Pandas组合两个分组依据，过滤并合并分组（计数）

Question

我有一个数据帧，我需要结合两个不同的分组和其中一个过滤。

 ID     EVENT      SUCCESS
 1       PUT          Y
 2       POST         Y
 2       PUT          N
 1       DELETE       Y 

下表是我希望数据的外观。首先对“EVENT”计数进行分组，第二个是对每个ID的成功次数（'Y'）进行计数

ID  PUT   POST  DELETE SUCCESS
 1   1     0       1      2
 2   1     1       0      1

我尝试了一些技术，我发现壁橱是两种不同的方法，产生以下结果

group_df = df.groupby(['ID', 'EVENT']) count_group_df = group_df.size().unstack()

这将产生以下“事件”计数

ID  PUT   POST  DELETE
 1   1     0       1      
 2   1     1       0      

对于过滤器的成功，我不知道我是否可以加入到“ID”的第一个集合中

 df_success = df.loc[df['SUCCESS'] == 'Y', ['ID', 'SUCCESS']]
 count_group_df_2 = df_success.groupby(['ID', 'SUCCESS'])


ID  SUCCESS
1      2
2      1

我需要把这些组合起来？

此外，我还想将“事件”的计数（例如PUT和POST）合并到一个列中。

Answer 1

pandas

pd.get_dummies(df.EVENT) \ 
  .assign(SUCCESS=df.SUCCESS.eq('Y').astype(int)) \
  .groupby(df.ID).sum().reset_index()

   ID  DELETE  POST  PUT  SUCCESS
0   1       1     0    1        2
1   2       0     1    1        1

numpy和pandas

f, u = pd.factorize(df.EVENT.values)
n = u.size
d = np.eye(n)[f]
s = (df.SUCCESS.values == 'Y').astype(int)
d1 = pd.DataFrame(
    np.column_stack([d, s]),
    df.index, np.append(u, 'SUCCESS')
)
d1.groupby(df.ID).sum().reset_index()

   ID  DELETE  POST  PUT  SUCCESS
0   1       1     0    1        2
1   2       0     1    1        1

---

计时
小数据

%%timeit
f, u = pd.factorize(df.EVENT.values)
n = u.size
d = np.eye(n)[f]
s = (df.SUCCESS.values == 'Y').astype(int)
d1 = pd.DataFrame(
    np.column_stack([d, s]),
    df.index, np.append(u, 'SUCCESS')
)
d1.groupby(df.ID).sum().reset_index()
1000 loops, best of 3: 1.32 ms per loop

%%timeit
df1 = df.groupby(['ID', 'EVENT']).size().unstack(fill_value=0)
df_success = (df['SUCCESS'] == 'Y').groupby(df['ID']).sum().astype(int)
pd.concat([df1, df_success],axis=1).reset_index()
100 loops, best of 3: 3.3 ms per loop

%%timeit
df1 = df.groupby(['ID', 'EVENT']).size().unstack(fill_value=0)
df_success = df.loc[df['SUCCESS'] == 'Y', 'ID'].value_counts().rename('SUCCESS')
pd.concat([df1, df_success],axis=1).reset_index()
100 loops, best of 3: 3.28 ms per loop

%timeit pd.get_dummies(df.EVENT).assign(SUCCESS=df.SUCCESS.eq('Y').astype(int)).groupby(df.ID).sum().reset_index()
100 loops, best of 3: 2.62 ms per loop

大数据

df = pd.DataFrame(dict(
        ID=np.random.randint(100, size=100000),
        EVENT=np.random.choice('PUT POST DELETE'.split(), size=100000),
        SUCCESS=np.random.choice(list('YN'), size=100000)
    ))

%%timeit
f, u = pd.factorize(df.EVENT.values)
n = u.size
d = np.eye(n)[f]
s = (df.SUCCESS.values == 'Y').astype(int)
d1 = pd.DataFrame(
    np.column_stack([d, s]),
    df.index, np.append(u, 'SUCCESS')
)
d1.groupby(df.ID).sum().reset_index()
100 loops, best of 3: 10.8 ms per loop

%%timeit
df1 = df.groupby(['ID', 'EVENT']).size().unstack(fill_value=0)
df_success = (df['SUCCESS'] == 'Y').groupby(df['ID']).sum().astype(int)
pd.concat([df1, df_success],axis=1).reset_index()
100 loops, best of 3: 17.7 ms per loop

%%timeit
df1 = df.groupby(['ID', 'EVENT']).size().unstack(fill_value=0)
df_success = df.loc[df['SUCCESS'] == 'Y', 'ID'].value_counts().rename('SUCCESS')
pd.concat([df1, df_success],axis=1).reset_index()
100 loops, best of 3: 17.4 ms per loop

%timeit pd.get_dummies(df.EVENT).assign(SUCCESS=df.SUCCESS.eq('Y').astype(int)).groupby(df.ID).sum().reset_index()
100 loops, best of 3: 16.8 ms per loop

Answer 2

使用^{}将它们合并在一起：

df1 = df.groupby(['ID', 'EVENT']).size().unstack(fill_value=0)
df_success = (df['SUCCESS'] == 'Y').groupby(df['ID']).sum().astype(int)
df = pd.concat([df1, df_success],axis=1)
print (df)
    DELETE  POST  PUT  SUCCESS
ID                            
1        1     0    1        2
2        0     1    1        1

使用^{}的另一个解决方案：

df1 = df.groupby(['ID', 'EVENT']).size().unstack(fill_value=0)
df_success = df.loc[df['SUCCESS'] == 'Y', 'ID'].value_counts().rename('SUCCESS')
df = pd.concat([df1, df_success],axis=1)
print (df)
    DELETE  POST  PUT  SUCCESS
ID                            
1        1     0    1        2
2        0     1    1        1

最后一种方法是将索引转换为列，并通过^{}+^{}删除列名ID：

df = df.reset_index().rename_axis(None, axis=1)
print (df)
   ID  DELETE  POST  PUT  SUCCESS
0   1       1     0    1        2
1   2       0     1    1        1