python - 比较两个字典并打印缺失或没有匹配项

Question

我是 python 新手，真的很想在这里得到一些指导。

我有两个几乎相同的词典 - First_Dict 和 Second_Dict

First_Dict = {"Texas": ["San Antonio", "Austin", "Houston", "Dallas"], 
         "California": ["San Diego", "Los Angeles", "San Francisco"],
        "Florida": ["Miami", "Orlando", "Jacksonville", "Naples"], 
         "Arizona": ["Phoenix", "Tucson"]}


Second_Dict = {"Texas": ["San Antonio, Austin, Houston"],
           "California": ["San Diego, Los Angeles, San Francisco"],
           "Florida": ["Miami", "Orlando", "Jacksonville"], "Illinois": 
          ["Chicago", "Naperville"]}

目标:我需要在以下流程中比较它们:

Compare keys
    if key match
        compare values
            if all values match
                break
            else:
                print the key and the corresponding missing value/s.
                    "Missing value/s on key "Florida" in the Second_Dict"
                        "Naples"

    if keys NOT match or missing
        print the unmatching/missing key and corresponding value/s.
            "Missing key and value/s on First_Dict"
                Illinois
                    Chicago
                    Naperville

            "Missing key and value/s on Second_Dict"
                Arizona
                    Phoenix
                    Tucson

到目前为止，我的代码并不多:) 抱歉，仍在学习。

for key, value in First_Dict.items() and Second_Dict.items():
    if key in First_Dict.keys() == Second_Dict.keys():
       for value in First_Dict.value() and Second_Dict.value :
          if value in First_Dict.value() == Second_Dict.value():
              break
          else:
              print(value)

Answer 1

我想您不仅想知道第一本词典与第二本词典的差异，而且还想知道第二本词典的差异。对我来说，一个好方法是按以下步骤分离控件:

1. 查找两个字典的公共(public)键。
2. 使用通用键计算两个词典中的值差异。
3. 用相对值指示缺失的键。

可能的代码:

#Step 1
#Determination of common keys 
first_keys = first_Dict.keys() #retrieve keys of the dictionary
second_keys = second_Dict.keys()
common_keys = [key for key in first_keys if key in second_keys]

#Step 2
#so now for common keys we look for differences in value and printing them
for common in common_keys:
  townsA = first_Dict[common]
  townsB = second_Dict[common]

  #with the first statement determine the cities that are in the second
  #dictionary but not in first one.
  #with the second the opposite 
  missingOnFirst = [town for town in townsB if town not in townsA]
  missingOnSecond = [town for town in townsA if town not in townsB]

  if missingOnFirst:
    print("Missing on {0} in first dictionary: \n\t{1}".format(common,"\n\t".join(missingOnFirst)))
  if missingOnSecond:
    print("Missing on {0} in second dictionary: \n\t{1}".format(common,"\n\t".join(missingOnSecond)))

#Step 3
#printing the missing keys:
#on First dictionary
print("\n")
print("Missing key and value/s on first dictionary")
for key in second_keys:
  if key not in common_keys:
    print("{0}:\n\t{1}".format(key,"\n\t".join(second_Dict[key])))
#on Second dictionary
print("Missing key and value/s on second dictionary")
for key in first_keys:
  if key not in common_keys:
    print("{0}:\n\t{1}".format(key,"\n\t".join(first_Dict[key])))