有点晚了，但仍然-我更喜欢在以下地方使用numpy:

import numpy as np
df['First Season'] = np.where(df['First Season'] > 1990, 1, df['First Season'])

df['First Season'].loc[(df['First Season'] > 1990)] = 1

奇怪的是没有人知道这个答案，你的代码中唯一缺少的部分是df之后的['First Season']，只需删除里面的花括号。

我们可以用以下语法更新df中的First Season列:

df['First Season'] = expression_for_new_values

要映射First Season中的值，我们可以使用pandas的.map()方法，语法如下:

data_frame(['column']).map({'initial_value_1':'updated_value_1','initial_value_2':'updated_value_2'})

你也可以使用mask替换满足条件的值:

df['First Season'].mask(lambda col: col > 1990, 1)

df.loc[df['First season'] > 1990, 'First Season'] = 1

解释:

df。Loc有两个参数，'行索引'和'列索引'。我们检查“First season”列下的每一行值是否大于1990，然后将其替换为1。

对于单一条件，即。('employrate') > 70)

       country        employrate alcconsumption
0  Afghanistan  55.7000007629394            .03
1      Albania  51.4000015258789           7.29
2      Algeria              50.5            .69
3      Andorra                            10.17
4       Angola  75.6999969482422           5.57

用这个:

df.loc[df['employrate'] > 70, 'employrate'] = 7

       country  employrate alcconsumption
0  Afghanistan   55.700001            .03
1      Albania   51.400002           7.29
2      Algeria   50.500000            .69
3      Andorra         nan          10.17
4       Angola    7.000000           5.57

因此这里的语法是:

df.loc[<mask>(here mask is generating the labels to index) , <optional column(s)> ]

对于多种情况，即。(df['employrate'] <=55) & (df['employrate'] > 50)

用这个:

df['employrate'] = np.where(
   (df['employrate'] <=55) & (df['employrate'] > 50) , 11, df['employrate']
   )

out[108]:
       country  employrate alcconsumption
0  Afghanistan   55.700001            .03
1      Albania   11.000000           7.29
2      Algeria   11.000000            .69
3      Andorra         nan          10.17
4       Angola   75.699997           5.57

因此这里的语法是:

 df['<column_name>'] = np.where((<filter 1> ) & (<filter 2>) , <new value>, df['column_name'])