使用Index.repeat乘以列差，加上1，重复DataFrame.loc，最后将计数器列乘以GroupBy.cumcount加到min列：

df1 = df.loc[df.index.repeat(df['max'].sub(df['min']) + 1)]
        
df1['value'] = df1['min'].add(df1.groupby(level=0).cumcount())
print (df1)
             ID  min  max  value
0   Product ABC    3    6      3
0   Product ABC    3    6      4
0   Product ABC    3    6      5
0   Product ABC    3    6      6
1  Product XXXX    8   10      8
1  Product XXXX    8   10      9
1  Product XXXX    8   10     10

创建从min到max的范围并分解它们（使用df.explode）：

df['value'] = df.apply(lambda x: range(x['min'], x['max'] + 1), axis=1)
df = df.explode('value', ignore_index=True)

ID  min  max value
0   Product ABC     3    6     3
1   Product ABC     3    6     4
2   Product ABC     3    6     5
3   Product ABC     3    6     6
4  Product XXXX     8   10     8
5  Product XXXX     8   10     9
6  Product XXXX     8   10    10

您可以使用Index.repeat复制行，然后使用groupby.cumcount复制新列：

rep = df['max'].sub(df['min']).add(1)

out = (df.loc[df.index.repeat(rep)]
         .assign(value=lambda d: d.groupby('ID').cumcount().add(d['min']))
      )

输出：

ID  min  max  value
0   Product ABC    3    6      3
0   Product ABC    3    6      4
0   Product ABC    3    6      5
0   Product ABC    3    6      6
1  Product XXXX    8   10      8
1  Product XXXX    8   10      9
1  Product XXXX    8   10     10

中间体：

ID  min  max  value  rep  cumcount
0   Product ABC    3    6      3    4         0
0   Product ABC    3    6      4    4         1
0   Product ABC    3    6      5    4         2
0   Product ABC    3    6      6    4         3
1  Product XXXX    8   10      8    3         0
1  Product XXXX    8   10      9    3         1
1  Product XXXX    8   10     10    3         2