df replace 不适用于 pandas 列中的分隔符

Question

我有一个df

technologies= {
    'Courses':["Spark,ABCD","PySpark","Hadoop","Python","Pandas"],
    'Fee' :[22000,25000,23000,24000,26000],
    'Duration':['30days','50days','30days', None,np.nan],
    'Discount':[1000,2300,1000,1200,2500]
          }
df = pd.DataFrame(technologies)
print(df)

我试图用 dict 值替换列值

dict = {"Spark" : 'S', "PySpark" : 'P', "Hadoop": 'H', "Python" : 'P', "Pandas": 'P'}
df2=df.replace({"Courses": dict})
print(df2)

但是带有分隔符的行即使存在值也不会被替换将其作为输出

      Courses    Fee Duration  Discount
0  Spark,ABCD  22000   30days      1000
1           P  25000   50days      2300
2           H  23000   30days      1000
3           P  24000     None      1200
4           P  26000      NaN      2500

但输出应该是

      Courses    Fee Duration  Discount
0      S,ABCD 22000   30days      1000
1           P  25000   50days      2300
2           H  23000   30days      1000
3           P  24000     None      1200
4           P  26000      NaN      2500

Answer 1

可能值得了解 regex 参数的工作原理，以便您将来可以利用它。 尽管如此，还是可以在,上拆分并分解，以便每行有一个单词。 然后，您可以替换原始索引并将其分组，然后重新连接到逗号分隔的字符串。

import pandas as pd
technologies= {
    'Courses':["Spark,ABCD","PySpark","Hadoop","Python","Pandas"],
    'Fee' :[22000,25000,23000,24000,26000],
    'Duration':['30days','50days','30days', None,np.nan],
    'Discount':[1000,2300,1000,1200,2500]
          }
df = pd.DataFrame(technologies)

d = {"Spark" : 'S', "PySpark" : 'P', "Hadoop": 'H', "Python" : 'P', "Pandas": 'P'}

df.Courses = (df.Courses.str.split(',').explode().replace(d)
                        .groupby(level=0).agg(','.join))

输出

  Courses    Fee Duration  Discount
0  S,ABCD  22000   30days      1000
1       P  25000   50days      2300
2       H  23000   30days      1000
3       P  24000     None      1200
4       P  26000      NaN      2500

Answer 2

方法一：确保所有复合词都在单词之前。 在字典中PySpark在Spark之前

d = {"PySpark" : 'P', "Spark" : 'S', "Hadoop": 'H', "Python" : 'P', "Pandas": 'P'}
df2 = df.replace({"Courses": d}, regex  = True)
print(df2)

  Courses    Fee Duration  Discount
0  S,ABCD  22000   30days      1000
1       P  25000   50days      2300
2       H  23000   30days      1000
3       P  24000     None      1200
4       P  26000      NaN      2500

方法2：将单词放在边界中：

new_dict = pd.DataFrame(d.items(), columns = ['keys', 'values'])
new_dict['keys'] = '\\b' + new_dict['keys'] + '\\b'
new_dict = new_dict.set_index('keys').to_dict()['values']
df3 = df.replace({"Courses": new_dict}, regex  = True)
df3

 Courses    Fee Duration  Discount
    0  S,ABCD  22000   30days      1000
    1       P  25000   50days      2300
    2       H  23000   30days      1000
    3       P  24000     None      1200
    4       P  26000      NaN      2500

Answer 3

这是一种专注于您要更改的列（ Courses ）的方法：

dct = {"Spark" : 'S', "PySpark" : 'P', "Hadoop": 'H', "Python" : 'P', "Pandas": 'P'}
df.Courses = df.Courses.transform(
    lambda x: x.str.split(',')).transform(
    lambda x: [dct[y] if y in dct else y for y in x]).str.join(',')

解释：

• 使用transform将列中的每个 csv 字符串值替换为列表
• 再次使用transform ，这次是使用字典dct替换值列表中的每个项目
• 使用Series.str.join将每个值的列表转换回 csv 字符串。

完整的测试代码：

import pandas as pd
import numpy as np
technologies= {
    'Courses':["Spark,ABCD","PySpark","Hadoop","Python","Pandas"],
    'Fee' :[22000,25000,23000,24000,26000],
    'Duration':['30days','50days','30days', None,np.nan],
    'Discount':[1000,2300,1000,1200,2500]
          }
df = pd.DataFrame(technologies)
print(df)

dct = {"Spark" : 'S', "PySpark" : 'P', "Hadoop": 'H', "Python" : 'P', "Pandas": 'P'}
df.Courses = df.Courses.transform(
    lambda x: x.str.split(',')).transform(
    lambda x: [dct[y] if y in dct else y for y in x]).str.join(',')
print(df)

输入：

      Courses    Fee Duration  Discount
0  Spark,ABCD  22000   30days      1000
1     PySpark  25000   50days      2300
2      Hadoop  23000   30days      1000
3      Python  24000     None      1200
4      Pandas  26000      NaN      2500

输出：

  Courses    Fee Duration  Discount
0  S,ABCD  22000   30days      1000
1       P  25000   50days      2300
2       H  23000   30days      1000
3       P  24000     None      1200
4       P  26000      NaN      2500