我有一个数组,里面有几天。每天都是一个对象,例如:
{day_year: "2012", day_month: "08", day_number: "03", day_name: "mon"}
我还使用以下方法将时间戳记属性添加到了每天的对象中:
function convertDays() {
var max_i = days.length;
for(var i = 0; i < max_i; i++) {
var tar_i = days[i];
tar_i.timestamp = new Date(tar_i.day_year, tar_i.day_month, tar_i.day_number);
}
}
数组中的日期是任意的,因此没有真正的逻辑。
现在,我想找到距给定日期最近的两天。因此,如果带天的数组包含
并且我搜索2012年8月11日,希望它返回2012年8月4日和2012年8月23日。
我尝试使用另一个问题的答案,看起来像这样:
function findClosest(a, x) {
var lo, hi;
for(var i = a.length; i--;) {
if(a[i] <= x && (lo === undefined || lo < a[i])) lo = a[i];
if(a[i] >= x && (hi === undefined || hi > a[i])) hi = a[i];
}
return [lo, hi];
}
但是,它返回unidentified
。
什么是最有效的方法(最少的处理器/内存密集型方法)?
编辑:“但是,结果如何“奇怪”?您能否提供代码和数据的示例?”
我现在正在使用以下内容来生成日期数组:
var full_day_array = [];
for(var i = 0; i < 10; i++) {
var d = new Date();
d.setDate(d.getDate() + i);
full_day_array.push({day_year: d.getFullYear().toString(), day_month: (d.getMonth() + 1).toString(), day_number: d.getDate().toString()});
}
奇怪的是,使用下面的代码,这仅适用于10个或更短日期的数组。每当我使用11个或更多日期的数组时,结果就会变得出乎意料。
例如:使用15个日期组成的数组,从2012年8月6日到2012年8月21日。如果我再调用findClosest(full_day_array, new Date("30/07/2012");
,则希望它返回{nextIndex: 0, prevIndex: -1}
。但是,它返回{nextIndex: 7, prevIndex: -1}
。为什么?
function findClosest(objects, testDate) {
var nextDateIndexesByDiff = [],
prevDateIndexesByDiff = [];
for(var i = 0; i < objects.length; i++) {
var thisDateStr = [objects[i].day_month, objects[i].day_number, objects[i].day_year].join('/'),
thisDate = new Date(thisDateStr),
curDiff = testDate - thisDate;
curDiff < 0
? nextDateIndexesByDiff.push([i, curDiff])
: prevDateIndexesByDiff.push([i, curDiff]);
}
nextDateIndexesByDiff.sort(function(a, b) { return a[1] < b[1]; });
prevDateIndexesByDiff.sort(function(a, b) { return a[1] > b[1]; });
var nextIndex;
var prevIndex;
if(nextDateIndexesByDiff.length < 1) {
nextIndex = -1;
} else {
nextIndex = nextDateIndexesByDiff[0][0];
}
if(prevDateIndexesByDiff.length < 1) {
prevIndex = -1;
} else {
prevIndex = prevDateIndexesByDiff[0][0];
}
return {nextIndex: nextIndex, prevIndex: prevIndex};
}
您可以通过自定义比较器功能轻松使用sort
function:
如果使用 var distancea = Math.abs(diffdate - a.the_date_object);
var distanceb = Math.abs(diffdate - b.the_date_object);
对象的数组而不是您自己定义的结构,则可以在O(N)中轻松实现:
var testDate = new Date(...);
var bestPrevDate = days.length;
var bestNextDate = days.length;
var max_date_value = Math.abs((new Date(0,0,0)).valueOf());
var bestPrevDiff = max_date_value;
var bestNextDiff = -max_date_value;
var currDiff = 0;
var i;
for(i = 0; i < days.length; ++i){
currDiff = testDate - days[i].the_date_object;
if(currDiff < 0 && currDiff > bestNextDiff){
// If currDiff is negative, then testDate is more in the past than days[i].
// This means, that from testDate's point of view, days[i] is in the future
// and thus by a candidate for the next date.
bestNextDate = i;
bestNextDiff = currDiff;
}
if(currDiff > 0 && currDiff < bestPrevDiff){
// If currDiff is positive, then testDate is more in the future than days[i].
// This means, that from testDate's point of view, days[i] is in the past
// and thus by a candidate for the previous date.
bestPrevDate = i;
bestPrevDiff = currDiff;
}
}
/* days[bestPrevDate] is the best previous date,
days[bestNextDate] is the best next date */
答案非常好,但是如果您想知道任一方向上最近的N个对象,我对如何处理这个问题很感兴趣。这是我的刺:
不管日期数组有多长,这都有效:
function newFindClosest(dates, testDate) {
var before = [];
var after = [];
var max = dates.length;
for(var i = 0; i < max; i++) {
var tar = dates[i];
var arrDate = new Date(tar.day_year, tar.day_month, tar.day_number);
// 3600 * 24 * 1000 = calculating milliseconds to days, for clarity.
var diff = (arrDate - testDate) / (3600 * 24 * 1000);
if(diff > 0) {
before.push({diff: diff, index: i});
} else {
after.push({diff: diff, index: i});
}
}
before.sort(function(a, b) {
if(a.diff < b.diff) {
return -1;
}
if(a.diff > b.diff) {
return 1;
}
return 0;
});
after.sort(function(a, b) {
if(a.diff > b.diff) {
return -1;
}
if(a.diff < b.diff) {
return 1;
}
return 0;
});
return {datesBefore: before, datesAfter: after};
}