Python–提取值总和大于 K 的字典
原文:https://www . geesforgeks . org/python-extract-dictionary-with-values-sum-大于-k/
给定一个字典列表，提取所有关键字总和超过 k 的字典。
输入:test _ list =[{“Gfg”:4、“is”:8、“best”:9 }、{“Gfg”:3、“is”:7、“best”:5 }]、K = 15 输出:[{“Gfg”:4、“is”:8、“best”:9 }] 解释 : 4 + 9 + 8 = 21。大于 K，因此返回。
输入:test _ list =[{“Gfg”:4、“is”:8、“best”:9 }、{“Gfg”:3、“is”:7、“best”:5 }]，K = 25 输出 : [] 解释:无字典带和> 25。
方法#1:使用循环
这是执行这项任务的粗暴方式。在这种情况下，我们对所有字典进行迭代，并提取每个字典的总和，超过 K，我们将它们过滤掉。
Python 3
# 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))
Output
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+求和()
这是完成这项任务的一种线性方式。在这种情况下，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单行的任务。
Python 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))
Output
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}]
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