这种操作非常适合lambda表达式。
类似这样的方法应该可以奏效：
df0['colX'] = df0.apply(lambda x: '-'.join(c for c in df0.columns if x[c] != 0), axis=1).replace('', 0)

• 首先它得到一个不为0的列的列表
• 用“-”连接这些列的名称
• 之后，用0填充空白名称

您可以尝试使用dot

df0['new'] = df0.ne(0).dot(df0.columns+'_').str[:-1]
df0
Out[9]: 
   points  assists  numbers                     new
0       0        0        0                        
1       0        0        0                        
2      -3        2        1  points_assists_numbers
3      16        0        6          points_numbers
4       0        1       10         assists_numbers
5       5       -7        5  points_assists_numbers
6      -3        0        8          points_numbers
7      14        6        7  points_assists_numbers

以下是一些选项：

#keep only the values that are not 0 and stack. reset the index so it can be joined by '-' on the index level.
df0['new'] = df0.where(df0.ne(0)).stack().reset_index(level=1).groupby(level=0)['level_1'].agg('-'.join)

或

#similar to the solution above, but multiplies the column value to avoid resetting index
df0['new'] = df0.ne(0).mul(df0.columns).where(lambda x: x.ne('')).stack().groupby(level=0).agg('-'.join)

或

#multiplies the column names by values that are not 0, and adds them together.
df0['new'] = df0.ne(0).dot(df0.columns + '-').str.rstrip('-')