英文:
New column based on last time row value equals some numbers in Pandas dataframe
问题
我已经将您提供的内容翻译成中文，以下是代码部分的中文翻译：
# 创建一个名为Recent_Predicted_Score的列，记录学生在前3名中的最近预测分数
df.sort_values(by=['Student_ID', 'Date'], ascending=[True, False], inplace=True)
lp1 = df['Predicted_Score'].where(df['Rank'].isin([1, 2, 3])).groupby(df['Student_ID']).bfill()
lp2 = df.groupby(['Student_ID', 'Rank'])['Predicted_Score'].shift(-1)
df = df.assign(Recent_Predicted_Score=lp1.mask(df['Rank'].isin([1, 2, 3]), lp2))
请注意，上述代码是用于创建"Recent_Predicted_Score"列的Python代码，用于记录学生在前三名中的最近预测分数。如果您有任何其他问题或需要进一步的帮助，请随时告诉我。
英文:
I have a dataframe sorted in descending order date that records the Rank of students in class and the predicted score.
Date Student_ID Rank Predicted_Score
4/7/2021 33 2 87
13/6/2021 33 4 88
31/3/2021 33 7 88
28/2/2021 33 2 86
14/2/2021 33 10 86
31/1/2021 33 8 86
23/12/2020 33 1 81
8/11/2020 33 3 80
21/10/2020 33 3 80
23/9/2020 33 4 80
20/5/2020 33 3 80
29/4/2020 33 4 80
15/4/2020 33 2 79
26/2/2020 33 3 79
12/2/2020 33 5 79
29/1/2020 33 1 70
I want to create a column called Recent_Predicted_Score that record the last predicted_score where that student actually ranks top 3. So the desired outcome looks like
Date Student_ID Rank Predicted_Score Recent_Predicted_Score
4/7/2021 33 2 87 86
13/6/2021 33 4 88 86
31/3/2021 33 7 88 86
28/2/2021 33 2 86 81
14/2/2021 33 10 86 81
31/1/2021 33 8 86 81
23/12/2020 33 1 81 80
8/11/2020 33 3 80 80
21/10/2020 33 3 80 80
23/9/2020 33 4 80 80
20/5/2020 33 3 80 79
29/4/2020 33 4 80 79
15/4/2020 33 2 79 79
26/2/2020 33 3 79 70
12/2/2020 33 5 79 70
29/1/2020 33 1 70
Here's what I have tried but it doesn't quite work, not sure if I am on the right track:
df.sort_values(by = ['Student_ID', 'Date'], ascending = [True, False], inplace = True)
lp1 = df['Predicted_Score'].where(df['Rank'].isin([1,2,3])).groupby(df['Student_ID']).bfill()
lp2 = df.groupby(['Student_ID', 'Rank'])['Predicted_Score'].shift(-1)
df = df.assign(Recent_Predicted_Score=lp1.mask(df['Rank'].isin([1,2,3]), lp2))
Thanks in advance.
答案1
得分: 2
尝试：
df['Date'] = pd.to_datetime(df['Date'], dayfirst=True)
df = df.sort_values(['Student_ID', 'Date'])
df['Recent_Predicted_Score'] = np.where(df['Rank'].isin([1, 2, 3]), df['Predicted_Score'], np.nan)
df['Recent_Predicted_Score'] = df.groupby('Student_ID', group_keys=False)['Recent_Predicted_Score'].apply(lambda x: x.ffill().shift().fillna(''))
df = df.sort_values(['Student_ID', 'Date'], ascending=[True, False])
print(df)
打印结果：
Date Student_ID Rank Predicted_Score Recent_Predicted_Score
0 2021-07-04 33 2 87 86.0
1 2021-06-13 33 4 88 86.0
2 2021-03-31 33 7 88 86.0
3 2021-02-28 33 2 86 81.0
4 2021-02-14 33 10 86 81.0
5 2021-01-31 33 8 86 81.0
6 2020-12-23 33 1 81 80.0
7 2020-11-08 33 3 80 80.0
8 2020-10-21 33 3 80 80.0
9 2020-09-23 33 4 80 80.0
10 2020-05-20 33 3 80 79.0
11 2020-04-29 33 4 80 79.0
12 2020-04-15 33 2 79 79.0
13 2020-02-26 33 3 79 70.0
14 2020-02-12 33 5 79 70.0
15 2020-01-29 33 1 70
英文:
Try:
df['Date'] = pd.to_datetime(df['Date'], dayfirst=True)
df = df.sort_values(['Student_ID', 'Date'])
df['Recent_Predicted_Score'] = np.where(df['Rank'].isin([1, 2, 3]), df['Predicted_Score'], np.nan)
df['Recent_Predicted_Score'] = df.groupby('Student_ID', group_keys=False)['Recent_Predicted_Score'].apply(lambda x: x.ffill().shift().fillna(''))
df = df.sort_values(['Student_ID', 'Date'], ascending = [True, False])
print(df)
Prints:
Date Student_ID Rank Predicted_Score Recent_Predicted_Score
0 2021-07-04 33 2 87 86.0
1 2021-06-13 33 4 88 86.0
2 2021-03-31 33 7 88 86.0
3 2021-02-28 33 2 86 81.0
4 2021-02-14 33 10 86 81.0
5 2021-01-31 33 8 86 81.0
6 2020-12-23 33 1 81 80.0
7 2020-11-08 33 3 80 80.0
8 2020-10-21 33 3 80 80.0
9 2020-09-23 33 4 80 80.0
10 2020-05-20 33 3 80 79.0
11 2020-04-29 33 4 80 79.0
12 2020-04-15 33 2 79 79.0
13 2020-02-26 33 3 79 70.0
14 2020-02-12 33 5 79 70.0
15 2020-01-29 33 1 70
答案2
得分: 1
让我们假设：
- 可能存在多个唯一的
Student_ID - 行按照 OP 指定的降序
Date排序，但不一定按Student_ID排序 - 我们希望保留原始数据框的索引
在满足这些假设的情况下，以下是实现您问题要求的方法：
df['Recent_Predicted_Score'] = df.loc[df['Rank'] <= 3, 'Predicted_Score']
df['Recent_Predicted_Score'] = ( df
.groupby('Student_ID', sort=False)
.apply(lambda group: group['Predicted_Score'].shift(-1).bfill())
['Recent_Predicted_Score'] )
解释：
- 创建一个新列
Recent_Predicted_Score，其中包含Rank在前三名的行的Predicted_Score，其他行为 NaN - 使用
groupby()根据Student_ID分组，将sort参数设置为 False 以提高性能（注意，groupby()保留每个组内行的顺序，具体而言，不影响现有的按Date降序排序的顺序） - 在每个组内，使用
shift(-1)和bfill()来获得Recent_Predicted_Score的期望结果。
以下是示例输入（包含两个不同的 Student_ID 值）和输出，已根据 Student_ID, Date 进行排序以便更轻松检查：
（示例输入和输出数据已省略）
希望这对您有所帮助。
英文:
Let's assume:
- there may be more than one unique
Student_ID - the rows are ordered by descending
Dateas indicated by OP, but may not be ordered byStudent_ID - we want to preserve the index of the original dataframe
Subject to these assumptions, here's a way to do what your question asks:
df['Recent_Predicted_Score'] = df.loc[df.Rank <= 3, 'Predicted_Score']
df['Recent_Predicted_Score'] = ( df
.groupby('Student_ID', sort=False)
.apply(lambda group: group.shift(-1).bfill())
['Recent_Predicted_Score'] )
Explanation:
- create a new column
Recent_Predicted_Scorecontaining thePredictedScorewhereRankis in the top 3 and NaN otherwise - use
groupby()onStudent_IDwith thesortargument set to False for better performance (note thatgroupby()preserves the order of rows within each group, specifically, not influencing the existing descending order byDate) - within each group, do
shift(-1)andbfill()to get the desired result forRecent_Predicted_Score.
Sample input (with two distinct Student_ID values):
Date Student_ID Rank Predicted_Score
0 2021-07-04 33 2 87
1 2021-07-04 66 2 87
2 2021-06-13 33 4 88
3 2021-06-13 66 4 88
4 2021-03-31 33 7 88
5 2021-03-31 66 7 88
6 2021-02-28 33 2 86
7 2021-02-28 66 2 86
8 2021-02-14 33 10 86
9 2021-02-14 66 10 86
10 2021-01-31 33 8 86
11 2021-01-31 66 8 86
12 2020-12-23 33 1 81
13 2020-12-23 66 1 81
14 2020-11-08 33 3 80
15 2020-11-08 66 3 80
16 2020-10-21 33 3 80
17 2020-10-21 66 3 80
18 2020-09-23 33 4 80
19 2020-09-23 66 4 80
20 2020-05-20 33 3 80
21 2020-05-20 66 3 80
22 2020-04-29 33 4 80
23 2020-04-29 66 4 80
24 2020-04-15 33 2 79
25 2020-04-15 66 2 79
26 2020-02-26 33 3 79
27 2020-02-26 66 3 79
28 2020-02-12 33 5 79
29 2020-02-12 66 5 79
30 2020-01-29 33 1 70
31 2020-01-29 66 1 70
Output:
Date Student_ID Rank Predicted_Score Recent_Predicted_Score
0 2021-07-04 33 2 87 86.0
1 2021-07-04 66 2 87 86.0
2 2021-06-13 33 4 88 86.0
3 2021-06-13 66 4 88 86.0
4 2021-03-31 33 7 88 86.0
5 2021-03-31 66 7 88 86.0
