Python - 提取值总和大于 K 的字典

给定一个字典列表，提取所有键总和超过 K 的字典。

Input : test_list = [{“Gfg” : 4, “is” : 8, “best” : 9}, {“Gfg” : 3, “is”: 7, “best” : 5}], K = 15
Output : [{‘Gfg’: 4, ‘is’: 8, ‘best’: 9}]
Explanation : 4 + 9 + 8 = 21. Greater than K, hence returned.

Input : test_list = [{“Gfg” : 4, “is” : 8, “best” : 9}, {“Gfg” : 3, “is”: 7, “best” : 5}], K = 25
Output : []
Explanation : No dictionary with sum > 25.

方法#1：使用循环

这是执行此任务的蛮力方式。在这里，我们迭代所有字典并提取每个字典的总和，如果超过 K，我们将它们过滤掉。

Python3

# Python3 code to demonstrate working of 
# Extract dictionaries with values sum greater than K
# Using 
  
# 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 = 15
  
# using loop to extract all dictionaries
res = []
for sub in test_list:
    sum = 0
    for key in sub:
        sum += sub[key]
    if sum > K:
        res.append(sub)
  
# printing result 
print("Dictionaries with summation greater than K : " + 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}]
Dictionaries with summation greater than K : [{'Gfg': 4, 'is': 8, 'best': 9}, {'Gfg': 3, 'is': 7, 'best': 6}]

方法 #2：使用列表理解 + sum()

这是可以执行此任务的一种线性方式。在这里，我们使用 sum() 执行求和任务，列表理解可用于执行将所有逻辑组合成一行的任务。

蟒蛇3

# Python3 code to demonstrate working of 
# Extract dictionaries with values sum greater than K
# Using list comprehension + sum()
  
# 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 = 15
  
# using one-liner to extract all the dictionaries
res = [sub for sub in test_list if sum(list(sub.values())) > K]
  
# printing result 
print("Dictionaries with summation greater than K : " + 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}]
Dictionaries with summation greater than K : [{'Gfg': 4, 'is': 8, 'best': 9}, {'Gfg': 3, 'is': 7, 'best': 6}]