Question

我希望能够为数据框分配PRNG。

我可以使用cat.codes或ngroup()

指定唯一ID

import pandas as pd
import random
import string

df1 = pd.DataFrame({'Name': ['John', 'Susie', 'Jack', 'Jill', 'John']})
df1['id'] = df1.groupby('Name').ngroup()
df1['idz'] = df1['Name'].astype('category').cat.codes

    Name    id  idz
0   John    2   2
1   Susie   3   3
2   Jack    0   0
3   Jill    1   1
4   John    2   2

我已使用this post中的函数逐行创建此唯一ID。

def id_generator(size=6, chars=string.ascii_uppercase + string.digits):
    return ''.join(random.SystemRandom().choice(chars) for _ in range(size))

df1['random id'] = df1['idz'].apply(lambda x : id_generator(3))

    Name    id  idz random id
0   John    2   2   118 #<--- Check Here
1   Susie   3   3   KGZ
2   Jack    0   0   KMQ
3   Jill    1   1   T2L
4   John    2   2   Q3F #<--- Check Here

但是我如何将两者结合在一起，以便约翰在这个小用例中会收到相同的ID？我希望喜欢，以避免因数据大小而导致长if ID not used, then ID, and if name has ID, use existing ID次循环。

Answer 1

gourpby + transform

df1['random id'] = df1.groupby('idz').idz.transform(lambda x : id_generator(3))
df1
Out[657]: 
    Name  id  idz random id
0   John   2    2       35P
1  Susie   3    3       6UU
2   Jack   0    0       XGF
3   Jill   1    1       5LC
4   John   2    2       35P

Answer 2

将其与＆＃34;作为前提可能不是最有效的选项＆＃34;。

我会首先找到每个唯一身份用户，为每个唯一身份用户生成随机ID。

# Finding unique users and storing in a new DataFrame
df_unique_users = pd.DataFrame({'Name':[x for x in set(df['Name'])]})

# Generating unique user ID's for length of data frame 
# By using a set you are guaranteed unique values. You just need to make sure
# you have enough permutations of the unique random_id so that your rand_set 
# will eventually be longer than your unique Names DataFrame. 

rand_set = set()
while(len(rand_set)<len(df_unique_users)):
    rand_set = rand_set.union([id_generator(3)])

df_unique_users['Rand_ID'] = rand_set

### Mapping the random ID's over to the original DataFrame
df = df.merge(df_unique_users, how='left', left_on='Name', right_on='Name')

您可以使用原始ID列而不是“名称”列来获取唯一值。

熊猫随机ID类别

2 个答案: