Question

我有一个熊猫数据框，其格式如下：

id,atr1,atr2,orig_date,fix_date
1,bolt,l,2000-01-01,nan
1,screw,l,2000-01-01,nan
1,stem,l,2000-01-01,nan
2,stem,l,2000-01-01,nan
2,screw,l,2000-01-01,nan
2,stem,l,2001-01-01,2001-01-01
3,bolt,r,2000-01-01,nan
3,stem,r,2000-01-01,nan
3,bolt,r,2001-01-01,2001-01-01
3,stem,r,2001-01-01,2001-01-01

结果如下：

id,atr1,atr2,orig_date,fix_date,failed_part_ind
1,bolt,l,2000-01-01,nan,0
1,screw,l,2000-01-01,nan,0
1,stem,l,2000-01-01,nan,0
2,stem,l,2000-01-01,nan,1
2,screw,l,2000-01-01,nan,0
2,stem,l,2001-01-01,2001-01-01,0
3,bolt,r,2000-01-01,nan,1
3,stem,r,2000-01-01,nan,1
3,bolt,r,2001-01-01,2001-01-01,0
3,stem,r,2001-01-01,2001-01-01,0

最欢迎任何提示或技巧！

Update2：

一种描述我需要完成的工作的更好方法是，在.groupby(['id','atr1','atr2'])中创建一个新的指标列，该列满足组中记录的以下条件：

(df['orig_date'] < df['fix_date'])

Answer 1

我认为这应该可行：

df['failed_part_ind'] = df.apply(lambda row: 1 if ((row['id'] == row['id']) &
                                                (row['atr1'] == row['atr1']) &
                                                (row['atr2'] == row['atr2']) &
                                                (row['orig_date'] < row['fix_date']))
                                            else 0, axis=1)

更新：我认为这是您想要的：

import numpy as np
def f(g):
    min_fix_date = g['fix_date'].min()
    if np.isnan(min_fix_date):
        g['failed_part_ind'] = 0
    else:
        g['failed_part_ind'] = g['orig_date'].apply(lambda d: 1 if d < min_fix_date else 0)
    return g

df.groupby(['id', 'atr1', 'atr2']).apply(lambda g: f(g))

带有布尔掩码的Python Groupby

1 个答案: