如何修复Python代码中的KeyError: (1, 0)错误？

原始代码 #

stats = ['ETT', 'EMT', 'MRR', 'MMR', 'EER', 'ATT', 'MMT' ,'EMM' ,'ERT' ,'AER' ,'AEE' ,'AMM' ,'EEM' ,'ART' ,'AAR' ,'AEM' ,'MRT','AAT','AMR','ARR','AMT','RTT','EMR','ERR','EET','RRT','AET','MTT','AAE','AAM','EEE','TTT',
'AAA','MMM','RRR']


def determine_rewards(number_of_look_up_table,current_row_index,action_index):
    if stats[number_of_look_up_table][current_row_index] == 'T' :
        reward = rewards[0,action_index]
    elif stats[number_of_look_up_table][current_row_index] == 'M' :
        reward = rewards[1,action_index]
    elif stats[number_of_look_up_table][current_row_index] == 'E' :
        reward = rewards[2,action_index]   
    elif stats[number_of_look_up_table][current_row_index] == 'R' :
        reward = rewards[3,action_index] 
    elif stats[number_of_look_up_table][current_row_index] == 'A' :
        reward = rewards[4,action_index]    
    return reward   

reward = determine_rewards(2,0,0)

注：rewards是5行20列的数值矩阵，调用函数时触发如下错误：

reward = determine_rewards(2,0,0)


KeyError                                  Traceback (most recent call last)
~\Anaconda3\lib\site-packages\pandas\core\indexes\base.py in get_loc(self, key, method, 
tolerance)
 2656             try:
-> 2657                 return self._engine.get_loc(key)
 2658             except KeyError:

pandas/_libs/index.pyx in pandas._libs.index.IndexEngine.get_loc()

pandas/_libs/index.pyx in pandas._libs.index.IndexEngine.get_loc()

pandas/_libs/hashtable_class_helper.pxi in pandas._libs.hashtable.PyObjectHashTable.get_item()

pandas/_libs/hashtable_class_helper.pxi in pandas._libs.hashtable.PyObjectHashTable.get_item()

KeyError: (1, 0)

During handling of the above exception, another exception occurred:

KeyError                                  Traceback (most recent call last)
~\AppData\Local\Temp/ipykernel_28528/3264022016.py in <module>
----> 1 reward = determine_rewards(2,0,0)

~\AppData\Local\Temp/ipykernel_28528/255015173.py in 
determine_rewards(number_of_look_up_table, current_row_index, action_index)
 20         reward = rewards[0,action_index]
 21     elif stats[number_of_look_up_table][current_row_index] == 'M' :
 ---> 22         reward = rewards[1,action_index]
      23     elif stats[number_of_look_up_table][current_row_index] == 'E' :
      24         reward = rewards[2,action_index]

~\Anaconda3\lib\site-packages\pandas\core\frame.py in __getitem__(self, key)
2925             if self.columns.nlevels > 1:
2926                 return self._getitem_multilevel(key)
-> 2927             indexer = self.columns.get_loc(key)
   2928             if is_integer(indexer):
   2929                 indexer = [indexer]

  ~\Anaconda3\lib\site-packages\pandas\core\indexes\base.py in get_loc(self, key, method, 
  tolerance)
  2657                 return self._engine.get_loc(key)
  2658             except KeyError:
  -> 2659                 return self._engine.get_loc(self._maybe_cast_indexer(key))
  2660         indexer = self.get_indexer([key], method=method, tolerance=tolerance)
  2661         if indexer.ndim > 1 or indexer.size > 1:

  pandas/_libs/index.pyx in pandas._libs.index.IndexEngine.get_loc()

  pandas/_libs/index.pyx in pandas._libs.index.IndexEngine.get_loc()

  pandas/_libs/hashtable_class_helper.pxi in 
  pandas._libs.hashtable.PyObjectHashTable.get_item()

  pandas/_libs/hashtable_class_helper.pxi in 
  pandas._libs.hashtable.PyObjectHashTable.get_item()

