使用基于字典键的信息创建 ID答案

【问题标题】：Create ID with information based on dict keys使用基于字典键的信息创建 ID
【发布时间】：2023-02-20 18:47:53
【问题描述】：

我有一个包含以下输入的大型数据框

client	type	country	potential
Client 1	Private	USA	low
Client 2	Private	Ireland	high
Client 3	Institutional	GB	mid
Client 4	Institutional	GB	mid
Client 5	Institutional	GB	mid

我想为每个客户创建一个 ID。身份证应该不是随机的（我尝试使用 uuid 包）但包含有关客户端和具有相同属性的单独客户端的信息。

ID_classification = {'type':{'A':'Private','B':'Institutional'},
                     'country':{'1':'USA','2':'GB','3':'Ireland'},
                     'potential':{'1':'low','2':'mid','3':'high'}}

ID 模式可能看起来像这样（尚未确定最终模式）

type.key-country.key-potential.key-unique_id

导致：

id client type country potential

A-1-1-1 Client 1 Private USA low

A-3-3-1 Client 2 Private Ireland high

B-2-2-1 Client 3 Institutional GB mid

B-2-2-2 Client 4 Institutional GB mid

B-2-2-3 Client 5 Institutional GB mid

提前致谢

id	client	type	country	potential
A-1-1-1	Client 1	Private	USA	low
A-3-3-1	Client 2	Private	Ireland	high
B-2-2-1	Client 3	Institutional	GB	mid
B-2-2-2	Client 4	Institutional	GB	mid
B-2-2-3	Client 5	Institutional	GB	mid

【问题讨论】：

标签： python pandas

【解决方案1】：

您可以使用：

# reorganize your mapping dictionary
# to have the key: value in correct order
mapper = {k1: {k: v for v, k in d.items()}
          for k1, d in ID_classification.items()}

# map all desired columns
df['id'] = df[list(mapper)].apply(lambda s: s.map(mapper[s.name])).agg('-'.join, axis=1)

# add unique id
df['id'] += '-' + df.groupby('id').cumcount().add(1).astype(str)

输出：

     client           type  country potential       id
0  Client 1        Private      USA       low  A-1-1-1
1  Client 2        Private  Ireland      high  A-3-3-1
2  Client 3  Institutional       GB       mid  B-2-2-1
3  Client 4  Institutional       GB       mid  B-2-2-2
4  Client 5  Institutional       GB       mid  B-2-2-3

【讨论】：

【解决方案2】：

使用：

#swap key value in inner dictionaries
d = {k:{v1:k1 for k1, v1 in v.items()} for k, v in ID_classification.items()}

#map columns by d with join together by -
s = pd.DataFrame([df[k].map(v) for k, v in d.items()]).agg('-'.join)

#added counter column by Series s
df['id'] = s + '-' + df.groupby(s).cumcount().add(1).astype(str)
print (df)
     client           type  country potential       id
0  Client 1        Private      USA       low  A-1-1-1
1  Client 2        Private  Ireland      high  A-3-3-1
2  Client 3  Institutional       GB       mid  B-2-2-1
3  Client 4  Institutional       GB       mid  B-2-2-2
4  Client 5  Institutional       GB       mid  B-2-2-3

【讨论】：