我是python的新手，真的很想在这里获得一些指导。
我有两个几乎相同的字典-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：我需要按照以下流程进行比较：
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，我的代码还不多：)很抱歉，还在学习。
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)
答案 0 :(得分：1)
我想您不仅想知道第一本字典与第二本字典的区别，反之亦然。 对我来说，一个好方法是按照以下步骤分离控件：
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])))
答案 1 :(得分：0)
您可以执行以下操作：
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"]}
for key, values in First_Dict.items():
if key in Second_Dict: # if key match
diff = [value for value in values if value not in Second_Dict[key]]
if not diff: # all values match
pass
else:
print("key: {}, missing values: {}".format(key, diff))
else:
print("key: {}, missing values: {}".format(key, values))
输出
key: Florida, missing values: ['Naples']
key: Texas, missing values: ['Dallas']
key: Arizona, missing values: ['Phoenix', 'Tucson']
第diff = [value for value in values if value not in Second_Dict[key]]行是list comprehension，当键匹配时，计算First_Dict和Second_Dict中的值之差。
更新
如果您需要两个区别，可以执行以下操作：
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"]}
for key, values in First_Dict.items():
if key in Second_Dict: # if key match
diff_first = [value for value in values if value not in Second_Dict[key]]
diff_second = [value for value in Second_Dict[key] if value not in values]
if not diff_first: # all values match
pass
else:
print("key: {}, missing values: {} in Second_Dict".format(key, diff_first))
if not diff_second:
pass
else:
print("key: {}, missing values: {} in First_Dict".format(key, diff_second))
else:
print("key: {}, missing values: {} in Second_Dict".format(key, values))
for key, values in Second_Dict.items():
if key not in First_Dict:
print("key: {}, missing values: {} in First_Dict".format(key, values))
输出
key: Texas, missing values: ['Dallas'] in Second_Dict
key: Florida, missing values: ['Naples'] in Second_Dict
key: Arizona, missing values: ['Phoenix', 'Tucson'] in Second_Dict
key: Illinois, missing values: ['Chicago', 'Naperville'] in First_Dict
第二个循环用于遍历Second_Dict中缺少的First_Dict中的键。
答案 2 :(得分：0)
首先声明两个空列表，一个用于存储丢失的键，第二个用于存储该键的值。
key_lst=[]
values_lst=[]
尝试此代码。
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","Dallas"],
"California": ["San Diego", "Los Angeles", "San Francisco"],
"Florida": ["Miami", "Orlando", "Jacksonville",], "Illinois":
["Chicago", "Naperville"]}
key_lst=[]
values_lst=[]
for key, value in First_Dict.items() and Second_Dict.items():
if key in First_Dict.keys() and Second_Dict.keys():
if key in Second_Dict.keys() and First_Dict.keys() :
continue
else:
key_lst.append(key)
else:
key_lst.append(key)
if value in First_Dict.values() and Second_Dict.values():
if value in Second_Dict.values() and First_Dict.values() :
continue
else:
values_lst.append(value)
else:
values_lst.append(value)
for key, value in Second_Dict.items() and First_Dict.items():
if key in First_Dict.keys() and Second_Dict.keys():
if key in Second_Dict.keys() and First_Dict.keys() :
continue
else:
key_lst.append(key)
else:
key_lst.append(key)
if value in First_Dict.values() and Second_Dict.values():
if value in Second_Dict.values() and First_Dict.values() :
continue
else:
values_lst.append(value)
else:
values_lst.append(value)
print("Missing Keys: ",key_lst[0],": Missing Values",values_lst[0])
print("Missing Keys: ",key_lst[1],": Missing Values",values_lst[1])
输出为
Missing Keys: Illinois : Missing Values ['Chicago', 'Naperville']
Missing Keys: Arizona : Missing Values ['Phoenix', 'Tucson']
如果有帮助，请标记答案。
答案 3 :(得分：0)
一组使用选项的更多选择：
mismatch = {}
missing_from_first = {}
missing_from_second = {}
for state in list(set([state for state in First_Dict] + [state for state in Second_Dict])):
set1 = set(First_Dict.get(state, []))
set2 = set(Second_Dict.get(state, []))
mismatch[state] = list(set1.union(set2) - set1.intersection(set2))
missing_from_first[state] = list(set2 - set1)
missing_from_second[state] = list(set1 - set2)
因此，打印出结果：
print mismatch
print missing_from_first
print missing_from_second
#=> {'Florida': ['Naples'], 'Arizona': ['Tucson', 'Phoenix'], 'California': [], 'Texas': ['Dallas'], 'Illinois': ['Naperville', 'Chicago']}
#=> {'Florida': [], 'Arizona': [], 'California': [], 'Texas': [], 'Illinois': ['Naperville', 'Chicago']}
#=> {'Florida': ['Naples'], 'Arizona': ['Tucson', 'Phoenix'], 'California': [], 'Texas': ['Dallas'], 'Illinois': []}
遍历结果以按需格式化打印。