📜  Python - 列表和字典中的常用键

📅  最后修改于: 2022-05-13 01:55:16.397000             🧑  作者: Mango

Python - 列表和字典中的常用键

给定字典和列表,提取所有常见的键和列表。

方法#1:使用列表理解

这是可以执行此任务的方法之一。在此,我们迭代所有字典和列表值,如果找到匹配项,则将它们添加到结果中。

Python3
# Python3 code to demonstrate working of 
# Common keys in list and dictionary
# Using list comprehension
  
# initializing dictionary
test_dict = {"Gfg": 3, "is" : 5, "best" : 9, "for" : 0, "geeks" : 3}
  
# printing original dictionary
print("The original dictionary is : " + str(test_dict))
  
# initializing test_list 
test_list = ["Gfg", "best", "geeks"]
  
# using in operator to check for match 
res = [ele for ele in test_dict if ele in test_list]
  
# printing result 
print("The required result : " + str(res))


Python3
# Python3 code to demonstrate working of 
# Common keys in list and dictionary
# Using set() + intersection()
  
# initializing dictionary
test_dict = {"Gfg": 3, "is" : 5, "best" : 9, "for" : 0, "geeks" : 3}
  
# printing original dictionary
print("The original dictionary is : " + str(test_dict))
  
# initializing test_list 
test_list = ["Gfg", "best", "geeks"]
  
# intersection() used to get Common elements 
res = set(test_list).intersection(set(test_dict))
  
# printing result 
print("The required result : " + str(list(res)))


输出
The original dictionary is : {'Gfg': 3, 'is': 5, 'best': 9, 'for': 0, 'geeks': 3}
The required result : ['Gfg', 'best', 'geeks']

方法#2:使用set()+intersection()

这是可以执行此任务的另一种方式。在此,我们将容器、列表和字典键都转换为 set(),然后相交以找到所需的匹配项。

蟒蛇3

# Python3 code to demonstrate working of 
# Common keys in list and dictionary
# Using set() + intersection()
  
# initializing dictionary
test_dict = {"Gfg": 3, "is" : 5, "best" : 9, "for" : 0, "geeks" : 3}
  
# printing original dictionary
print("The original dictionary is : " + str(test_dict))
  
# initializing test_list 
test_list = ["Gfg", "best", "geeks"]
  
# intersection() used to get Common elements 
res = set(test_list).intersection(set(test_dict))
  
# printing result 
print("The required result : " + str(list(res))) 
输出
The original dictionary is : {'Gfg': 3, 'is': 5, 'best': 9, 'for': 0, 'geeks': 3}
The required result : ['best', 'geeks', 'Gfg']