Question

当前，我正在使用以下代码进行替换，这有点麻烦：

df1['CompanyA'] = df1['CompanyA'].str.replace('.','')
df1['CompanyA'] = df1['CompanyA'].str.replace('-','')
df1['CompanyA'] = df1['CompanyA'].str.replace(',','')
df1['CompanyA'] = df1['CompanyA'].str.replace('ltd','limited')
df1['CompanyA'] = df1['CompanyA'].str.replace('&','and')
df1['Address1A'] = df1['Address1A'].str.replace('.','')
df1['Address1A'] = df1['Address1A'].str.replace('-','')
df1['Address1A'] = df1['Address1A'].str.replace('&','and')
df1['Address1A'].str.replace(r'\brd\b', 'road')
df1['Address2A'] = df1['Address2A'].str.replace('.','')
df1['Address2A'] = df1['Address2A'].str.replace('-','')
df1['Address2A'] = df1['Address2A'].str.replace('&','and')
df1['Address2A'].str.replace(r'\brd\b', 'road')

为了便于即时更改，我的理想情况是：

df1['CompanyA'] = df1['CompanyA'].str.replace(('&','and'), ('.', ''), ('-','')....)
df1['Address1A'] = df1['Address1A'].str.replace(('&','and'), ('.', ''), ('-','')....)
df1['Address2A'] = df1['Address2A'].str.replace(('&','and'), ('.', ''), ('-','')....)

这样一来，我可以输入/更改要替换的特定列的内容，而无需调整多行代码。

这有可能吗？

Answer 1

您可以创建字典并将其传递给功能replace()，而无需多次链接或命名该功能。

replacers = {',':'','.':'','-':'','ltd':'limited'} #etc....
df1['CompanyA'] = df1['CompanyA'].replace(replacers)

Answer 2

您可以链接替换项：

Shareable

Answer 3

替换功能也接受值作为字典。您可以执行以下操作：

df1.replace({'CompanyA' : { '&' : 'and', '.': '' , '-': ''}},regex=True)

Answer 4

您可以使用字典来映射每一列的字符：

to_replace = {'.': '',
              ',': '',
              'foo': 'bar'
             }

for k, v in to_replace.items():
    df1['CompanyA'] = df1['CompanyA'].str.replace(k, v)

Answer 5

您最有可能使用pd.Dataframe，所以我建议制作通用卸妆液

def remover(row, replaces):
    for k,v in replacers.items():
        if k in row:
            row = row.replace(k, v)
    return row      


replacers = {',' : "",
         '.':'',
         '-':'',
         'ltd':'limited'
        }

for column in df.columns:
    df[column] = df[column].apply(lambda row: remover(row, replacers))

或者您可以指定要修改的特定列名称

如何使用str.replace一次替换多对？

5 个答案: