用np.where按条件填充np.where

Question

我有一个样本df：

id           email1       email2            output
 1       abc@gmail.com    123@gmail.com   random_email1@gmail.com
 1       xyz@gmail.com    234@gmail.com   random_email1@gmail.com
 1            NaN          NaN            random_email1@gmail.com
 2       a123@gmail.com    NaN            random_email2@gmail.com
 2       b123@gmail.com    NaN            random_email2@gmail.com
 2           NaN          lol@gmail.com   random_email2@gmail.com
 3           NaN           NaN            random_email3@gmail.com
 4           NaN          lolz@gmail.com   random_email3@gmail.com

我的主要目标是基于多个条件覆盖output列。如果email1有多个唯一的电子邮件，则用email1_broken覆盖所有的输出并使用相应的If。email2也是如此，但如果两者都有多个唯一的电子邮件，email2优先，因此是output = email2_broken。最后，如果当前的output列都有一个id和一个唯一的电子邮件，那么我们将保留电子邮件。

企图：

df['output'] = np.where(df.groupby('id')['email1'].nunique() > 1, 'email1_broken',df['output'])

df['output'] = np.where(df.groupby('id')['email2'].nunique() > 1, 'email2_broken',df['output'])

想要的df：

id           email1       email2            output
 1       abc@gmail.com    123@gmail.com   email2_broken
 1       xyz@gmail.com    234@gmail.com   email2_broken
 1            NaN          NaN            email2_broken
 2       a123@gmail.com    NaN            email1_broken
 2       b123@gmail.com    NaN            email1_broken
 2           NaN          lol@gmail.com   email1_broken
 3           NaN           NaN            random_email3@gmail.com
 4           NaN          lolz@gmail.com  random_email3@gmail.com

样本数据：

import pandas as pd
import numpy as np

cols = ['id','email1','email2', 'output']

data = [
[1   ,  'abc@gmail.com' ,   '123@gmail.com'    , 'random_email1@gmail.com'],
[1   ,  'xyz@gmail.com'  ,  '234@gmail.com'    , 'random_email1@gmail.com'],
[1   ,      np.nan      ,      np.nan          , 'random_email1@gmail.com'],
[2   ,   'a123@gmail.com',     np.nan          , 'random_email2@gmail.com'],
[2   ,   'b123@gmail.com',     np.nan          , 'random_email2@gmail.com'],
[2   ,     np.nan        ,   'lol@gmail.com'   , 'random_email2@gmail.com'],
[3   ,     np.nan        ,     np.nan          , 'random_email3@gmail.com'],
[4   ,     np.nan        ,   'lolz@gmail.com'  , 'random_email3@gmail.com']]

df = pd.DataFrame(data, columns=cols)

Answer 1

你们非常亲密：

df['output'] = np.where(df.groupby('id')['email1'].transform('nunique') > 1, 'email1_broken',df['output'])
df['output'] = np.where(df.groupby('id')['email2'].transform('nunique') > 1, 'email2_broken',df['output'])

使用转换获得形状相同的布尔数组。

Answer 2

您可以使用np.select (在涉及多个条件时等效于numpy.where )和transform('nunique')。

g = df.groupby('id')

df['output'] = np.select(
 [g['email2'].transform('nunique').gt(1),
  g['email1'].transform('nunique').gt(1)],
 ['email2_broken', 'email1_broken'],
 df['output'])

print(df)

输出：

  id          email1          email2                   output
0   1   abc@gmail.com   123@gmail.com            email2_broken
1   1   xyz@gmail.com   234@gmail.com            email2_broken
2   1             NaN             NaN            email2_broken
3   2  a123@gmail.com             NaN            email1_broken
4   2  b123@gmail.com             NaN            email1_broken
5   2             NaN   lol@gmail.com            email1_broken
6   3             NaN             NaN  random_email3@gmail.com
7   4             NaN  lolz@gmail.com  random_email3@gmail.com