重塑數據框並為每一行應用計算

Question

我有一個數據框，如下所示：

df=pd.DataFrame({ 'family' : ["A","A","B","B"],
  'V1' : [5,5,40,10,],
  'V2' :[50,10,180,20],
  'gr_0' :["all","all","all","all"],
  'gr_1' :["m1","m1","m2","m3"],
  'gr_2' :["m12","m12","m12","m9"],
  'gr_3' :["NO","m14","m15","NO"]
                           })

我想通過以下方式對其進行轉換：

df_new=pd.DataFrame({ 'family' : ["A","A","A","A","B","B","B","B","B","B"],
  'gr' : ["all","m1","m12","m14","all","m2","m3","m12","m9","m15"],
  "calc(sumV2/sumV1)":[6,6,6,2,4,4.5,2,4.5,2,4.5]            
               })

  family   gr  calc(sumV2/sumV1)
0      A  all                6.0
1      A   m1                6.0
2      A  m12                6.0
3      A  m14                2.0
4      B  all                4.0
5      B   m2                4.5
6      B   m3                2.0
7      B  m12                4.5
8      B   m9                2.0
9      B  m15                4.5

為了達到df_new：

1. 我希望這些行按“系列” X“ gr_”列的每個唯一值對齊。
2. 對於每一行，分別計算df_new中所示的sum（V2）/ sum（V1）。

我是python的新手。 對我來說，這的軟編碼似乎很復雜。 最好，我不想在此df_new中列出“否”記錄，但它也可以保留在輸出中。

Answer 1

您可以這樣做：

df_new = df.melt(id_vars=['family','V1','V2']).groupby(['family','value'])
                .apply(lambda x: x.V2.sum()/x.V1.sum())
                .reset_index(name='calc(sumV2/sumV1)')
df_new = df_new[df_new.value != 'NO'].reset_index(drop=True)

print(df_new)

     family value  calc(sumV2/sumV1)
0      A    all           6.0
1      A    m1            6.0
2      A    m12           6.0
3      A    m14           2.0
4      B    all           4.0
5      B    m12           4.5
6      B    m15           4.5
7      B    m2            4.5
8      B    m3            2.0
9      B    m9            2.0

Answer 2

melt + groupby ：

v = df.melt(id_vars=['family','V1','V2'], value_name='gr')
w = v.loc[v.gr != 'NO']
x = w.groupby(['family', 'gr']).sum()

(x.V2 / x.V1).reset_index(name='calc(sumV2/sumV1)')

  family   gr  calc(sumV2/sumV1)
0      A  all                6.0
1      A   m1                6.0
2      A  m12                6.0
3      A  m14                2.0
4      B  all                4.0
5      B  m12                4.5
6      B  m15                4.5
7      B   m2                4.5
8      B   m3                2.0
9      B   m9                2.0

---

與此答案類似的方法，但具有完全矢量化的優點，避免了apply

---

性能 ：

a = np.random.randint(1, 1000, (1_000_000, 7))
df = pd.DataFrame(a, columns=['family', 'V1', 'V2', 'gr_0', 'gr_1', 'gr_2', 'gr_3'])   
df[['gr_0', 'gr_1', 'gr_2', 'gr_3']] = df[['gr_0', 'gr_1', 'gr_2', 'gr_3']].astype(str)

%%timeit
v = df.melt(id_vars=['family','V1','V2'], value_name='gr')
w = v.loc[v.gr != 'NO']
x = w.groupby(['family', 'gr']).sum()    
(x.V2 / x.V1).reset_index(name='calc(sumV2/sumV1)')

2.71 s ± 32.9 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%%timeit
df_new = (df.melt(id_vars=['family','V1','V2']).groupby(['family','value'])
                 .apply(lambda x: x.V2.sum()/x.V1.sum())
                 .reset_index(name='calc(sumV2/sumV1)'))
df_new = df_new[df_new.value != 'NO'].reset_index(drop=True)

5min 24s ± 3.35 s per loop (mean ± std. dev. of 7 runs, 1 loop each)