python - 使用数据框中的两列组合创建字典列，然后计算具有公共(public)键的两列值的比率

Question

我有以下格式的数据框:

Id  Name_prev   Weight_prev Name_now    Weight_now
1   [1,3,4,5]   [10,34,67,37]   [1,3,5] [45,76,12]
2   [10,3,40,5] [100,134,627,347]   [10,40,5]   [34,56,78]
3   [1,30,4,50] [11,22,45,67]   [1,30,50]   [12,45,78]
4   [1,7,8,9]   [32,54,76,98]   [7,8,9] [34,12,32]

我想创建两个新变量:

1. Name_prev 和 Name_now 的并集:这是 Name_prev 和 Name_now 字段的交集，可以使用 set 操作来完成两列，我能够计算出相同的值。

2. Name_prev 和 Name_now 的比率:这是与 (Name_prev 和 Name_now)。

预期输出:

Id  Union of Name_prev and Name_now Ratio of Name_prev and Name_now
1   [1,3,5]                         [10/45, 34/76,37/12]
2   [10,40,5]                       [100/34,627/56,347/78]
3   [1,30,50]                       [11/12,22/45,67/78]
4   [7,8,9]                         [54/34,76/12,98/32]

我正在尝试通过将 Name_prev 和 Weigth_prev 组合为键、值对来创建类似字典的结构，并对 Name_now 和Weight_now 然后采用常用键的比率，但卡住了......

Answer 1

用途:

a, b = [],[]
for n1, n2, w1, w2 in zip(df['Name_prev'], df['Name_now'], 
                          df['Weight_prev'], df['Weight_now']):
    #get intersection of lists
    n = [val for val in n1 if val in n2]
    #get indices by enumerate and select weights   
    w3 = [w1[i] for i, val in enumerate(n1) if val in n2]
    w4 = [w2[i] for i, val in enumerate(n2) if val in n1]
    #divide each value in list
    w = [i/j for i, j in zip(w3, w4)]
    a.append(n)
    b.append(w)

df = df.assign(name=a, weight=b)
print (df)
   Id       Name_prev           Weight_prev     Name_now    Weight_now  \
0   1    [1, 3, 4, 5]      [10, 34, 67, 37]    [1, 3, 5]  [45, 76, 12]   
1   2  [10, 3, 40, 5]  [100, 134, 627, 347]  [10, 40, 5]  [34, 56, 78]   
2   3  [1, 30, 4, 50]      [11, 22, 45, 67]  [1, 30, 50]  [12, 45, 78]   
3   4    [1, 7, 8, 9]      [32, 54, 76, 98]    [7, 8, 9]  [34, 12, 32]   

          name                                             weight  
0    [1, 3, 5]  [0.2222222222222222, 0.4473684210526316, 3.083...  
1  [10, 40, 5]  [2.9411764705882355, 11.196428571428571, 4.448...  
2  [1, 30, 50]  [0.9166666666666666, 0.4888888888888889, 0.858...  
3    [7, 8, 9]     [1.588235294117647, 6.333333333333333, 3.0625]  

如果需要删除原始列，请使用 DataFrame.pop :

a, b = [],[]
for n1, n2, w1, w2 in zip(df.pop('Name_prev'), df.pop('Name_now'), 
                          df.pop('Weight_prev'), df.pop('Weight_now')):
    n = [val for val in n1 if val in n2]
    w3 = [w1[i] for i, val in enumerate(n1) if val in n2]
    w4 = [w2[i] for i, val in enumerate(n2) if val in n1]
    w = [i/j for i, j in zip(w3, w4)]
    a.append(n)
    b.append(w)

df = df.assign(name=a, weight=b)
print (df)
   Id         name                                             weight
0   1    [1, 3, 5]  [0.2222222222222222, 0.4473684210526316, 3.083...
1   2  [10, 40, 5]  [2.9411764705882355, 11.196428571428571, 4.448...
2   3  [1, 30, 50]  [0.9166666666666666, 0.4888888888888889, 0.858...
3   4    [7, 8, 9]     [1.588235294117647, 6.333333333333333, 3.0625]

编辑:

在 pandas 中使用列表始终不是矢量化的，因此最好先展平列表，合并，并在必要时聚合列表:

from itertools import chain

df_prev = pd.DataFrame({
    'Name' : list(chain.from_iterable(df['Name_prev'].values.tolist())), 
    'Weight_prev' : list(chain.from_iterable(df['Weight_prev'].values.tolist())), 
    'Id' : df['Id'].values.repeat(df['Name_prev'].str.len())
})

print (df_prev)
    Name  Weight_prev  Id
0      1           10   1
1      3           34   1
2      4           67   1
3      5           37   1
4     10          100   2
5      3          134   2
6     40          627   2
7      5          347   2
8      1           11   3
9     30           22   3
10     4           45   3
11    50           67   3
12     1           32   4
13     7           54   4
14     8           76   4
15     9           98   4

<小时/>

df_now = pd.DataFrame({
    'Name' : list(chain.from_iterable(df['Name_now'].values.tolist())), 
    'Weight_now' : list(chain.from_iterable(df['Weight_now'].values.tolist())), 
    'Id' : df['Id'].values.repeat(df['Name_now'].str.len())
})

print (df_now)
    Name  Weight_now  Id
0      1          45   1
1      3          76   1
2      5          12   1
3     10          34   2
4     40          56   2
5      5          78   2
6      1          12   3
7     30          45   3
8     50          78   3
9      7          34   4
10     8          12   4
11     9          32   4

<小时/>

df = df_prev.merge(df_now, on=['Id','Name'])
df['Weight'] = df['Weight_prev'] / df['Weight_now']
print (df)
    Name  Weight_prev  Id  Weight_now     Weight
0      1           10   1          45   0.222222
1      3           34   1          76   0.447368
2      5           37   1          12   3.083333
3     10          100   2          34   2.941176
4     40          627   2          56  11.196429
5      5          347   2          78   4.448718
6      1           11   3          12   0.916667
7     30           22   3          45   0.488889
8     50           67   3          78   0.858974
9      7           54   4          34   1.588235
10     8           76   4          12   6.333333
11     9           98   4          32   3.062500

df = df.groupby('Id')['Name','Weight'].agg(list).reset_index()
print (df)
   Id         Name                                             Weight
0   1    [1, 3, 5]  [0.2222222222222222, 0.4473684210526316, 3.083...
1   2  [10, 40, 5]  [2.9411764705882355, 11.196428571428571, 4.448...
2   3  [1, 30, 50]  [0.9166666666666666, 0.4888888888888889, 0.858...
3   4    [7, 8, 9]     [1.588235294117647, 6.333333333333333, 3.0625]