获取满足条件的单元格的列名称列表，并为 df 中的每一行存储在单独的列中答案

【问题标题】：Get list of columns names with cells satisfying a condition and store in separate column for each row in df获取满足条件的单元格的列名称列表，并为 df 中的每一行存储在单独的列中
【发布时间】：2023-01-16 19:31:18
【问题描述】：

我有一个 df，其结构类似于下面的数据，其中包含更多列和更多行。如何获得预期的结果。我的代码有一个错误——我认为列表不能保存在 df 单元格中？如果可能，如何避免循环？

data = [[0, 0, 1], [0, 1, 0], [1, 0, 0], [1, 0, 1], [0, 0, 0]]

df = pd.DataFrame(data, columns=["choice_a", "choice_b", "choice_c"])

Expected result

       choice_a  choice_b    choice_c  choices
index                  
0        0       0           1         ['c']
1        0       1           0         ['b']
2        1       0           0         ['a']
3        1.      0.          1         ['a','b']
4        0.      0.          0         NA

我的代码

df['choices']=0
for i in np.arange(df.shape[0]):
    choice_list = []
    for j in np.arange(len(df.columns)):
        if df.iloc[i,j]==1:
            choice_list.append(df.columns[j].split('_')[1])
    df.iloc[i,4]=choice_list

我得到的错误

---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
/var/folders/65/3mqr9fpn37jf2xt2pxbcgp_w0000gn/T/ipykernel_1513/2279334138.py in <module>
      5         if main_dataset.iloc[i,j]==1:
      6             choice_list.append(main_dataset.columns[j].split('_')[1])
----> 7     main_dataset.iloc[i,5]=choice_list

~/opt/anaconda3/lib/python3.9/site-packages/pandas/core/indexing.py in __setitem__(self, key, value)
    714 
    715         iloc = self if self.name == "iloc" else self.obj.iloc
--> 716         iloc._setitem_with_indexer(indexer, value, self.name)
    717 
    718     def _validate_key(self, key, axis: int):

~/opt/anaconda3/lib/python3.9/site-packages/pandas/core/indexing.py in _setitem_with_indexer(self, indexer, value, name)
   1689         if take_split_path:
   1690             # We have to operate column-wise
-> 1691             self._setitem_with_indexer_split_path(indexer, value, name)
   1692         else:
   1693             self._setitem_single_block(indexer, value, name)

~/opt/anaconda3/lib/python3.9/site-packages/pandas/core/indexing.py in _setitem_with_indexer_split_path(self, indexer, value, name)
   1744                     return self._setitem_with_indexer((pi, info_axis[0]), value[0])
   1745 
-> 1746                 raise ValueError(
   1747                     "Must have equal len keys and value "
   1748                     "when setting with an iterable"

ValueError: Must have equal len keys and value when setting with an iterable

【问题讨论】：

标签： python pandas

【解决方案1】：

利用：

cols = df.columns.str.split('_').str[1].to_numpy()
df['choices'] = [list(cols[x == 1]) for x in df.to_numpy()]

df['choices'] = df['choices'].where(df['choices'].astype(bool))
print (df)
   choice_a  choice_b  choice_c choices
0         0         0         1     [c]
1         0         1         0     [b]
2         1         0         0     [a]
3         1         0         1  [a, c]
4         0         0         0     NaN

【讨论】：