Pandas将随机字符串分配给每个组作为新列

Question

我们有一个类似的数据框

Out[90]: 
   customer_id                 created_at
0     11492288 2017-03-15 10:20:18.280437
1      8953727 2017-03-16 12:51:00.145629
2     11492288 2017-03-15 10:20:18.284974
3     11473213 2017-03-09 14:15:22.712369
4      9526296 2017-03-14 18:56:04.665410
5      9526296 2017-03-14 18:56:04.662082

我想在这里创建一个新列,基于customer_id组,分配给每个组的8个字符的随机字符串.

例如,输出看起来像

Out[90]: 
   customer_id                 created_at     code
0     11492288 2017-03-15 10:20:18.280437 nKAILfyV
1      8953727 2017-03-16 12:51:00.145629 785Vsw0b
2     11492288 2017-03-15 10:20:18.284974 nKAILfyV
3     11473213 2017-03-09 14:15:22.712369 dk6JXq3u
4      9526296 2017-03-14 18:56:04.665410 1WESdAsD
5      9526296 2017-03-14 18:56:04.662082 1WESdAsD

我习惯了R和dplyr,使用它们编写这个转换非常容易.我在Pandas寻找类似的东西:

library(dplyr)
library(stringi)

df %>%
  group_by(customer_id) %>%
  mutate(code = stri_rand_strings(1, 8))

我可以找出随机字符部分.只是好奇Pandas groupby在这种情况下是如何工作的.

谢谢!

Answer 1

import random
from string import ascii_letters, digits
chars = list(ascii_letters + digits)

choose = lambda x, k=8: ''.join(random.choices(chars, k=k))
df.assign(code=df.groupby('customer_id').transform(choose))

   customer_id                  created_at      code
0     11492288  2017-03-15 10:20:18.280437  S5HtmbeN
1      8953727  2017-03-16 12:51:00.145629  MMfFFn8U
2     11492288  2017-03-15 10:20:18.284974  S5HtmbeN
3     11473213  2017-03-09 14:15:22.712369  4VsKmDZ5
4      9526296  2017-03-14 18:56:04.665410  VhQfu2Rf
5      9526296  2017-03-14 18:56:04.662082  VhQfu2Rf

灵感来自@Wen的使用 pd.util.testing.rands_array

f, u = pd.factorize(df.customer_id.values)

df.assign(code=pd.util.testing.rands_array(8, u.size)[f])

   customer_id                  created_at      code
0     11492288  2017-03-15 10:20:18.280437  tSuQbTBm
1      8953727  2017-03-16 12:51:00.145629  qmCl6NEX
2     11492288  2017-03-15 10:20:18.284974  tSuQbTBm
3     11473213  2017-03-09 14:15:22.712369  Wsa3lNxh
4      9526296  2017-03-14 18:56:04.665410  jBfXS2Nk
5      9526296  2017-03-14 18:56:04.662082  jBfXS2Nk

Answer 2

在熊猫(R's mutate)是transform

df['code']=df.groupby('customer_id').transform(lambda x:pd.util.testing.rands_array(8,1))
df
Out[314]: 
   customer_id  created_at      code
0     11492288  2017-03-15  L6Odf65d
1      8953727  2017-03-16  fwLpgLnt
2     11492288  2017-03-15  L6Odf65d
3     11473213  2017-03-09  AuSUPnJ9
4      9526296  2017-03-14  U1AiLyx0
5      9526296  2017-03-14  U1AiLyx0

编辑(来自cᴏʟᴅsᴘᴇᴇᴅ):df.groupby('customer_id').customer_id.transform(lambda x:pd.util.testing.rands_array(8,1))

你的R代码也有一些改进,

Match=data.frame(A=unique(df$customer_id),B=replicate(length(unique(df$year)), stri_rand_strings(1, 8)))
df$Code=Match$B[match(df$customer_id,Match$A)]