6 2021-02-28 33 2 86 81.0
7 2021-02-28 66 2 86 81.0
8 2021-02-14 33 10 86 81.0
9 2021-02-14 66 10 86 81.0
10 2021-01-31 33 8 86 81.0
11 2021-01-31 66 8 86 81.0
12 2020-12-23 33 1 81 80.0
13 2020-12-23 66 1 81 80.0
14 2020-11-08 33 3 80 80.0
15 2020-11-08 66 3 80 80.0
16 2020-10-21 33 3 80 80.0
17 2020-10-21 66 3 80 80.0
18 2020-09-23 33 4 80 80.0
19 2020-09-23 66 4 80 80.0
20 2020-05-20 33 3 80 79.0
21 2020-05-20 66 3 80 79.0
22 2020-04-29 33 4 80 79.0
23 2020-04-29 66 4 80 79.0
24 2020-04-15 33 2 79 79.0
25 2020-04-15 66 2 79 79.0
26 2020-02-26 33 3 79 70.0
27 2020-02-26 66 3 79 70.0
28 2020-02-12 33 5 79 70.0
29 2020-02-12 66 5 79 70.0
30 2020-01-29 33 1 70 NaN
31 2020-01-29 66 1 70 NaN
Output sorted by Student_ID, Date for easier inspection:
Date Student_ID Rank Predicted_Score Recent_Predicted_Score
0 2021-07-04 33 2 87 86.0
2 2021-06-13 33 4 88 86.0
4 2021-03-31 33 7 88 86.0
6 2021-02-28 33 2 86 81.0
8 2021-02-14 33 10 86 81.0
10 2021-01-31 33 8 86 81.0
12 2020-12-23 33 1 81 80.0
14 2020-11-08 33 3 80 80.0
16 2020-10-21 33 3 80 80.0
18 2020-09-23 33 4 80 80.0
20 2020-05-20 33 3 80 79.0
22 2020-04-29 33 4 80 79.0
24 2020-04-15 33 2 79 79.0
26 2020-02-26 33 3 79 70.0
28 2020-02-12 33 5 79 70.0
30 2020-01-29 33 1 70 NaN
1 2021-07-04 66 2 87 86.0
3 2021-06-13 66 4 88 86.0
5 2021-03-31 66 7 88 86.0
7 2021-02-28 66 2 86 81.0
9 2021-02-14 66 10 86 81.0
11 2021-01-31 66 8 86 81.0
13 2020-12-23 66 1 81 80.0
15 2020-11-08 66 3 80 80.0
17 2020-10-21 66 3 80 80.0
19 2020-09-23 66 4 80 80.0
21 2020-05-20 66 3 80 79.0
23 2020-04-29 66 4 80 79.0
25 2020-04-15 66 2 79 79.0
27 2020-02-26 66 3 79 70.0
29 2020-02-12 66 5 79 70.0
31 2020-01-29 66 1 70 NaN
答案3
得分: 1
将排名大于3的分数进行屏蔽，然后按学生ID分组，向后填充以传播最后一个预测分数
c = 'Recent_Predicted_Score'
df[c] = df['Predicted_Score'].mask(df['Rank'] > 3)
df[c] = df.groupby('Student_ID')[c].apply(lambda s: s.shift(-1).bfill())
结果
Date Student_ID Rank Predicted_Score Recent_Predicted_Score
0 4/7/2021 33 2 87 86.0
1 13/6/2021 33 4 88 86.0
2 31/3/2021 33 7 88 86.0
3 28/2/2021 33 2 86 81.0
4 14/2/2021 33 10 86 81.0
5 31/1/2021 33 8 86 81.0
6 23/12/2020 33 1 81 80.0
7 8/11/2020 33 3 80 80.0
8 21/10/2020 33 3 80 80.0
9 23/9/2020 33 4 80 80.0
10 20/5/2020 33 3 80 79.0
11 29/4/2020 33 4 80 79.0
12 15/4/2020 33 2 79 79.0
13 26/2/2020 33 3 79 70.0
14 12/2/2020 33 5 79 70.0
15 29/1/2020 33 1 70 NaN
注意：确保你的数据帧按日期降序排序。
英文:
Mask the scores where rank is greater than 3 then group the masked column by Student_ID and backward fill to propagate the last predicted score
c = 'Recent_Predicted_Score'
df[c] = df['Predicted_Score'].mask(df['Rank'].gt(3))
df[c] = df.groupby('Student_ID')[c].apply(lambda s: s.shift(-1).bfill())
Result
Date Student_ID Rank Predicted_Score Recent_Predicted_Score
0 4/7/2021 33 2 87 86.0
1 13/6/2021 33 4 88 86.0
2 31/3/2021 33 7 88 86.0
3 28/2/2021 33 2 86 81.0
4 14/2/2021 33 10 86 81.0
5 31/1/2021 33 8 86 81.0
6 23/12/2020 33 1 81 80.0
7 8/11/2020 33 3 80 80.0
8 21/10/2020 33 3 80 80.0
9 23/9/2020 33 4 80 80.0
10 20/5/2020 33 3 80 79.0
11 29/4/2020 33 4 80 79.0
12 15/4/2020 33 2 79 79.0
13 26/2/2020 33 3 79 70.0
14 12/2/2020 33 5 79 70.0
15 29/1/2020 33 1 70 NaN
Note: Make sure your dataframe is sorted on Date in descending order.
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