熊猫 - 案例何时&大熊猫默认

时间:2018-03-12 05:23:37

标签: python pandas

我在python中有以下case语句,

pd_df['difficulty'] = 'Unknown'
pd_df['difficulty'][(pd_df['Time']<30) & (pd_df['Time']>0)] = 'Easy'
pd_df['difficulty'][(pd_df['Time']>=30) & (pd_df['Time']<=60)] = 'Meduim'
pd_df['difficulty'][pd_df['Time']>60] = 'Hard'

但是当我运行代码时,它会抛出一个错误。

A value is trying to be set on a copy of a slice from a DataFrame

1 个答案:

答案 0 :(得分:6)

选项1
为了提高性能,请使用嵌套的np.where条件。对于条件,您可以使用pd.Series.between,并相应地插入默认值。

pd_df['difficulty'] = np.where(
     pd_df['Time'].between(0, 30, inclusive=False), 
    'Easy', 
     np.where(
        pd_df['Time'].between(0, 30, inclusive=False), 'Medium', 'Unknown'
     )
)

选项2
同样,使用np.select,这为添加条件提供了更多空间:

pd_df['difficulty'] = np.select(
    [
        pd_df['Time'].between(0, 30, inclusive=False), 
        pd_df['Time'].between(30, 60, inclusive=True)
    ], 
    [
        'Easy', 
        'Medium'
    ], 
    default='Unknown'
)

选项3
另一个高性能解决方案涉及loc

pd_df['difficulty'] = 'Unknown'
pd_df.loc[pd_df['Time'].between(0, 30, inclusive=False), 'difficulty'] = 'Easy'
pd_df.loc[pd_df['Time'].between(30, 60, inclusive=True), 'difficulty'] = 'Medium'