Python - 如果不存在特定值，则从字典列表中删除字典

给定一个字典列表，删除所有没有 K 作为值的字典。

例子：

Input : test_list = [{“Gfg” : 4, “is” : 8, “best” : 9}, {“Gfg” : 3, “is”: 7, “best” : 5}], K = 7 
Output : [{‘Gfg’: 4, ‘is’: 8, ‘best’: 9}] 
Explanation : Resultant dictionary doesn’t contain 7 as any element.

Input : test_list = [{“Gfg” : 4, “is” : 7, “best” : 9}, {“Gfg” : 3, “is”: 7, “best” : 5}], K = 7 
Output : [] 
Explanation : All contain 7 as element in List. 
 

方法 #1：使用循环 + values()

这是可以执行此任务的方法之一。在这种情况下，我们使用循环执行迭代所有字典的任务，并使用 values() 检查值的存在。

Python3

# Python3 code to demonstrate working of 
# Remove dictionary if K value not present
# Using loop + values()
  
# initializing lists
test_list = [{"Gfg" : 4, "is" : 8, "best" : 9},
             {"Gfg" : 5, "is": 8, "best" : 1},
             {"Gfg" : 3, "is": 7, "best" : 6}, 
             {"Gfg" : 3, "is": 7, "best" : 5}]
  
# printing original list
print("The original list : " + str(test_list))
  
# initializing K 
K = 7
  
res = []
  
# using loop to check for K element 
for sub in test_list:
    if K not in list(sub.values()):
        res.append(sub)
  
# printing result 
print("Filtered dictionaries : " + str(res))

输出：

The original list : [{‘Gfg’: 4, ‘is’: 8, ‘best’: 9}, {‘Gfg’: 5, ‘is’: 8, ‘best’: 1}, {‘Gfg’: 3, ‘is’: 7, ‘best’: 6}, {‘Gfg’: 3, ‘is’: 7, ‘best’: 5}]
Filtered dictionaries : [{‘Gfg’: 4, ‘is’: 8, ‘best’: 9}, {‘Gfg’: 5, ‘is’: 8, ‘best’: 1}]

方法#2：使用列表理解

这是可以执行此任务的另一种方式。在这里，我们使用列表理解使用单行提取所有值。使用 values() 提取值。

蟒蛇3

# Python3 code to demonstrate working of 
# Remove dictionary if K value not present
# Using list comprehension
  
# initializing lists
test_list = [{"Gfg" : 4, "is" : 8, "best" : 9},
             {"Gfg" : 5, "is": 8, "best" : 1},
             {"Gfg" : 3, "is": 7, "best" : 6}, 
             {"Gfg" : 3, "is": 7, "best" : 5}]
  
# printing original list
print("The original list : " + str(test_list))
  
# initializing K 
K = 7
  
res = []
  
# using one-liner to extract dicts with NO K value
# using not in operator to check presence
res = [sub for sub in test_list if K not in list(sub.values())]
  
# printing result 
print("Filtered dictionaries : " + str(res))

输出：

The original list : [{‘Gfg’: 4, ‘is’: 8, ‘best’: 9}, {‘Gfg’: 5, ‘is’: 8, ‘best’: 1}, {‘Gfg’: 3, ‘is’: 7, ‘best’: 6}, {‘Gfg’: 3, ‘is’: 7, ‘best’: 5}]
Filtered dictionaries : [{‘Gfg’: 4, ‘is’: 8, ‘best’: 9}, {‘Gfg’: 5, ‘is’: 8, ‘best’: 1}]