将多个熊猫列合并为新列

Question

我有一个数据框，其中的一些列指示是否看到了一组调查问题。例如：

Q1_Seen    Q2_Seen    Q3_Seen    Q4_Seen
    Q1a        nan        nan        nan
    nan        Q2a        nan        nan
    nan        nan        Q3d        nan
    nan        Q2c        nan        nan

我想将这些列折叠为一列，假设Q_Seen，其形式为：

Q_Seen
   Q1a
   Q2a
   Q3d
   Q2c

请注意，每一行都是互斥的：如果其中一列中有值，则所有其他列都是 NaN。

我尝试用pd.concat这样做，但它似乎没有产生正确的结果。

Answer 1

试试这个：

df['Q_Seen'] = df.stack().values
>>> df
Q1_Seen    Q2_Seen    Q3_Seen     Q4_Seen     Q_Seen
    Q1a        nan        nan         nan        Q1a
    nan        Q2a        nan         nan        Q2a
    nan        nan        Q3d         nan        Q3d
    nan        Q2c        nan         nan        Q2c

Answer 2

使用按列max() -- 即 max(axis=1) -- 将允许您将所有值折叠到一列中：

In [1]: import pandas as pd
In [2]: df = pd.DataFrame({"Q1_Seen": ['Q1a', None, None, None], "Q2_Seen": [None, "Q2a", None, "Q2c"], "Q3_Seen": [None, None, "Q3d", None],"Q4_Seen": [None, None, None, None]})
In [3]: df
Out[3]: 
  Q1_Seen Q2_Seen Q3_Seen Q4_Seen
0     Q1a    None    None    None
1    None     Q2a    None    None
2    None    None     Q3d    None
3    None     Q2c    None    None
In [4]: df['Q_Seen'] = df.max(axis=1)
In [5]: df
Out[5]: 
  Q1_Seen Q2_Seen Q3_Seen Q4_Seen Q_Seen
0     Q1a    None    None    None    Q1a
1    None     Q2a    None    None    Q2a
2    None    None     Q3d    None    Q3d
3    None     Q2c    None    None    Q2c

Answer 3

以下内容对我有用：

df = pd.DataFrame({'Q1': [1, None, None], 'Q2': [None, 2, None], 'Q3': [None, None, 3]})
df['Q'] = df.concat([df['Q1'], df['Q2'], df['Q3']]).dropna()

可能有一个更优雅的解决方案，但这是我首先想到的。