如何使用 python 定义函数来集成列表？

Question

我正在尝试定义一个遍历三种列表列表的函数，每个列表具有以下结构：

sentiment_labels =  [['B_S', 'O', 'O', 'O'], ['O', 'O', 'B_S', 'O']]
aspect_labels =     [['O', 'B_A', 'O', 'O'], ['O', 'O', 'O', 'B_A']]
modifier_labels =   [['O', 'O', 'BM', 'O'], ['O', 'O', 'O', 'O']]

# those lists contain 'B_A', 'I_S', 'B_S', 'I_S', 'BM', 'IM', and 'O' labels (IOB Tagging)

目标结果必须像：

labels = [['B_S', 'B_A', 'BM', 'O'], ['O', 'O', 'B_S', 'B_A'] ]

为此，我定义了以下函数：

# define function to integrate sentiments, aspects, and modifiers lists

def integrate_labels(sentiments_lists, aspects_lists, modifiers_lists):
  all_labels = []
  integrated_label = []

  # iterate through each list, then iterate through each element 
  for sentiment_list, aspect_list, modifier_list in zip(sentiments_lists, aspects_lists, modifiers_lists):
    
    for sentiment, aspect, modifier in zip(sentiment_list, aspect_list, modifier_list):

      # if the element is a sentiment label append it to the integrated_label list
      if sentiment != 'O':
        integrated_label.append(sentiment)

      # if the element is an aspect label append it to the integrated_label list
      elif aspect != 'O':
        integrated_label.append(aspect)

      # if the element is a modifier label append it to the integrated_label list
      elif modifier != 'O':
        integrated_label.append(modifier)

      else:
        integrated_label.append('O')
        
    # now append each integrated_label list to all_labels list
    all_labels.append(integrated_label)
  
  return all_labels

但我得到了这个结果：

integrate_labels(sentiment_labels, aspect_labels, modifier_labels)

[['B_S', 'B_A', 'BM', 'O', 'O', 'O', 'B_S', 'B_A'],
 ['B_S', 'B_A', 'BM', 'O', 'O', 'O', 'B_S', 'B_A']]

如何更改 integrate_labels 函数以获得 目标结果？

提前致谢！

Answer 1

改为

# now append each integrated_label list to all_labels list
all_labels.append(integrated_label.copy())
integrated_label = []

如何使用 python 定义函数来集成列表？

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