NaN
  7. 5   12.0  2.0    8.0
  8. 6   23.0  5.0    9.0
  9. 7    2.0  3.0    7.0
  10. 8    1.0  5.0    5.0
  11. 9    NaN  NaN    NaN
  12. 10   NaN  NaN    NaN
  13. 11   3.0  5.0    5.0
英文:

Here is another way:

pd.Index.union() has a sort option, so when reindexing, it should be in the correct order

  1. m = df['two'].ne(0)
  2. df.reindex(df.loc[m].index.union(df.loc[m.diff().ne(0) & ~m].index.repeat(2) + .5)).reset_index(drop=True)

Output:

  1.      one  two  three
  2. 0    1.0  4.0    4.0
  3. 1    3.0  5.0    5.0
  4. 2    5.0  7.0    5.0
  5. 3    NaN  NaN    NaN
  6. 4    NaN  NaN    NaN
  7. 5   12.0  2.0    8.0
  8. 6   23.0  5.0    9.0
  9. 7    2.0  3.0    7.0
  10. 8    1.0  5.0    5.0
  11. 9    NaN  NaN    NaN
  12. 10   NaN  NaN    NaN
  13. 11   3.0  5.0    5.0

huangapple
  • 本文由 发表于 2023年5月11日 19:31:27
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  • pandas
  • python
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