您可以使用pd.Series的apply和str.strip方法：In [13]: df
Out[13]:
a b c
0 dog quick the
1 lazy lazy fox
2 brown quick dog
3 quick the over
4 brown over lazy
5 fox brown quick
6 quick fox the
7 dog jumped the
8 lazy brown the
9 dog lazy the
In [14]: df = df + "@"
In [15]: df
Out[15]:
a b c
0 [email protected] [email protected] [email protected]
1 [email protected] [email protected] [email protected]
2 [email protected] [email protected] [email protected]
3 [email protected] [email protected] [email protected]
4 [email protected] [email protected] [email protected]
5 [email protected] [email protected] [email protected]
6 [email protected] [email protected] [email protected]
7 [email protected] [email protected] [email protected]
8 [email protected] [email protected] [email protected]
9 [email protected] [email protected] [email protected]
In [16]: df = df.apply(lambda S:S.str.strip('@'))
In [17]: df
Out[17]:
a b c
0 dog quick the
1 lazy lazy fox
2 brown quick dog
3 quick the over
4 brown over lazy
5 fox brown quick
6 quick fox the
7 dog jumped the
8 lazy brown the
9 dog lazy the
注意，您的方法不起作用，因为在for循环中执行以下分配时：row = row.str.rstrip('@')
这只是将row.str.strip的结果赋给名称row，而不改变DataFrame。对于所有python对象和简单的名称分配，这是相同的行为：In [18]: rows = [[1,2,3],[4,5,6],[7,8,9]]
In [19]: print(rows)
[[1, 2, 3], [4, 5, 6], [7, 8, 9]]
In [20]: for row in rows:
...: row = ['look','at','me']
...:
In [21]: print(rows)
[[1, 2, 3], [4, 5, 6], [7, 8, 9]]
要实际更改基础数据结构，需要使用mutator方法：In [22]: rows
Out[22]: [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
In [23]: for row in rows:
...: row.append("LOOKATME")
...:
In [24]: rows
Out[24]: [[1, 2, 3, 'LOOKATME'], [4, 5, 6, 'LOOKATME'], [7, 8, 9, 'LOOKATME']]
注意，对于mutator方法，slice赋值只是语法糖：In [26]: rows
Out[26]: [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
In [27]: for row in rows:
...: row[:] = ['look','at','me']
...:
...:
In [28]: rows
Out[28]: [['look', 'at', 'me'], ['look', 'at', 'me'], ['look', 'at', 'me']]
这类似于基于pandasloc或iloc的赋值。