  KeyError: (1, 0)

错误原因 #

rewards是pandas的DataFrame对象，你误用了numpy数组的索引语法rewards[行,列]。DataFrame会把(1,0)当成列名去查找，而你的数据中没有这个列名，因此触发KeyError。

修复方案 #

方案1：使用DataFrame官方索引方法.iloc

.iloc是DataFrame中基于整数位置获取元素的标准方式，替换原有索引语法即可：

stats = ['ETT', 'EMT', 'MRR', 'MMR', 'EER', 'ATT', 'MMT' ,'EMM' ,'ERT' ,'AER' ,'AEE' ,'AMM' ,'EEM' ,'ART' ,'AAR' ,'AEM' ,'MRT','AAT','AMR','ARR','AMT','RTT','EMR','ERR','EET','RRT','AET','MTT','AAE','AAM','EEE','TTT',
'AAA','MMM','RRR']


def determine_rewards(number_of_look_up_table,current_row_index,action_index):
    char = stats[number_of_look_up_table][current_row_index]
    if char == 'T' :
        reward = rewards.iloc[0, action_index]
    elif char == 'M' :
        reward = rewards.iloc[1, action_index]
    elif char == 'E' :
        reward = rewards.iloc[2, action_index]   
    elif char == 'R' :
        reward = rewards.iloc[3, action_index] 
    elif char == 'A' :
        reward = rewards.iloc[4, action_index]    
    return reward   

reward = determine_rewards(2,0,0)

方案2：将DataFrame转为numpy数组

如果习惯numpy的索引逻辑，可以先把rewards转成numpy数组，之后就能继续用[行,列]的语法：

# 先转换为numpy数组
rewards = rewards.to_numpy()

stats = ['ETT', 'EMT', 'MRR', 'MMR', 'EER', 'ATT', 'MMT' ,'EMM' ,'ERT' ,'AER' ,'AEE' ,'AMM' ,'EEM' ,'ART' ,'AAR' ,'AEM' ,'MRT','AAT','AMR','ARR','AMT','RTT','EMR','ERR','EET','RRT','AET','MTT','AAE','AAM','EEE','TTT',
'AAA','MMM','RRR']


def determine_rewards(number_of_look_up_table,current_row_index,action_index):
    char = stats[number_of_look_up_table][current_row_index]
    if char == 'T' :
        reward = rewards[0,action_index]
    elif char == 'M' :
        reward = rewards[1,action_index]
    elif char == 'E' :
        reward = rewards[2,action_index]   
    elif char == 'R' :
        reward = rewards[3,action_index] 
    elif char == 'A' :
        reward = rewards[4,action_index]    
    return reward   

reward = determine_rewards(2,0,0)

可选优化：用字典简化分支逻辑 #

可以用字典映射字符和行索引，替代多个elif，让代码更简洁：

stats = ['ETT', 'EMT', 'MRR', 'MMR', 'EER', 'ATT', 'MMT' ,'EMM' ,'ERT' ,'AER' ,'AEE' ,'AMM' ,'EEM' ,'ART' ,'AAR' ,'AEM' ,'MRT','AAT','AMR','ARR','AMT','RTT','EMR','ERR','EET','RRT','AET','MTT','AAE','AAM','EEE','TTT',
'AAA','MMM','RRR']

# 字符到行索引的映射表
char_to_row = {'T':0, 'M':1, 'E':2, 'R':3, 'A':4}

def determine_rewards(number_of_look_up_table,current_row_index,action_index):
    char = stats[number_of_look_up_table][current_row_index]
    row_idx = char_to_row[char]
    # 若用numpy数组则改为 rewards[row_idx, action_index]
    return rewards.iloc[row_idx, action_index]

reward = determine_rewards(2,0,0)

内容的提问来源于stack exchange，提问作者znb