python - 如何在 Pandas 中填充行不为 NULL 的列名？

Question

我有一个数据框如下；

df_d = {'consolidate_elem': {0: np.nan,
  1: np.nan,
  2: np.nan,
  3: np.nan,
  4: np.nan,
  5: np.nan,
  6: np.nan,
  7: np.nan,
  8: np.nan,
  9: np.nan,
  10: np.nan},
 'A': {0: np.nan,
  1: np.nan,
  2: '6/9/1972',
  3: '9/4/1943',
  4: '1/29/1944',
  5: '7/31/1965',
  6: '4/5/1979',
  7: np.nan,
  8: '3/17/2000',
  9: '3/18/2000',
  10: '3/17/2000'},
 'B': {0: np.nan,
  1: np.nan,
  2: np.nan,
  3: 'Yes',
  4: np.nan,
  5: np.nan,
  6: np.nan,
  7: np.nan,
  8: np.nan,
  9: np.nan,
  10: np.nan},
 'C': {0: np.nan,
  1: np.nan,
  2: np.nan,
  3: np.nan,
  4: 'Yes',
  5: 'Yes',
  6: np.nan,
  7: 'Yes',
  8: 'Yes',
  9: 'Yes',
  10: 'Yes'},
 'D': {0: '11100',
  1: '11721',
  2: np.nan,
  3: '55626',
  4: '5,562,655,626.00',
  5: np.nan,
  6: np.nan,
  7: '33050',
  8: '33050',
  9: '33050',
  10: np.nan}}

在这里，我想用列名填充列consolidate_elem，其行在上面的数据框中不是NAN。预期输出如下；

因此，在第一行中，我们在 D 列中有一个值，因此在 consolidate_elem 中将其填充为 D。

类似地，在索引 -4 中，它与 A;C;D 一样，因为它们具有值。

Answer 1

使用DataFrame.dot用于通过比较非缺失值创建的掩码进行矩阵乘法 DataFrame.notna省略第一列:

df = pd.DataFrame(df_d)

df['consolidate_elem'] = df.iloc[:, 1:].notna().dot(df.columns[1:] + ';').str.strip(';')
print (df)
   consolidate_elem          A    B    C                 D
0                 D        NaN  NaN  NaN             11100
1                 D        NaN  NaN  NaN             11721
2                 A   6/9/1972  NaN  NaN               NaN
3             A;B;D   9/4/1943  Yes  NaN             55626
4             A;C;D  1/29/1944  NaN  Yes  5,562,655,626.00
5               A;C  7/31/1965  NaN  Yes               NaN
6                 A   4/5/1979  NaN  NaN               NaN
7               C;D        NaN  NaN  Yes             33050
8             A;C;D  3/17/2000  NaN  Yes             33050
9             A;C;D  3/18/2000  NaN  Yes             33050
10              A;C  3/17/2000  NaN  Yes               NaN

或者按名称删除列:

df['consolidate_elem'] = (df.drop('consolidate_elem', axis=1).notna()
                            .dot(df.columns.drop('consolidate_elem') + ';')
                            .str.strip(';'))