python之如何在 python 中动态组合集合

我有一个字典列表，需要遍历它们并检查已经存在的键。我已经实现了一个 python 代码来手动计算分数，如下所示。在我的代码中，我在每次迭代中手动组合以前字典中的键。迭代将从 dict11 开始。

如何更改此代码以自动遍历动态数量的字典，以及在每次迭代中如何动态组合键？

 
dict10 = {'A': 1, 'C': 2} 
dict11 = {'B': 3, 'C': 4} 
dict12 = {'A': 5, 'E': 6, 'F': 7} 
dict13 = {'G': 8, 'E': 9} 
 
exist_score = 0 
 
for key in dict11.keys() & dict10.keys(): 
    exist_score += dict11[key] 
 
for key in dict12.keys() & set(dict11.keys()).union(set(dict10.keys())): 
    exist_score += dict12[key] 
 
for key in dict13.keys() & set(dict12.keys()).union(set(dict11.keys()).union(set(dict10.keys()))): 
    exist_score += dict13[key] 
 
print(exist_score) 
 

请您参考如下方法:

首先，您需要将它变成可以放入循环中的东西。 dict11、dict12 和 dict13 也会发生同样的事情:

# same definition for dict10, dict11, dict12, dict13 
 
exist_score = 0 
seen_keys = set(dict10.keys()) 
 
for key in dict11.keys(): 
  if key in seen_keys: 
    exist_score += dict11[key] 
seen_keys.update(dict11.keys()) 
 
for key in dict12.keys(): 
  if key in seen_keys: 
    exist_score += dict12[key] 
seen_keys.update(dict12.keys()) 
 
for key in dict13.keys(): 
  if key in seen_keys: 
    exist_score += dict13[key] 
seen_keys.update(dict13.keys()) 

这应该与您的脚本执行相同的操作。现在你可以把它放到一个循环中......

# same definition for dict10, dict11, dict12, dict13 
 
exist_score = 0 
seen_keys = set(dict10.keys()) 
other_dicts = [dict11, dict12, dict13] 
 
for d in other_dicts: 
  for key in d.keys(): 
    if key in seen_keys: 
      exist_score += d[key] 
  seen_keys.update(d.keys())