Python - 两个字典值列表的交叉映射

给定两个具有列表值的字典，通过检查值-键链接来执行第一个列表的键与其他列表的值的映射。

Input : test_dict1 = {“Gfg” : [4, 10], “Best” : [8, 6], “is” : [9, 3]}, test_dict2 = {6 : [15, 9], 8 : [6, 3], 7 : [9, 8], 9 : [10, 11]} Output : {‘Best’: [6, 3, 15, 9], ‘is’: [10, 11]} Explanation : “Best” has 8 and 6, which are mapped to 6, 3 and 15, 9 hence output for that key. Input : test_dict1 = {“Gfg” : [4, 10], “Best” : [18, 16], “is” : [9, 3]}, test_dict2 = {6 : [15, 9], 8 : [6, 3], 7 : [9, 8], 9 : [10, 11]} Output : {‘is’: [10, 11]} Explanation : Only 9 present as possible key.

方法 #1：使用循环 + setdefault() + extend()

上述功能的组合可以用来解决这个问题。在此，我们使用 get() 执行获取具有值的匹配键的任务，并使用 setdefault 构造空列表以进行映射。

Python3

# Python3 code to demonstrate working of
# Cross mapping of Two dictionary value lists
# Using loop + setdefault() + extend()
 
# initializing dictionaries
test_dict1 = {"Gfg" : [4, 7], "Best" : [8, 6], "is" : [9, 3]}
test_dict2 = {6 : [15, 9], 8 : [6, 3], 7 : [9, 8], 9 : [10, 11]}
 
# printing original lists
print("The original dictionary 1 is : " + str(test_dict1))
print("The original dictionary 2 is : " + str(test_dict2))
 
res = {}
 
# getting all values of first dictionary
for key, val in test_dict1.items():
    for key1 in val:
         
        # getting result with default value list and extending
        # according to value obtained from get()
        res.setdefault(key, []).extend(test_dict2.get(key1, []))
 
# printing result
print("The constructed dictionary : " + str(res))

输出

The original dictionary 1 is : {'Gfg': [4, 7], 'Best': [8, 6], 'is': [9, 3]}
The original dictionary 2 is : {6: [15, 9], 8: [6, 3], 7: [9, 8], 9: [10, 11]}
The constructed dictionary : {'Gfg': [9, 8], 'Best': [6, 3, 15, 9], 'is': [10, 11]}

方法#2：使用列表理解+字典理解

这是可以解决此问题的另一种方法。在此，我们使用列表推导提取所有映射，然后通过交叉映射提取的值来构造新字典。

Python3

# Python3 code to demonstrate working of
# Cross mapping of Two dictionary value lists
# Using list comprehension + dictionary comprehension
 
# initializing dictionaries
test_dict1 = {"Gfg" : [4, 7], "Best" : [8, 6], "is" : [9, 3]}
test_dict2 = {6 : [15, 9], 8 : [6, 3], 7 : [9, 8], 9 : [10, 11]}
 
# printing original lists
print("The original dictionary 1 is : " + str(test_dict1))
print("The original dictionary 2 is : " + str(test_dict2))
 
# using internal and external comprehension to
# solve problem
res = {key: [j for i in val if i in test_dict2 for j in test_dict2[i]]
      for key, val in test_dict1.items()}
 
# printing result
print("The constructed dictionary : " + str(res))

输出

The original dictionary 1 is : {'Gfg': [4, 7], 'Best': [8, 6], 'is': [9, 3]}
The original dictionary 2 is : {6: [15, 9], 8: [6, 3], 7: [9, 8], 9: [10, 11]}
The constructed dictionary : {'Gfg': [9, 8], 'Best': [6, 3, 15, 9], 'is': [10, 11]}