使用concat的一个选项：

x = [16,17,18,19]

last_max = int(df.loc['s'].index.max())

tmp = pd.DataFrame({'I1': x})
tmp.index = (tmp.index+last_max).astype(str)

out = (pd.concat([df, pd.concat([tmp], keys='s')])
         .sort_index(level=0, sort_remaining=False, kind='stable')
      )

输出：

I1   I2
c 1  10  5.0
  2  22  1.0
  3  33  1.0
  4  44  1.0
  5  55  1.0
s 1  11  0.0
  2  12  1.0
  3  13  2.0
  4  14  3.0
  5  15  4.0
  5  16  NaN
  6  17  NaN
  7  18  NaN
  8  19  NaN

@cs95建议的替代方案：

pd.concat([df,
           pd.DataFrame(x, columns=['I1'],
                        index=pd.MultiIndex.from_product([['s'],np.arange(1,len(x)+1)
                                                          + int(df.loc['s'].index[-1])
                                                         ]))
          ]).sort_index()

可以使用reindex为索引的新值插入新索引，然后使用loc将新值分配给I1

x = [16, 17, 18, 19]
s_last = int(df.loc["s"].index[-1])
new_idx = pd.MultiIndex.from_product(
    [["s"], map(str, range(s_last + 1, s_last + len(x) + 1))]
)
df = df.reindex(df.index.union(new_idx))
df.loc[new_idx, "I1"] = x

print(df)

       I1   I2
c 1  10.0  5.0
  2  22.0  1.0
  3  33.0  1.0
  4  44.0  1.0
  5  55.0  1.0
s 1  11.0  0.0
  2  12.0  1.0
  3  13.0  2.0
  4  14.0  3.0
  5  15.0  4.0
  6  16.0  NaN
  7  17.0  NaN
  8  18.0  NaN
  9  19.0  NaN