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大熊猫:检查特定的一天是否以固定的时间间隔在索引中,如果没有,在输入前标记为某物?

pandas python

Sid · 技术社区 · 6 年前

我对大熊猫还是很陌生,只是发现我在之前跟踪的过程中犯了一个错误。

 df_date
         Date        day
0  2016-05-26   Thursday
1  2016-05-27     Friday
2  2016-05-30     Monday
3  2016-05-31    Tuesday
4  2016-06-01  Wednesday
5  2016-06-02   Thursday
6  2016-06-03     Friday
7  2016-06-06     Monday
8  2016-06-07    Tuesday
9  2016-06-08  Wednesday
10 2016-06-09   Thursday
11 2016-06-10     Friday
12 2016-06-13     Monday
13 2016-06-14    Tuesday
14 2016-06-15  Wednesday
15 2016-06-16   Thursday
16 2016-06-17     Friday
17 2016-06-20     Monday
18 2016-06-21    Tuesday
19 2016-06-22  Wednesday
20 2016-06-24     Friday
21 2016-06-27     Monday
22 2016-06-28    Tuesday
23 2016-06-29  Wednesday

大约有600多行。

我想做什么

制作专栏 'Exit' 如果在哪里 thursday 不在一周内,星期三变为E,如果星期三不在,那么星期二。

我尝试了一个for循环,但我似乎不能把这个正确。

预期输出:

 df_date
         Date        day  Exit
0  2016-05-26   Thursday  E
1  2016-05-27     Friday  
2  2016-05-30     Monday
3  2016-05-31    Tuesday
4  2016-06-01  Wednesday
5  2016-06-02   Thursday  E
6  2016-06-03     Friday
7  2016-06-06     Monday
8  2016-06-07    Tuesday
9  2016-06-08  Wednesday
10 2016-06-09   Thursday  E
11 2016-06-10     Friday
12 2016-06-13     Monday
13 2016-06-14    Tuesday
14 2016-06-15  Wednesday
15 2016-06-16   Thursday  E
16 2016-06-17     Friday
17 2016-06-20     Monday
18 2016-06-21    Tuesday
19 2016-06-22  Wednesday  E
20 2016-06-24     Friday
21 2016-06-27     Monday
22 2016-06-28    Tuesday
23 2016-06-29  Wednesday  E

我在评论中添加了这个,但也应该在这里:

如果星期四不存在,那么就在它之前记录。

所以如果星期三也不在,那么星期二

如果星期二也不是星期一,如果星期一不是星期五。星期六和星期天永远不会有记录。

2 回复 | 直到 6 年前

yatu Sayali Sonawane 6 年前

以下是解决方案:

ix = df.groupby(pd.Grouper(key='Date', freq='W')).Date
       .apply(lambda x: (x.dt.dayofweek <= 3)[::-1].idxmax()).values
df.loc[ix,'Exit'] = 'E'
df.fillna('')

      Date        day     Exit
0  2016-05-26   Thursday    E
1  2016-05-27     Friday     
2  2016-05-30     Monday     
3  2016-05-31    Tuesday     
4  2016-06-01  Wednesday     
5  2016-06-02   Thursday    E
6  2016-06-03     Friday     
7  2016-06-06     Monday     
8  2016-06-07    Tuesday     
9  2016-06-08  Wednesday     
10 2016-06-09   Thursday    E
11 2016-06-10     Friday     
12 2016-06-13     Monday     
13 2016-06-14    Tuesday     
14 2016-06-15  Wednesday     
15 2016-06-16   Thursday    E
16 2016-06-17     Friday     
17 2016-06-20     Monday     
18 2016-06-21    Tuesday     
19 2016-06-22  Wednesday     
20 2016-06-23   Thursday    E
21 2016-06-24     Friday     
22 2016-06-27     Monday     
23 2016-06-28    Tuesday     
24 2016-06-29  Wednesday    E

jpp 6 年前

你可以使用 dt.week 和 dt.weekday 您的属性 datetime 系列。然后使用 groupby + max 为了你需要的逻辑。这可能比顺序的平等检查更有效。

df['Date'] = pd.to_datetime(df['Date'])

# add week and weekday series
df['Week'] = df['Date'].dt.week
df['Weekday'] = df['Date'].dt.weekday.where(df['Date'].dt.weekday.isin([1, 2, 3]))

df['Exit'] = np.where(df['Weekday'] == df.groupby('Week')['Weekday'].transform('max'),
                      'E', '')

结果

我已经离开了helper列,所以解决方案的工作方式是明确的。这些可以很容易地去除。

print(df)

         Date        day  Week  Weekday Exit
0  2016-05-26   Thursday    21      3.0    E
1  2016-05-27     Friday    21      NaN     
2  2016-05-30     Monday    22      NaN     
3  2016-05-31    Tuesday    22      1.0     
4  2016-06-01  Wednesday    22      2.0     
5  2016-06-02   Thursday    22      3.0    E
6  2016-06-03     Friday    22      NaN     
7  2016-06-06     Monday    23      NaN     
8  2016-06-07    Tuesday    23      1.0     
9  2016-06-08  Wednesday    23      2.0     
10 2016-06-09   Thursday    23      3.0    E
11 2016-06-10     Friday    23      NaN     
12 2016-06-13     Monday    24      NaN     
13 2016-06-14    Tuesday    24      1.0     
14 2016-06-15  Wednesday    24      2.0     
15 2016-06-16   Thursday    24      3.0    E
16 2016-06-17     Friday    24      NaN     
17 2016-06-20     Monday    25      NaN     
18 2016-06-21    Tuesday    25      1.0     
19 2016-06-22  Wednesday    25      2.0    E
20 2016-06-24     Friday    25      NaN     
21 2016-06-27     Monday    26      NaN     
22 2016-06-28    Tuesday    26      1.0     
23 2016-06-29  Wednesday    26      2.0    E