统计字典列表中所有键对应值的出现频次

以下几种Python方法可实现需求：

• 基础循环计数
  初始化空字典，遍历每个字典的所有值，逐个更新计数：

  data = [
      {'migs_ba': 'O', 'migs_eu': 'O', 'migs_org': 'O', 'migs_pl': 'O', 'migs_vi': 'O',
       'mimag': 'EM', 'mimarks_c': 'O', 'mimarks_s': 'EM', 'mims': 'EM', 'misag': 'EM', 'miuvig': 'EM'},
      {'migs_ba': 'O', 'migs_eu': 'O', 'migs_org': 'O', 'migs_pl': 'O', 'migs_vi': 'O',
       'mimag': 'EM', 'mimarks_c': 'O', 'mimarks_s': 'EM', 'mims': 'EM', 'misag': 'EM', 'miuvig': 'EM'}
  ]
  
  counts = {}
  for d in data:
      for value in d.values():
          counts[value] = counts.get(value, 0) + 1
  
  print(counts)
  # 输出: {'O': 12, 'EM': 10}
  

• 使用collections.defaultdict简化计数逻辑
  借助defaultdict自动处理默认值，减少代码判断：

  from collections import defaultdict
  
  counts = defaultdict(int)
  for d in data:
      for value in d.values():
          counts[value] += 1
  
  result = dict(counts)
  print(result)
  # 输出: {'O': 12, 'EM': 10}
  

• 使用collections.Counter快速统计
  先将所有值合并为一个列表，再用Counter一键统计：

  from collections import Counter
  
  all_values = [value for d in data for value in d.values()]
  counts = Counter(all_values)
  
  print(dict(counts))
  # 输出: {'O': 12, 'EM': 10}
  

内容的提问来源于stack exchange，提问作者Mark Miller