Python程序将字典中具有相似值的键分组

给定一个值作为列表的字典。将所有具有相似值的键组合在一起。

Input : test_dict = {“Gfg”: [5, 6], “is”: [8, 6, 9], “best”: [10, 9], “for”: [5, 2], “geeks”: [19]} 
Output : [[‘Gfg’, ‘is’, ‘for’], [‘is’, ‘Gfg’, ‘best’], [‘best’, ‘is’], [‘for’, ‘Gfg’]] 
Explanation : Gfg has 6, is has 6, and for has 5 which is also in Gfg hence, grouped.

Input : test_dict = {“Gfg”: [6], “is”: [8, 6, 9], “best”: [10, 9], “for”: [5, 2]} 
Output : [[‘Gfg’, ‘is’], [‘is’, ‘Gfg’, ‘best’], [‘best’, ‘is’]] 
Explanation : Gfg has 6, is has 6, hence grouped. 
 

方法：使用循环 + any() + len()

以上功能的组合可以解决这个问题。在这里，我们使用 any() 检查每个键是否与任何值匹配，len() 用于检查匹配键是否超过 1。迭代部分使用循环进行。

Python3

# Python3 code to demonstrate working of 
# Group keys with similar values in Dictionary
# Using loop + any() + len()
  
# initializing dictionary
test_dict = {"Gfg": [5, 6], "is": [8, 6, 9], 
             "best": [10, 9], "for": [5, 2], 
             "geeks": [19]}
  
# printing original dictionary
print("The original dictionary is : " + str(test_dict))
  
res = []
for key in test_dict:
    chr = [key]
    for ele in test_dict:
          
        # check with other keys 
        if key != ele:
              
            # checking for any match in values
            if any(i == j for i in test_dict[key] for j in test_dict[ele]):
                chr.append(ele)
    if len(chr) > 1:
        res.append(chr)
  
# printing result 
print("The dictionary after values replacement : " + str(res))

输出：

The original dictionary is : {‘Gfg’: [5, 6], ‘is’: [8, 6, 9], ‘best’: [10, 9], ‘for’: [5, 2], ‘geeks’: [19]}

The dictionary after values replacement : [[‘Gfg’, ‘is’, ‘for’], [‘is’, ‘Gfg’, ‘best’], [‘best’, ‘is’], [‘for’, ‘Gfg’]]