英文:
Get a weighted value between 0 and 1 based on zero being min value and 1 max value
问题
让我们假设我有一组数字：Set(6) { 0, 2.58, 2.74, 2.75, 4.12, 5.5 }。它们的总和为17.69。
我知道我可以通过将每个数字除以总和来获得介于0和1之间的加权/百分比值，这将产生以下结果：
{
'0': 0,
'2.58': 0.14584511023176935,
'2.74': 0.15488976823063877,
'2.75': 0.1554550593555681,
'4.12': 0.2328999434708875,
'5.5': 0.3109101187111362
}
然而，我正在寻找的结果更像是这样的：
{
'0': 0,
'2.58': 约为0.5,
'2.74': 稍微超过0.5,
'2.75': 稍微超过0.5,
'4.12': 约为0.8,
'5.5': 1
}
因此，数字越接近集合中的最大数字，它就越接近1。
我确信我在这里漏掉了一些简单的东西。
为了提供背景信息，我需要能够在包含这些数字的方框周围显示一个淡化的颜色，因此我正在寻找一个介于0和1之间的值，以便我可以使用CSS将其分配给不透明度。
英文:
Let's set I have a set of numbers: Set(6) { 0, 2.58, 2.74, 2.75, 4.12, 5.5 }. These sum to 17.69.
I know I can get a weighted / percentage based value between 0 and 1 by dividing each number by the total, which produces:
{
'0': 0,
'2.58': 0.14584511023176935,
'2.74': 0.15488976823063877,
'2.75': 0.1554550593555681,
'4.12': 0.2328999434708875,
'5.5': 0.3109101187111362
}
The result I am looking for though, is more like this:
{
'0': 0,
'2.58': around 0.5,
'2.74': just over 0.5,
'2.75': just over 0.5,
'4.12': around 0.8,
'5.5': 1
}
So the closer a number is to the higher number in the set, the closer to 1 it is.
I'm sure I'm missing something simple here.
For context, I need to be able to show a faded color around the box containing these numbers so I'm looking for a value between 0 and 1 so that I can assign that to the opacity using CSS.
答案1
得分: 1
mySet = new Set([0, 2.58, 2.74, 2.75, 4.12, 5.5]);
myArray = Array.from(mySet);
max = myArray.reduce((prev, curr) => {
if (prev === null) {
return curr;
} else if (curr > prev) {
return curr;
} else {
return prev;
}
}, null);
resultAsArray = myArray.map(item => item / max);
resultsAsObject = myArray.reduce((prev, curr, i) => {
return Object.assign(prev, { [curr]: resultAsArray[i] });
}, {})
如果你想将类似 0.4982734923847 的数字处理为较短的形式，
你可以这样做，例如：{[curr]: resultAsArray[i].toFixed(2)}。
英文:
mySet = new Set([0, 2.58, 2.74, 2.75, 4.12, 5.5]);
myArray = Array.from(mySet);
max = myArray.reduce((prev, curr) => {
if (prev === null) {
return curr;
} else if (curr > prev) {
return curr;
} else {
return prev;
}
}, null);
resultAsArray = myArray.map(item => item/max);
resultsAsObject = myArray.reduce((prev, curr, i) => {
return Object.assign(prev, {[curr]: resultAsArray[i]});
}, {})
If you want to handle numbers like 0.4982734923847 as shorter things,
you can do, for example, {[curr]: resultAsArray[i].toFixed(2)}
答案2
得分: 0
对于数组中的每个元素：
- 减去输入数组的最小值
- 除以（输入数组的最大值减去最小值）
- 如果需要，四舍五入。
例如，输入：[7, 13, 17, 22]
- 减去7得：[0, 6, 10, 15]
- 除以15得：[0, 6/15, 10/15, 15/15] = [0, 0.4, 0.666667, 1]
英文:
For every element in your array:
- subtract min(input array)
- divide by (max(input array) - min(input array))
- round if you like.
E.g., inputs: [7,13, 17, 22]
- subtraction by 7 yields: [0, 6, 10, 15]
- division by 15 yields: [0, 6/15, 10/15, 15/15] = [0, 0.4, 0.666667, 1]
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