Question

我有以下数据集：

HID     Score   Decile_Name Result
2089    62      4th decile  1
897     47      2nd decile  0
85      55      3rd decile  0
8       74      7th decile  1
23      31      1st decile  1
5657    77      8th decile  1
52      85      9th decile  0
781     63      6th decile  0
565     42      1st decile  0
456     62      4th decile  1
12      89      10th decile 1
56      85      9th decile  1

#Create a DataFrame
df1 = {
     'HID':[2089,897,85,8,23,5657,52,781,565,456,12,56],
    'Score':[62,74,31,77,85,63,42,62,89,85],
    'Decile_Name':['4th decile','7th decile','1st decile','8th decile','9th decile','6th decile','1st decile','4th decile','10th decile','9th decile'],
    'Result' :[1,1,1,1,0,0,0,1,1,1]
]}



df1 = pd.DataFrame(df1,columns=['HID','Score','Decile_Name','Result'])

这将捕获每个学生的主题得分和相应分数的十分位。它还可以记录学生是否通过或失败（结果）

我想计算每个十分位数（Result％）和整体（在整个数据集中）中Result = 1的比例。预期输出：

Attribute Level         Result %    num_of_stu  
Score - All Categories  0.5         12 # This captures the values for the whole df(df1).
Score - 1st Decile      0.5         2
Score - 2nd Decile      0           1
Score - 3rd Decile      0           1
...
Score - 9th Decile      0.5         2
Score - 10th Decile     1           1

有人可以帮我吗？

Answer 1

如果0列中只有1和Result值的解决方案：

首先通过agg进行聚合，然后使用extract将argsort的索引值按整数排序，创建新的摘要DataFrame并对其进行append：

df1 = df.groupby('Decile_Name').agg({'Result':'mean', 'HID':'size'})
df1 = df1.iloc[df1.index.str.extract('(\d+)', expand=False).astype(int).argsort()]

df2 = pd.DataFrame({'Result': [df['Result'].mean()],
                    'HID': [len(df)]}, index=['All Categories'])

d = {'Result':'Result %','HID':'num_of_stu'}
df1 = df2.append(df1).rename(columns=d)
print (df1)
                Result %  num_of_stu
All Categories  0.583333          12
1st decile      0.500000           2
2nd decile      0.000000           1
3rd decile      0.000000           1
4th decile      1.000000           2
6th decile      0.000000           1
7th decile      1.000000           1
8th decile      1.000000           1
9th decile      0.500000           2
10th decile     1.000000           1

一般解决方案-仅为1值创建boolena蒙版：

df['Result1'] = df['Result'] == 1
df1 = df.groupby('Decile_Name').agg({'Result1':'mean', 'HID':'size'})
df1 = df1.iloc[df1.index.str.extract('(\d+)', expand=False).astype(int).argsort()]

df2 = pd.DataFrame({'Result1': [df['Result1'].mean()],
                  'HID': [len(df)]}, index=['All Categories'])

d = {'Result1':'Result %','HID':'num_of_stu'}
df1 = df2.append(df1).rename(columns=d)
print (df1)
                Result %  num_of_stu
All Categories  0.583333          12
1st decile      0.500000           2
2nd decile      0.000000           1
3rd decile      0.000000           1
4th decile      1.000000           2
6th decile      0.000000           1
7th decile      1.000000           1
8th decile      1.000000           1
9th decile      0.500000           2
10th decile     1.000000           1

Answer 2

#build mean of Results grouped by Decile Name
result_df = df1[['Decile_Name','Result']].groupby(['Decile_Name']).mean()

#build count of Students grouped by Decile Name
students_df = df1[['Decile_Name','HID']].groupby(['Decile_Name']).count()

#merge the two dataframes
merged_df = pd.concat([result_df, students_df], axis=1)

#Add the sum for all studends as Index "All Students"
merged_df.loc["All Studends"] = [df1[['Result']].mean()["Result"], df1[['HID']].count()["HID"]]

#print 
print(merged_df)

结果：

                 Result     HID
Decile_Name         
10th decile     1.000000    1.0
1st decile  0.500000    2.0
2nd decile  0.000000    1.0
3rd decile  0.000000    1.0
4th decile  1.000000    2.0
6th decile  0.000000    1.0
7th decile  1.000000    1.0
8th decile  1.000000    1.0
9th decile  0.500000    2.0
All Studends    0.583333    12.0

在python

2 个答案: