为什么lambda没有工作属性“值”？

Question

这个df2'CLINE_TYPE‘

Increase_FALSE
Decrease
Increase_FALSE
Increase_SUPERPOSITION
Decrease_FALSE
Increase
Increase_SUPERPOSITION
Decrease_FALSE
Increase
Increase_SUPERPOSITION

这一职能：

def nearest(lst, target):
  return min(lst, key=lambda x: abs(x-target))

我需要从lamda获得一个值，我得到了一个错误：'AttributeError：'str‘对象没有属性值’value‘

所有代码：

import pandas as pd
import numpy as np
import re

def nearest(lst, target):
  return min(lst, key=lambda x: abs(x-target))

    df2['res']=np.nan
    
    df2['res'] = df2['CLINE_TYPE'].apply(lambda x: nearest(df2.loc[df2['CLINE_TYPE'].str.contains('Increase')].index.to_list(), x.name)
                                        if bool(re.match(r'Decrease', x.values))==True else
                                        nearest(df2.loc[df2['CLINE_TYPE'].str.contains('Decrease')].index.to_list(), x.name))
    
    print(df2[['CLINE_TYPE','res']])

查找列表中最近的最小索引

Answer 1

Series.apply将串联中的值传递给函数，因此通常它没有任何属性。由于您希望访问行索引，所以您需要的是DataFrame.apply。

df2['res'] = df2.apply(lambda row: nearest(df2.loc[df2['CLINE_TYPE'].str.contains('Increase')].index.to_list(), row.name)
                       if bool(re.match(r'Decrease', row['CLINE_TYPE']))==True else
                       nearest(df2.loc[df2['CLINE_TYPE'].str.contains('Decrease')].index.to_list(), row.name), axis=1)

print(df2[['CLINE_TYPE','res']])

               CLINE_TYPE  res
0          Increase_FALSE    1
1                Decrease    0
2          Increase_FALSE    1
3  Increase_SUPERPOSITION    4
4          Decrease_FALSE    3
5                Increase    4
6  Increase_SUPERPOSITION    7
7          Decrease_FALSE    6
8                Increase    7
9  Increase_SUPERPOSITION    7