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 Pandas 来实现这一点。
最佳答案
选项0
super 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
像外科医生一样使用
np.where
的广义解决方案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
设置为null,然后stack
设置为null否则我们可能会
# 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 - - -