基于时间轴递增的局部最小图像过滤数据帧
问题描述
编辑:
我有以下学生数据框,显示他们在不同日期的考试成绩(已排序):
df = pd.DataFrame({'student': 'A A A B B B B C C'.split(),
'exam_date':[datetime.datetime(2013,4,1),datetime.datetime(2013,6,1),
datetime.datetime(2013,7,1),datetime.datetime(2013,9,2),
datetime.datetime(2013,10,1),datetime.datetime(2013,11,2),
datetime.datetime(2014,2,2),datetime.datetime(2013,7,1),
datetime.datetime(2013,9,2),],
'score': [15, 17, 32, 22, 28, 24, 33, 33, 15]})
print(df)
student exam_date score
0 A 2013-04-01 15
1 A 2013-06-01 17
2 A 2013-07-01 32
3 B 2013-09-02 22
4 B 2013-10-01 28
5 B 2013-11-02 24
6 B 2014-02-02 33
7 C 2013-07-01 33
8 C 2013-09-02 15
我只需要保留分数从局部最小值增加了10以上的那些行。
例如,对于学生A
,局部最小值为15
,而分数在下一个最新数据中增加到32
,因此我们将保留该值。
对于学生B
,分数不会从局部极小值增加超过10
。28-22
和33-24
均小于10
。
对于学生C
,局部最小值为15
,但分数在此之后不会增加,因此我们将删除该分数。
我正在尝试以下脚本:
out = df[df['score'] - df.groupby('student', as_index=False)['score'].cummin()['score']>= 10]
print(out)
2 A 2013-07-01 32
6 B 2014-02-02 33 #--Shouldn't capture this as it's increased by `9` from local minima of `24`
所需输出:
student exam_date score
2 A 2013-07-01 32
# For A, score of 32 is increased by 17 from local minima of 15
做这件事最聪明的方式是什么?如有任何建议,我们将不胜感激。谢谢!
解决方案
假定您的数据帧已按日期排序:
highest_score = lambda x: x['score'] - x['score'].mask(x['score'].gt(x['score'].shift())).ffill() > 10
out = df[df.groupby('student').apply(highest_score).droplevel(0)]
print(out)
# Output
student exam_date score
2 A 2013-07-01 32
关注lambda函数
让我们修改您的数据帧并提取一个学生以避免groupby
:
>>> df = df[df['student'] == 'B']
student exam_date score
3 B 2013-09-02 22
4 B 2013-10-01 28
5 B 2013-11-02 24
6 B 2014-02-02 33
# Step-1: find row where value is not a local minima
>>> df['score'].gt(df['score'].shift())
3 False
4 True
5 False
6 True
Name: score, dtype: bool
# Step-2: hide non local minima values
>>> df['score'].mask(df['score'].gt(df['score'].shift()))
3 22.0
4 NaN
5 24.0
6 NaN
Name: score, dtype: float64
# Step-3: fill forward local minima values
>>> df['score'].mask(df['score'].gt(df['score'].shift()))
3 22.0
4 22.0
5 24.0
6 24.0
Name: score, dtype: float64
# Step-4: check if the condition is True
>>> df['score'] - df['score'].mask(df['score'].gt(df['score'].shift())) > 10
3 False
4 False
5 False
6 False
Name: score, dtype: bool
相关文章