如果所有行的列中只有一个值，则在Pandas数据框中折叠行

Question

我有以下DF

         col1  |  col2   | col3   | col4   | col5  | col6
    0    -     |   15.0  |  -     |  -     |   -   |  -
    1    -     |   -     |  -     |  -     |   -   |  US
    2    -     |   -     |  -     |  Large |   -   |  -
    3    ABC1  |   -     |  -     |  -     |   -   |  -
    4    -     |   -     |  24RA  |  -     |   -   |  -
    5    -     |   -     |  -     |  -     |   345 |  -

我想将行折叠成一个如下

    output DF:
         col1  |  col2    | col3   | col4   | col5  | col6
    0    ABC1  |   15.0   |  24RA  |  Large |   345 |  US

我不想遍历列，但想使用熊猫来实现这一点。

Answer 1

选项0
超级简单

pd.concat([pd.Series(df[c].dropna().values, name=c) for c in df], axis=1)

   col1  col2  col3   col4   col5 col6
0  ABC1  15.0  24RA  Large  345.0   US

---

每列可以处理多个值吗？
我们当然可以！

df.loc[2, 'col3'] = 'Test'

   col1  col2  col3   col4   col5 col6
0  ABC1  15.0  Test  Large  345.0   US
1   NaN   NaN  24RA    NaN    NaN  NaN

选项1
使用np.where像外科医生一样的通用解决方案

v = df.values
i, j = np.where(np.isnan(v))

s = pd.Series(v[i, j], df.columns[j])

c = s.groupby(level=0).cumcount()
s.index = [c, s.index]
s.unstack(fill_value='-')  # <-- don't fill to get NaN

   col1  col2  col3   col4 col5 col6
0  ABC1  15.0  24RA  Large  345   US

---

df.loc[2, 'col3'] = 'Test'

v = df.values
i, j = np.where(np.isnan(v))

s = pd.Series(v[i, j], df.columns[j])

c = s.groupby(level=0).cumcount()
s.index = [c, s.index]
s.unstack(fill_value='-')  # <-- don't fill to get NaN

   col1  col2  col3   col4 col5 col6
0  ABC1  15.0  Test  Large  345   US
1     -     -  24RA      -    -    -

---

选项2
mask 制作null然后stack摆脱它们

否则我们可以

# This should work even if `'-'` are NaN
# but you can skip the `.mask(df == '-')`
s = df.mask(df == '-').stack().reset_index(0, drop=True)
c = s.groupby(level=0).cumcount()
s.index = [c, s.index]
s.unstack(fill_value='-')

   col1  col2  col3   col4 col5 col6
0  ABC1  15.0  Test  Large  345   US
1     -     -  24RA      -    -    -