Question

我试图将电子邮件清除功能应用于列，并将结果记录在单独的列中。

我不确定如何使用.apply()将函数应用于两列，但这是我尝试过的方法：

首先设置数据框，并列出常见电子邮件错误的字典：

import pandas as pd

df = pd.DataFrame({'emails':['jim@gmailcom','bob@gmail.com','mary@gmaicom','bobby@gmail.com'],
                   'result':['','','','']})

df

    emails          result
0   jim@gmailcom    
1   bob@gmail.com   
2   mary@gmaicom    
3   bobby@gmail.com 

# common mistakes:

correct_domain = {'gmailcom': 'gmail.com',
 'gmaicom': 'gmail.com',
 'gmaillom': 'gmail.com',
 'gmalcom': 'gmail.com'}

现在，我想浏览电子邮件，并用正确的域替换拼写错误的域。例如。 gmailcom-> gmail.com

def clean_emails(x):

    # for each domain(key) in this dict ( e.g. 'gmailcom':'gmail.com')
    for mistake in correct_domain:  

        # if incorrect domain ('gmailcom') is in the email we're checking
        if mistake  in x['emails']:

            # replace it with the dict value which is the correctly formatted domain ('gmail.com')
            x['emails'] = x['emails'].replace(mistake ,correct_domain[mistake ])

            # record result
            x['result'] = 'email cleaned'

        else:
            x['result'] = 'no cleaning needed'

然后，当我应用此功能时，我什么也没得到：

df.apply(clean_emails,axis=1)

0    None
1    None
2    None
3    None
dtype: object

我尝试使用return进行混合，但是无法为单独的列找出两个单独的返回值。

我想要的结果，电子邮件已经清理，结果记录到result：

    emails          result
0   jim@gmail.com    'email cleaned'    
1   bob@gmail.com   'no cleaning needed'    
2   mary@gmail.com    'email cleaned'   
3   bobby@gmail.com 'no cleaning needed'

编辑：我以为在函数的末尾添加return x会返回新编辑的行，但是电子邮件没有被清除。

    emails  result
0   jim@gmail.com   email cleaned
1   bob@gmail.com   no cleaning needed
2   mary@gmaicom    no cleaning needed
3   bobby@gmail.com no cleaning needed

Answer 1

使用Series.str.contains检查是否需要用numpy.where进行按条件列清洁，然后使用Series.str.replace进行回调以仅用字典替换必要的行：

pat = '|'.join(correct_domain.keys())
m = df['emails'].str.contains(pat, na=False)
df['result'] = np.where(m, 'email cleaned', 'no cleaning needed')
df.loc[m, 'emails'] = (df.loc[m, 'emails']
                         .str.replace(pat, lambda x: correct_domain[x.group()], regex=True))

print (df)
            emails              result
0    jim@gmail.com       email cleaned
1    bob@gmail.com  no cleaning needed
2   mary@gmail.com       email cleaned
3  bobby@gmail.com  no cleaning needed

Answer 2

为什么不是两线制：

df['result'] = df['emails'].str.contains('|'.join(correct_domain.keys()).map({0:'email cleaned', 1:'no cleaning needed'})
df['emails'] = df['emails'].str.replace('|'.join(correct_domain.keys()),list(correct_domain.values())[0])

现在：

print(df)

将会是：

            emails              result
0    jim@gmail.com       email cleaned
1    bob@gmail.com  no cleaning needed
2   mary@gmail.com       email cleaned
3  bobby@gmail.com  no cleaning needed

Answer 3

我一直在想，我已经看到您已经提供了许多解决方案。按照您的逻辑，我们可以像这样到达那里：


    df = pd.DataFrame({'emails':['jim@gmailcom','bob@gmail.com','mary@gmaicom','bobby@gmail.com']})

    regexExp = [r'gmailcom$', r'gmaicom$', r'gmaillom', r'gmalcom']

    df2 = df.replace(regex=regexExp, value='gmail.com')

    result = []
    for dfLines, df2Lines in zip(df.itertuples(),df2.itertuples()):
        if df2Lines.emails != dfLines.emails:
            result.append('email cleaned')
        else:
            result.append('no cleaning needed')

    df2['result'] = result

    print(df2)

将操作应用于返回“无”的两列

3 个答案: