如果选择列输出的第一个值是单值-这里是字符串。所以不能使用标量的Pandas函数。

df = pd.DataFrame({'cc':['a','b']})
                         
print (df.cc.iloc[0])
a

如果确实需要一个元素Series，则使用双[]：

a = df.cc.iloc[[0]]
print (a)
0    a
Name: cc, dtype: object

一个元素DataFrame：

a = df[['cc']].iloc[[0]]
print (a)
  cc
0  a

由DataFrame构造函数生成：

a = df.cc.iloc[0]
print (a)
a

df1 = pd.DataFrame({'col':[a]})
print (df1)
  col
0   a

如果填充cc列的每个值完全新的DataFrame这不是最佳做法，但如果选择第一个值返回DataFrame：

def qwe(y,z): 
    df = pd.DataFrame({"one":range(1,10), "two":range(2,11), "three":range(3,12)}) 
    df["four"] = (df.one * (y+z)) + (df.two * (y+z)) + (df.three * (y+z)) 
    return df 

fr = pd.DataFrame(columns=["a", "b"]) 

for aa in range(1, 5): 
    for bb in range(1, 5): 
        x = [aa,bb] 
        fr.loc[len(fr)] = x

#for each value in cc is set DataFrame
fr["cc"] = fr.apply(lambda x: qwe(x['a'], x['b']), axis=1) 
print(fr.cc)
0        one  two  three  four
0    1    2      3   ...
1        one  two  three  four
0    1    2      3   ...
2        one  two  three  four
0    1    2      3   ...
3        one  two  three  four
0    1    2      3   ...
4        one  two  three  four
0    1    2      3   ...
5        one  two  three  four
0    1    2      3   ...
6        one  two  three  four
0    1    2      3   ...
7        one  two  three  four
0    1    2      3   ...
8        one  two  three  four
0    1    2      3   ...
9        one  two  three  four
0    1    2      3   ...
10       one  two  three  four
0    1    2      3   ...
11       one  two  three  four
0    1    2      3   ...
12       one  two  three  four
0    1    2      3   ...
13       one  two  three  four
0    1    2      3   ...
14       one  two  three  four
0    1    2      3   ...
15       one  two  three  four
0    1    2      3   ...
Name: cc, dtype: object

print(fr.cc.iloc[0])
   one  two  three  four
0    1    2      3    12
1    2    3      4    18
2    3    4      5    24
3    4    5      6    30
4    5    6      7    36
5    6    7      8    42
6    7    8      9    48
7    8    9     10    54
8    9   10     11    60

编辑：如果需要从fr.cc列加入每个DataFrame，请使用concat和列表解析：

out = pd.concat([x for x in fr.cc])
print(out)
    one  two  three  four
0     1    2      3    12
1     2    3      4    18
2     3    4      5    24
3     4    5      6    30
4     5    6      7    36
..  ...  ...    ...   ...
4     5    6      7   144
5     6    7      8   168
6     7    8      9   192
7     8    9     10   216
8     9   10     11   240

[144 rows x 4 columns]

如果混合cc列- Dataframe 与字符串：

out = pd.concat([x for x in fr.cc if isinstance(x, pd.DataFrame)])
print(out)
    one  two  three  four
0     1    2      3    12
1     2    3      4    18
2     3    4      5    24
3     4    5      6    30
4     5    6      7    36
..  ...  ...    ...   ...
4     5    6      7   144
5     6    7      8   168
6     7    8      9   192
7     8    9     10   216
8     9   10     11   240

[144 rows x 4 columns]