我正在尝试将1D数组分解为2D数组，其中子数组的长度不同。这种变化应遵循高斯曲线（或丘形）。因此，假设我们制作的2D数组变量名为gaussianCurve。 gaussianCurve [0]和gaussianCurve [n]中的数组长度为1，而gaussianCurve [n / 2]则为参数“ maxArrayLength”提供的最大值。这迫使gaussianCurve索引的数量变为可变的。
说我有以下伪代码：
function (oneDimentionalArray, maxArrayLength) {
// oneDimentionalArray is ["A","B","C","D","E","F","G","H","I","J","K"]
// maxArrayLength is 5
// Currently working like this (i.e. "batches"):
// return [["A","B","C","D","E"],["F","G","H","I","J"],["K"]]
// would LIKE it to work like this
gaussianCurve = []
gaussianCurve.push(["A"])
gaussianCurve.push(["B", "C"])
gaussianCurve.push(["D", "E", "F", "G", "H"])
gaussianCurve.push(["I", "J"])
gaussianCurve.push(["K"])
return gaussianCurve
}
我为什么要这样的东西？进度条。
他们没有表明我正在立即取得进步
这是因为必须先完成第一个作业，然后钢筋才能移动
他们放慢速度达到95％以上，有时甚至会坚持100％
只是烦人
欢迎任何建议。我只是在脑海中看不到答案。
编辑：我觉得它的措词很差，所以我改写了它。
... gaussianCurve [0] .length和gaussianCurve [gaussianCurve.length-1] .length将为1，而gaussianCurve [gaussianCurve.length / 2] .length将达到“ maxArrayLength”。
输入：
function gaussianRefactor(["A","B","C","D","E","F","G","H","I","J","K"], 1)
function gaussianRefactor(["A","B","C","D","E","F","G","H","I","J","K"], 2)
function gaussianRefactor(["A","B","C","D","E","F","G","H","I","J","K"], 4)
function gaussianRefactor(["A","B","C","D","E","F","G","H","I","J","K"], 8)
function gaussianRefactor(["A","B","C","D","E","F","G","H","I","J","K"], 16)
输出：
[["A"],["B"],["C"],["D"],["E"],["F"],["G"],["H"],["I"],["J"],["K"]]
[["A"],["B","C"],["D","E"],["F","G"],["H","I"],["J"],["K"]]
[["A"],["B","C","D"],["E","F","G","H"],["I","J","K"]]
[["A"],["B","C","D","E","F","G","H","I"],["J","K"]]
[["A","B","C","D","E","F","G","H","I","J","K"]]
内部数组的长度不能超过maxArrayLength的长度
最佳答案
我给了它一个快速的镜头，它似乎起作用。一些潜在的改进：
输入检查功能
它将所有可能的剩余值放入中间仓。对于偶数总数的仓，它将受益于一些平衡。之后，尝试根据输入数据中的原始索引对每个bin进行排序可能是一件好事，因为现在事情可能最终会乱序。但是，如果这只是为了使进度条具有非线性分布的作业，则顺序可能无关紧要。
function probability(s, m, x) {
var eExp = -Math.pow(x - m, 2) /
(2 * Math.pow(s, 2));
return 1/(Math.sqrt(2*Math.PI) * s) *
Math.pow(Math.E, eExp);
}
function gassianArray(input, nBins) {
// first try to determine a reasonable value of s so that the outer bins have a value
var s = 0.1;
var sMax = 10;
var m = (nBins - 1) / 2.0;
var outerBinMinimum = 1 / input.length;
var p = 0;
while (true && s <= sMax) {
p = probability(s, m, 0);
if (p >= outerBinMinimum) {
break;
} else {
s += 0.1;
}
}
// holds arrays
var output = [];
// holds desired array sizes
var outputLengths = [];
// fill these based on probability density
for (var b=0; b<nBins; b++) {
var n = Math.floor(probability(s, m, b) * input.length);
output.push([]);
outputLengths.push(n);
}
// fill arrays from outside, leaving extra values for the middle
var midIndex = Math.floor(m);
// left side
for (var i=0; i<midIndex; i++) {
for (var j=0; j<outputLengths[i]; j++) {
output[i].push(input.shift());
}
}
// right side
for (var i=nBins-1; i>=midIndex; i--) {
for (var j=0; j<outputLengths[i]; j++) {
output[i].push(input.pop());
}
output[i].reverse();
}
// whatever remains goes in the "middle"
while (input.length !== 0) {
output[midIndex].unshift(input.pop());
}
return output;
}
var input = ["A","B","C","D","E","F","G","H","I","J","K"];
var n = 5;
console.log(gassianArray(input, n));
/*
[ [ 'A' ],
[ 'B', 'C' ],
[ 'E', 'D', 'F', 'G', 'H' ],
[ 'I', 'J' ],
[ 'K' ] ]
*/
var input = ["A","B","C","D","E","F","G","H","I","J","K","L","M","N","O","P","Q","R","S","T","U","V","W","X","Y","Z"];
var n = 6;
console.log(gassianArray(input, n));
/*
[ [ 'A' ],
[ 'B', 'C', 'D', 'E' ],
[ 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N' ],
[ 'O', 'P', 'Q', 'R', 'S', 'T', 'U' ],
[ 'V', 'W', 'X', 'Y' ],
[ 'Z' ] ]
*/关于javascript - 一维-> 2D数组WITH/法线曲线子数组长度，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/55943151/