替换系列熊猫中同一列中的字符串

Question

事情可能乍一看并不容易理解，但要循序渐进......

这是我大约 50000 行数据框 df_answers_clean 的开始

使用len(set(df_answers_clean['Race'])) 后，我得到了 98 个独特的职位，这对于我未来的分类来说太多了。 我举了一个例子——他们的 25 个职位列表如下：

['Native American, Pacific Islander, or Indigenous Australian; South Asian; White or of European descent', 'Hispanic or Latino/Latina; South Asian', 'East Asian; Hispanic or Latino/Latina',
 'East Asian', 'Black or of African descent; East Asian; South Asian; White or of European descent',
 'Black or of African descent; East Asian; Hispanic or Latino/Latina; Middle Eastern; Native American, Pacific Islander, or Indigenous Australian; South Asian; White or of European descent',
 'Hispanic or Latino/Latina; White or of European descent', 'White or of European descent; I prefer not to say', 'South Asian; White or of European descent', 'White or of European descent',
 'Hispanic or Latino/Latina', 'Black or of African descent; I don’t know; I prefer not to say',
 'Native American, Pacific Islander, or Indigenous Australian; White or of European descent; I don’t know', 'East Asian; White or of European descent; I don’t know', 'Native American, Pacific Islander, or Indigenous Australian', 'South Asian; White or of European descent; I don’t know',
 'Black or of African descent; Middle Eastern; White or of European descent; I don’t know',
 'Hispanic or Latino/Latina; Middle Eastern; White or of European descent',
 'Middle Eastern; White or of European descent',
 'Middle Eastern; South Asian']

我用很多行代码清理了这个烂摊子：

df_answers_clean['Race'] = df_answers_clean['Race'].str.replace('^Black or of African descent[\s\S]*', 'Black or of African descent')    
df_answers_clean['Race'] = df_answers_clean['Race'].str.replace('^East Asian[\s\S]*', 'East Asian')    
df_answers_clean['Race'] = df_answers_clean['Race'].str.replace('^Hispanic or Latino/Latina[\s\S]*', 'Hispanic or Latino/Latina')
df_answers_clean['Race'] = df_answers_clean['Race'].str.replace('^Middle Eastern[\s\S]*', 'Middle Eastern')
df_answers_clean['Race'] = df_answers_clean['Race'].str.replace('^Native American, Pacific Islander, or Indigenous Australian[\s\S]*', 'Native American, Pacific Islander, or Indigenous Australian')    
df_answers_clean['Race'] = df_answers_clean['Race'].str.replace('^South Asian[\s\S]*', 'South Asian')    
df_answers_clean['Race'] = df_answers_clean['Race'].str.replace('^White or of European descent[\s\S]*', 'White or of European descent')    
df_answers_clean['Race'] = df_answers_clean['Race'].str.replace('^I don’t know[\s\S]*', 'No data')    
df_answers_clean['Race'] = df_answers_clean['Race'].str.replace('^I prefer not to say[\s\S]*', 'No data')

结果是唯一的组，现在才对后面的分类任务有用：

{'Black or of African descent',
 'East Asian',
 'Hispanic or Latino/Latina',
 'Middle Eastern',
 'Native American, Pacific Islander, or Indigenous Australian',
 'No data',
 'South Asian',
 'White or of European descent'}

正如我所说 - 它可以工作，但许多重复的代码行不实用/不实用。

我这样做的另一个想法是列出我的最终结果 (race_names_change) 并通过 for-loop 放置所有内容：

race_names_change = ['Black or of African descent', 'East Asian', 'Hispanic or Latino/Latina', 'Middle Eastern', 'South Asian', 'Native American, Pacific Islander, or Indigenous Australian', 'White or of European descent']

for i in race_names_change:
    replace_string = str('^'+ i +'[\s\S]*')
    df_answers_clean['Race'].str.replace('replace_string', i, regex=True)

但不幸的是它不起作用 - 列表与开头相同（98个位置）。

也许循环代码或任何其他方式（映射、应用）有问题？

感谢您的建议。

Answer 1

如果您可以创建具有所需正则表达式模式和相应输出的字典，那么您可以简单地使用pd.Series.replace()

d = {
    'pattern1':'output1',
    'pattern2':'output2'
    }

df_answers_clean['Race'].replace(d, regex=True)

注意，pd.Series.str.replace() 与 pd.Series.replace() 不同

---

试试这个 -

race_names_change = ['Black or of African descent', 'East Asian', 'Hispanic or Latino/Latina', 'Middle Eastern', 'South Asian', 'Native American, Pacific Islander, or Indigenous Australian', 'White or of European descent']

d = {}
for i in race_names_change:
    replace_string = str('^'+ i +'[\s\S]*')
    replace_string
    d.update({replace_string:i})
    
df_answers_clean['Race'].replace(d, regex=True)