python - 如何替换匹配条件的选定行的列中的字符串

Question

我有一个数据帧，需要从中替换 sense 列中的所有字符串，其中该字符串恰好所属。我可以应用以下代码来执行相同的操作

pd[pd['sense'] == 'pertain']['sense'].str.replace('pertain','pertaining')

但这不会对原始数据框生效。如何确保效果在原始数据集上发生，并且列中的所有 pertain 都将更改为 pertaining

 | affix    | word           |   sense       | meaning                         |
0   ical    neuroanatomical     pertaining   of or pertaining to [[neuroanatomy]]
1   ical    neuroanatomical     pertain      of or pertaining to [[neuroanatomy]]
2   ical    biotechnological    pertaining   of or pertaining to [[biotechnology]]
3   ical    biotechnological    pertain      of or pertaining to [[biotechnology]]
4   ical    educological        relating     relating to [[educology]].

所以理想情况下，操作后我的输出应如下所示

 | affix    | word           |   sense       | meaning                         |
0   ical    neuroanatomical     pertaining   of or pertaining to [[neuroanatomy]]
1   ical    neuroanatomical     pertaining   of or pertaining to [[neuroanatomy]]
2   ical    biotechnological    pertaining   of or pertaining to [[biotechnology]]
3   ical    biotechnological    pertaining   of or pertaining to [[biotechnology]]
4   ical    educological        relating     relating to [[educology]].

Answer 1

您可以将 loc 与赋值一起使用，在其中传递一个逻辑向量来指示应替换列 sense 值的行(假设您的数据框名为df):

df.loc[df.sense == "pertain", "sense"] = "pertaining"

df
#  affix                word         sense   meaning
#0  ical     neuroanatomical    pertaining  of or pertaining to [[neuroanatomy]]
#1  ical     neuroanatomical    pertaining  of or pertaining to [[neuroanatomy]]
#2  ical    biotechnological    pertaining  of or pertaining to [[biotechnology]]
#3  ical    biotechnological    pertaining  of or pertaining to [[biotechnology]]
#4  ical        educological    relating    relating to [[educology]].

<小时/>

另一种方法是仅在 sense 列上使用 replace，然后将其分配回原始数据框，从而替换 sense 列修改后的版本:

df['sense'] = df.sense.replace("pertain", "pertaining")

<小时/>

综上所述，您的方法也有效，您只需将其分配回原始数据框，如上面的第二个选项。