Question

我的数据如下所示。如果'Missing'不是'Married'，我想用'no_of_children'替换marital_status nan。

>cust_data_df[['marital_status','no_of_children']]
>

    marital_status  no_of_children

0   Married           NaN
1   Married           NaN
2   Missing            1
3   Missing            2
4   Single            NaN
5   Single            NaN
6   Married           NaN
7   Single            NaN
8   Married           NaN
9   Married           NaN
10  Single            NaN

这是我尝试过的：

cust_data_df.loc[cust_data_df['no_of_children'].notna()==True, 'marital_status'].replace({'Missing':'Married'},inplace=True)

但这没有做任何事情。

Answer 1

为避免chained assignments而分配替换值：

m = cust_data_df['no_of_children'].notna()
d = {'Missing':'Married'}
cust_data_df.loc[m, 'marital_status'] = cust_data_df.loc[m, 'marital_status'].replace(d)

如果需要设置所有值：

cust_data_df.loc[m, 'marital_status'] = 'Married'

编辑：

感谢@ Quickbeam2k1进行解释：

cust_data_df.loc [cust_data_df ['no_of_children']。notna（）==真的，'marital_status'] 只是一个没有引用的新对象。在那里替换，将保持原始对象不变

熊猫替换即使使用inplace = True也不会替换值

1 个答案: