比较两个 defaultidict 并只保留满足条件的记录

Question

我有一个 df，我想以 id 为键的方式将其转换为字典，然后获取字典列表作为值：

d = {'id': [1,1,1,1,2,2,3,3,3,4,4,4,4],
     'label':['A','A','B','G','A','BB','C','C','A','BB','B','AA','AA']
    ,'amount':[2,-12,12,-12,5,-5,2,3,5,3,3,10,-10]}
df = pd.DataFrame(d)

d = defaultdict(lambda: defaultdict(list))   

#only append the negative amounts
for index,row in df.iterrows():
    if row["amount"] < 0:
        d[row["id"]][row["amount"]].append(
            { "id": row["id"],
                "label": row["label"]})
print(d)       
Out: defaultdict(<function __main__.<lambda>()>,
            {1: defaultdict(list,
                           {-12: [{'id': 1, 'description': 'A'},
                           {'id': 1, 'description': 'G'}]}),
             2: defaultdict(list, {-5: [{'id': 2, 'description': 'BB'}]}),
             4: defaultdict(list, {-10: [{'id': 4, 'description': 'AA'}]})})

d2 = defaultdict(lambda: defaultdict(list)) 
#only append the positive amounts

for index,row in df.iterrows():
    account_id = row["id"]
    amount = row["amount"]
    
    if amount > 0:
        d2[account_id][amount].append( { "id": row["id"],
                "label": row["label"]})
print(d2)

Out: defaultdict(<function __main__.<lambda>()>,
            {1: defaultdict(list,
                         {2: [{'id': 1, 'description': 'A'}],
                         12: [{'id': 1, 'description': 'B'}]}),
             2: defaultdict(list, {5: [{'id': 2, 'description': 'A'}]}),
             3: defaultdict(list,
                         {2: [{'id': 3, 'description': 'C'}],
                          3: [{'id': 3, 'description': 'C'}],
                          5: [{'id': 3, 'description': 'A'}]}),
             4: defaultdict(list,
                         {3: [{'id': 4, 'description': 'BB'},
                           {'id': 4, 'description': 'B'}],
                          10: [{'id': 4, 'description': 'AA'}]})})

我如何比较两个字典，以便我得到包含同一用户的匹配正数和负数的记录，以便我的字典看起来像下面这样？我只想使用字典而不是 pandas operations.d

defaultdict(<function __main__.<lambda>()>,
            {1: defaultdict(list,
                         {-12: [{'id': 1, 'description': 'A'},
                               {'id': 1, 'description': 'G'}], 
                           12: [{'id': 1, 'description': 'B'}]},
             2: defaultdic (list, {-5: [{'id': 2, 'description': 'BB'},
                                    5: {'id': 2, 'description': 'A'}]}),
             4: defaultdict(list, {-10: [{'id': 4, 'description': 'AA'}],
                                   10: [{'id': 4, 'description': 'AA'}]}))

Answer 1

所以您基本上是想过滤每个 id 以正整数和负整数形式存在的金额？我建议在将其转换为字典之前在 pandas 中对其进行过滤。您可以按 id 进行分组，然后通过比较哪些数量也存在于相同的负 Series:

中来过滤组

new_df = df.groupby('id').apply(lambda g: g[g['amount'].isin(g['amount']*-1)]).reset_index(drop=True)

输出：

	id	label	amount
0	1	A	-12
1	1	B	12
2	1	G	-12
3	2	A	5
4	2	BB	-5
5	4	AA	10
6	4	AA	-10

然后您可以随心所欲地导出整个 df:

d = defaultdict(lambda: defaultdict(list))   

for index,row in new_df.iterrows():
    d[row["id"]][row["amount"]].append(
        { "id": row["id"],
            "label": row["label"]})

如果您只想比较字典，可以迭代一个字典，比较另一个字典中是否存在负值，然后将两个值写入新字典：

new_dict = {}

for item in d:
    for i in d[item]:
        if i*-1 in [i for item in d2 for i in d2[item]]:
            new_dict[item] = {i:d[item][i], -i:d2[item][-i]}

结果：

{1: {-12: [{'id': 1, 'label': 'A'}, {'id': 1, 'label': 'G'}],
  12: [{'id': 1, 'label': 'B'}]},
 2: {-5: [{'id': 2, 'label': 'BB'}], 5: [{'id': 2, 'label': 'A'}]},
 4: {-10: [{'id': 4, 'label': 'AA'}], 10: [{'id': 4, 'label': 'AA'}]}}

比较两个 defaultidict 并只保留满足条件的记录

compare two defaultidicts and keep only records that satisfy a condition

python

python-3.x

list

dictionary

defaultdict