具有动态键的对象的变平数组，仅与“ _abs”的对象相加的值以关键名称

发布于 2025-02-05 11:19:43 字数 1105 浏览 1 评论 0原文

我有一系列看起来像这样的对象：

[{"A_abs":4,"A_perc":2, "B_abs":4,"B_perc":2, ...},{"A_abs":4,"A_perc":2, "B_abs":4,"B_perc":2, ...},{"A_abs":4,"A_perc":2, "B_abs":4,"B_perc":2, ...},{"A_abs":4,"A_perc":2, "B_abs":4,"B_perc":2, ...}]

每个对象中的元素可以n元素，每个对象都由_perc和_abs值组成。我想将数组仅减少到1个条目，然后将每个_abs值总结，_perc可以是其他任何东西，例如“ - ”。

预期的结果是：

{"A_abs": 16, "A_perc": "-" "B_abs": 16, "B_perc": "-"  "C_abs": x, "C_perc": "-", ...}

解决方案这个问题并没有完全指出我知道如何使函数仅总和“ _abs”值。

如何适应此函数仅总和_abs值，而对对象的条目有多少个条目？

var data = [{"A_abs":4,"A_perc":2, "B_abs":4,"B_perc":2},{"A_abs":4,"A_perc":2, "B_abs":4,"B_perc":2},{"A_abs":4,"A_perc":2, "B_abs":4,"B_perc":2},{"A_abs":4,"A_perc":2, "B_abs":4,"B_perc":2}],
    result = data.reduce((r, o) => (Object.entries(o).forEach(([k, v]) => r[k] = (r[k] || 0) + v), r), {});

console.log(result);

原文

I have an array of objects that looks like this:

[{"A_abs":4,"A_perc":2, "B_abs":4,"B_perc":2, ...},{"A_abs":4,"A_perc":2, "B_abs":4,"B_perc":2, ...},{"A_abs":4,"A_perc":2, "B_abs":4,"B_perc":2, ...},{"A_abs":4,"A_perc":2, "B_abs":4,"B_perc":2, ...}]

There can be up to n elements in each object, each one consists of a _perc and an _abs value.
I want to reduce the array to only 1 entry and sum each _abs value, the _perc can be anything else, like "-".

An expected result would be:

{"A_abs": 16, "A_perc": "-" "B_abs": 16, "B_perc": "-"  "C_abs": x, "C_perc": "-", ...}

The solution of this question does not fully statisfy me because I don't know how to adapt the function to only sum the "_abs" values.

How to adapt this function to only sum the _abs values, regardles of how many entries the object has?

var data = [{"A_abs":4,"A_perc":2, "B_abs":4,"B_perc":2},{"A_abs":4,"A_perc":2, "B_abs":4,"B_perc":2},{"A_abs":4,"A_perc":2, "B_abs":4,"B_perc":2},{"A_abs":4,"A_perc":2, "B_abs":4,"B_perc":2}],
    result = data.reduce((r, o) => (Object.entries(o).forEach(([k, v]) => r[k] = (r[k] || 0) + v), r), {});

console.log(result);

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呆萌少年 2025-02-12 11:19:43

您可以检查key 以 _abs结束，并且仅当它添加值时，您只需分配> -的值即可。

const data = [
    { A_abs: 4, A_perc: 2, B_abs: 4, B_perc: 2 },
    { A_abs: 4, A_perc: 2, B_abs: 4, B_perc: 2 },
    { A_abs: 4, A_perc: 2, B_abs: 4, B_perc: 2 },
    { A_abs: 4, A_perc: 2, B_abs: 4, B_perc: 2 },
  ],
  result = data.reduce(
    (r, o) => (
      Object.entries(o).forEach(([k, v]) =>
        k.endsWith("_abs") ? (r[k] = (r[k] || 0) + v) : (r[k] = "-")
      ),
      r
    ),
    {}
  );

console.log(result);

You can check if the key ends with _abs and only if it does you add the values else you simply assign a value of -.

const data = [
    { A_abs: 4, A_perc: 2, B_abs: 4, B_perc: 2 },
    { A_abs: 4, A_perc: 2, B_abs: 4, B_perc: 2 },
    { A_abs: 4, A_perc: 2, B_abs: 4, B_perc: 2 },
    { A_abs: 4, A_perc: 2, B_abs: 4, B_perc: 2 },
  ],
  result = data.reduce(
    (r, o) => (
      Object.entries(o).forEach(([k, v]) =>
        k.endsWith("_abs") ? (r[k] = (r[k] || 0) + v) : (r[k] = "-")
      ),
      r
    ),
    {}
  );

console.log(result);

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