JS 将 2 个数组及其单个索引中的单个值合并为 1 个数组中的单个索引

Question

我想知道如何使用下面的代码实现上述目标。因此，例如，当数据被连接时，我将在新数组中的索引中组合结果类似于 [2.62, 460]。当用户单击按钮时，以下两个函数都通过事件侦听器调用。任何帮助将不胜感激，谢谢。

var mouseDistance = new Array();
var timers = new Array();
var combinedresults = new Array();

//THIS FUNCTION CALCULATES THE DISTANCE MOVED
function printMousePos(e) {
    var lastSeenAt = {
        x: null,
        y: null
    };
    var cursorX = e.clientX;
    var cursorY = e.clientY;

    var math = Math.round(Math.sqrt(Math.pow(lastSeenAt.y - cursorY, 2) +
        Math.pow(lastSeenAt.x - cursorX, 2)));
    mouseDistance.push(math);
}

function stopCount() {
    clearTimeout(t);
    timer_is_on = 0;
    timers.push(t);
}

Answer 1

您可以将 lastSeenAt 的 init 移出函数。并在函数末尾分配新值。

要获得组合结果，请使用mouseDistance 和timers 中的较短者，并将数据推送到combinedresults 应该可以工作。

var mouseDistance = new Array();
var timers = new Array();
var combinedresults = new Array();
// Init with both is null.
var lastSeenAt = {
  x: null,
  y: null
};

//THIS FUNCTION CALCULATES THE DISTANCE MOVED
function printMousePos(e) {
  var cursorX = e.clientX;
  var cursorY = e.clientY;

  // Don't calculate when x, y is null, which is the first time.
  // Or you can give lastSeen some other initValue rather than (null, null).
  if (lastSeenAt.x !== null) {
    var math = Math.round(Math.sqrt(Math.pow(lastSeenAt.y - cursorY, 2) +
        Math.pow(lastSeenAt.x - cursorX, 2)));
    mouseDistance.push(math);      
  }

  // Keep the x,y value.
  lastSeenAt.x = cursorX;
  lastSeenAt.y = cursorY;
}

function stopCount() {
    clearTimeout(t);
    timer_is_on = 0;
    timers.push(t);
}

// get combinedresults 
function getCombinedResult() {
  // Get the shorter length.
  var length = Math.min(mouseDistance.length, timers.length);
  var i;

  // 
  for (i = 0; i < length; ++i) {
    combinedresults[i] = [timers[i], mouseDistance[i]];
  }
}