在Python中使用Counts来减去字典列表

发布于 2025-01-19 17:41:09 字数 895 浏览 0 评论 0原文

我已经计算出如何使用“计数器”将字典列表与下面的代码添加列表

from collections import Counter
a = [{'num': 'star1', 'count': 1},
     {'num': 'star2', 'count': 3}]
b = [{'num': 'star1', 'count': 7},
      {'num': 'star2', 'count': 2},
      {'num': 'star3', 'count': 1}]
joint = sum((Counter({elem['num']: elem['count']}) for elem in a + b), Counter())
[{'num': num, 'count': counts} for num, counts in joint .items()]

，但是，当我尝试减去错误时，我会遇到错误。例如：

from collections import Counter
a = [{'num': 'star1', 'count': 1},
     {'num': 'star2', 'count': 3}]
b = [{'num': 'star1', 'count': 7},
      {'num': 'star2', 'count': 2},
      {'num': 'star3', 'count': 1}]
joint = sum((Counter({elem['num']: elem['count']}) for elem in a - b), Counter())
[{'num': num, 'count': counts} for num, counts in joint .items()]

有人知道这项工作吗？还是我如何解决这个问题？

我尝试在计数器中使用减法功能，但似乎仍然不起作用

原文

I've worked out how to use "counter" to add lists of dictionarys with the code below

from collections import Counter
a = [{'num': 'star1', 'count': 1},
     {'num': 'star2', 'count': 3}]
b = [{'num': 'star1', 'count': 7},
      {'num': 'star2', 'count': 2},
      {'num': 'star3', 'count': 1}]
joint = sum((Counter({elem['num']: elem['count']}) for elem in a + b), Counter())
[{'num': num, 'count': counts} for num, counts in joint .items()]

However, when I try to subtract I get an error. For example:

from collections import Counter
a = [{'num': 'star1', 'count': 1},
     {'num': 'star2', 'count': 3}]
b = [{'num': 'star1', 'count': 7},
      {'num': 'star2', 'count': 2},
      {'num': 'star3', 'count': 1}]
joint = sum((Counter({elem['num']: elem['count']}) for elem in a - b), Counter())
[{'num': num, 'count': counts} for num, counts in joint .items()]

Is anyone aware of a work around to this? Or how I could approach this issue?

I've tried using the subtract function in counter but it still doesn't seem to work

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一抹苦笑 2025-01-26 17:41:09

我认为计数器不接受否定计数，但是如果您不介意没有负数，这将起作用（仅分别进行两个计数器）：

from collections import Counter
a = [{'num': 'star1', 'count': 1},
     {'num': 'star2', 'count': 3}]
b = [{'num': 'star1', 'count': 7},
      {'num': 'star2', 'count': 2},
      {'num': 'star3', 'count': 1}]
joint = sum((Counter({elem['num']: elem['count']}) for elem in a), Counter())
joint -= sum((Counter({elem['num']: elem['count']}) for elem in b), Counter())
[{'num': num, 'count': counts} for num, counts in joint .items()]

它输出：

[{'num': 'star2', 'count': 1}]

其他值（Star1和Star3）都会少少比0分别达到1-7 = -6 = -6 = -1

，您可以不使用计数器并这样做（这支持负数和您想要的任何操作）：

a_info = {d['num']:d['count'] for d in a}
b_info = {d['num']:d['count'] for d in b}
[{'num':item, 'count': a_info.get(item, 0)-b_info.get(item, 0)} for item in set(a_info)|set(b_info)]

I don't think Counter accept negative counts, but if you don't mind about not having negative counts, this will work (just doing the two Counters separately):

from collections import Counter
a = [{'num': 'star1', 'count': 1},
     {'num': 'star2', 'count': 3}]
b = [{'num': 'star1', 'count': 7},
      {'num': 'star2', 'count': 2},
      {'num': 'star3', 'count': 1}]
joint = sum((Counter({elem['num']: elem['count']}) for elem in a), Counter())
joint -= sum((Counter({elem['num']: elem['count']}) for elem in b), Counter())
[{'num': num, 'count': counts} for num, counts in joint .items()]

It outputs:

[{'num': 'star2', 'count': 1}]

The other values (star1 and star3) would both be less than 0 because they come to 1-7=-6 and 0-1=-1 respectively

Alternatively you can just not use counters and do it like this (this supports negative numbers and whatever operation you want):

a_info = {d['num']:d['count'] for d in a}
b_info = {d['num']:d['count'] for d in b}
[{'num':item, 'count': a_info.get(item, 0)-b_info.get(item, 0)} for item in set(a_info)|set(b_info)]

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