如何获取将python字典的总和列为元组？

发布于 2025-02-07 21:47:49 字数 370 浏览 6 评论 0原文

我的python词典被列为如下所示；在这里，每个元组的第一个值描绘了组号，每个字典描述了与组内部的每个数字相关的生成值（ie，{'1'：0，'3'：1，'2'：1，，'303'：3}

[('1',{'1': 0,'3': 1,'2': 1,'5': 2,'303': 3}), ('2',{'4': 0,'5': 1,'7': 1,'5': 2,'303': 2}), ('3',{'1': 0,'3': 0,'2': 0,'5': 2,'303': 3})]

是否有任何方法来获取每个组的总和

[('1', 7), ('2', 6), ('3', 5)]

'5'： 2 这种方法是如此昂贵。

原文

I have python dictionary listed as tuples as follows; Here the first value of the each tuple depicts the group number and each dictionary depicts the generated value related to the each number inside the group, (i.e., {'1': 0,'3': 1,'2': 1,'5': 2,'303': 3}.

[('1',{'1': 0,'3': 1,'2': 1,'5': 2,'303': 3}), ('2',{'4': 0,'5': 1,'7': 1,'5': 2,'303': 2}), ('3',{'1': 0,'3': 0,'2': 0,'5': 2,'303': 3})]

Is there any method to get the sum of the each group by getting the sum of the each group, (i.e.,

[('1', 7), ('2', 6), ('3', 5)]

I tried to iterate over the each value and get the sum but that method is so expensive.

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

丘比特射中我 2025-02-14 21:47:49

一个简单的列表理解应该如此

>>> [(k, sum(dic.values())) for k, dic in tuples]

A simple list comprehension should so

>>> [(k, sum(dic.values())) for k, dic in tuples]

回复收藏 0 原文

等往事风中吹 2025-02-14 21:47:49

lst = [
    ("1", {"1": 0, "3": 1, "2": 1, "5": 2, "303": 3}),
    ("2", {"4": 0, "3": 1, "7": 1, "5": 2, "303": 2}),
    ("3", {"1": 0, "3": 0, "2": 0, "5": 2, "303": 3}),
]

sum_of_each_group = [(i, sum(d.values())) for (i, d) in lst]

print(sum_of_each_group)

# output: [('1', 7), ('2', 6), ('3', 5)]

lst = [
    ("1", {"1": 0, "3": 1, "2": 1, "5": 2, "303": 3}),
    ("2", {"4": 0, "3": 1, "7": 1, "5": 2, "303": 2}),
    ("3", {"1": 0, "3": 0, "2": 0, "5": 2, "303": 3}),
]

sum_of_each_group = [(i, sum(d.values())) for (i, d) in lst]

print(sum_of_each_group)

# output: [('1', 7), ('2', 6), ('3', 5)]

回复收藏 0 原文

~没有更多了~