从字典列表中提取具有给定键的字典的Python程序

给定字典列表，任务是编写一个Python程序，该程序仅提取包含特定给定键值的那些字典。

Input : test_list = [{‘gfg’ : 2, ‘is’ : 8, ‘good’ : 3}, {‘gfg’ : 1, ‘for’ : 10, ‘geeks’ : 9}, {‘love’ : 3}], key= “gfg”
Output : [{‘gfg’: 2, ‘is’: 8, ‘good’: 3}, {‘gfg’ : 1, ‘for’ : 10, ‘geeks’ : 9}] 
Explanation : gfg is present in first two dictionaries, hence extracted.
 

Input : test_list = [{‘gfg’ : 2, ‘is’ : 8, ‘good’ : 3}, {‘gfg’ : 1, ‘for’ : 10, ‘geeks’ : 9}, {‘love’ : 3, ‘gfgs’ : 4}], key = “good”
Output : [{‘gfg’: 2, ‘is’: 8, ‘good’: 3}] 
Explanation : good is present in 1st dictionary, hence extracted. 
 

方法 1：使用列表理解和keys()

在此，我们使用in 运算符测试密钥是否存在，使用 key() 提取密钥。列表理解用于迭代不同的字典。

例子：

Python3

# Python3 code to demonstrate working of
# Extract Dictionaries with given Key
# Using list comprehension + keys()
  
# initializing list
test_list = [{'gfg': 2, 'is': 8, 'good': 3},
             {'gfg': 1, 'for': 10, 'geeks': 9},
             {'love': 3}]
  
# printing original list
print("The original list is : " + str(test_list))
  
# initializing key
key = 'gfg'
  
# checking for key using in operator
# keys() used to get keys
res = [sub for sub in test_list if key in list(sub.keys())]
  
# printing result
print("The filtered Dictionaries : " + str(res))

输出：

The original list is : [{‘gfg’: 2, ‘is’: 8, ‘good’: 3}, {‘gfg’: 1, ‘for’: 10, ‘geeks’: 9}, {‘love’: 3}]

The filtered Dictionaries : [{‘gfg’: 2, ‘is’: 8, ‘good’: 3}, {‘gfg’: 1, ‘for’: 10, ‘geeks’: 9}]

方法 2：使用filter()和lambda

在这里，我们使用 filter() 执行过滤任务，并且使用 lambda函数将逻辑注入过滤。 in运算符用于检查特定键的存在。

例子：

蟒蛇3

# Python3 code to demonstrate working of
# Extract Dictionaries with given Key
# Using filter() + lambda
  
# initializing list
test_list = [{'gfg': 2, 'is': 8, 'good': 3},
             {'gfg': 1, 'for': 10, 'geeks': 9},
             {'love': 3, 'gfg': 4}]
  
# printing original list
print("The original list is : " + str(test_list))
  
# initializing key
key = 'good'
  
# checking for key using in operator
# keys() used to get keys
# filter() + lambda used to perform fileration
res = list(filter(lambda sub: key in list(sub.keys()), test_list))
  
# printing result
print("The filtered Dictionaries : " + str(res))

输出：

The original list is : [{‘gfg’: 2, ‘is’: 8, ‘good’: 3}, {‘gfg’: 1, ‘for’: 10, ‘geeks’: 9}, {‘love’: 3, ‘gfg’: 4}]

The filtered Dictionaries : [{‘gfg’: 2, ‘is’: 8, ‘good’: 3}]