不如@jezrael的解决方案那么快.但优雅:-)

Not as fast as @jezrael's solution. But elegant :-)

apply与pd.Series

df.a.apply(pd.Series)

   0  1  2  3  4  5
0  0  1  2  3  4  5
1  0  1  2  3  4  5

或

df.a.apply(pd.Series, index=list('abcdef'))

   a  b  c  d  e  f
0  0  1  2  3  4  5
1  0  1  2  3  4  5