我正在构建一个工具,通过最近的时间戳连接多个流。流可能不同步,因此我将最后n
(可能是> = 500)项存储在固定大小的循环缓冲区中。我想使用sortedIndex
(非search
)来查找项目在缓冲区中的位置。我需要这个索引才能在时间戳之前和之后找到流项目。
处理尖端问题的边缘情况并不重要,我不在乎是否返回数组外的索引或0
以获取最大值。昨晚我正在玩这个实现这个,但无法弄清楚是否正在实施。
功能合约如下,基于_.sortedIndex
(implementation)
/**
* Binary search for finding the closest computed (by iterator) value
* in some sorted circular array
*
* @param {Array} array a circular array-like
* @param {Object} value to search for an index
* @param {Function} iterator to compute the compare value of an item
*/
function sortedIndex(array, value, iterator) {
var low = 0,
high = array.length;
while (low != high && iterator(array[low]) > iterator(array[high])) {
// The binary search I failed to implement
}
return low;
}
一些测试用例(再次随意处理角落情况): http://jsbin.com/yusilaba/1/edit
function identity(x) {
return x;
}
function property(prop) {
return function(x) {
return x[prop];
};
}
test('sortedIndex should work on simple case', function(t) {
var array = [1, 2, 3, 4];
equal(sortedIndex(array, 2, identity), 1, 'equal case sorts towards left');
equal(sortedIndex(array, 2.5, identity), 2);
equal(sortedIndex(array, 10, identity), 0);
equal(sortedIndex(array, -10, identity), 3);
array = [{a: 1}, {a: 2}, {a: 3}, {a: 4}];
equal(sortedIndex(array, {a: 2}, property('a')), 1);
equal(sortedIndex(array, {a: 2.5}, property('a')), 2);
equal(sortedIndex(array, {a: 10}, property('a')), 0);
equal(sortedIndex(array, {a: -10}, property('a')), 3);
});
test('sortedIndex should work on circular collections', function() {
var array = [2, 3, 4, 1, 1.5];
equal(sortedIndex(array, 2, identity), 0, 'equal case sorts towards left');
equal(sortedIndex(array, 2.5, identity), 1);
equal(sortedIndex(array, 10, identity), 3);
equal(sortedIndex(array, -10, identity), 2);
equal(sortedIndex(array, 5, identity), 4);
equal(sortedIndex(array, 3.5, identity), 3);
array = [{a: 2}, {a: 3}, {a: 4}, {a: 1}, {a: 1.5}];
equal(sortedIndex(array, {a: 2}, property('a')), 0, 'equal case sorts towards left');
equal(sortedIndex(array, {a: 2.5}, property('a')), 1);
equal(sortedIndex(array, {a: 10}, property('a')), 3);
equal(sortedIndex(array, {a: -10}, property('a')), 2);
});
编辑---这是我完成的版本https://github.com/trevnorris/cbuffer/pull/14
sortedIndex : function(value, comparitor, context) {
var low = this.start,
high = this.size - 1;
// Tricky part is finding if its before or after the pivot
// we can get this info by checking if the target is less than
// the last item. After that it's just a typical binary search.
if (low && comparitor.call(context, value, this.data[high]) > 0) {
low = 0, high = this.end;
}
while (low < high) {
var mid = (low + high) >>> 1;
if (comparitor.call(context, value, this.data[mid]) > 0) low = mid + 1;
else high = mid;
}
// http://stackoverflow.com/a/18618273/1517919
return (((low - this.start) % this.size) + this.size) % this.size;
}
答案 0 :(得分:3)
这是我想出的(它通过了你的测试用例)。基本上,它在数组排序时进行正常的二进制搜索。当它是圆形时(例如:[2,3,4,1]),它找到了枢轴(它是圆开始的位置的索引,因此在该示例中索引3,对应于数组中的4,将是枢轴),然后二进制搜索数据透视图所在的数组部分。
function findPivot(arr, low, high, iterable){
// base cases
if (high < low) return -1;
if (high == low) return low;
var mid = Math.floor((low + high)/2);
if (mid < high && iterable(arr[mid]) > iterable(arr[mid + 1]))
return mid;
if (mid > low && iterable(arr[mid]) < iterable(arr[mid - 1]))
return (mid-1);
if (iterable(arr[low]) >= iterable(arr[mid]))
return findPivot(arr, low, mid-1, iterable);
else
return findPivot(arr, mid + 1, high, iterable);
}
function binarySearch(arr, low, high, val, iterable)
{
if (high < low)
return low;
var mid = Math.floor((low + high)/2);
if (iterable(val) == iterable(arr[mid]))
return mid;
if (iterable(val) > iterable(arr[mid]))
return binarySearch(arr, (mid + 1), high, val, iterable);
else
return binarySearch(arr, low, (mid -1), val, iterable);
}
function sortedIndex(array, value, iterable) {
var arr_size = array.length;
var pivot = findPivot(array, 0, arr_size-1, iterable);
if (pivot == -1) {
if(iterable(array[arr_size-1]) < iterable(value)){
return 0;
} else if(iterable(array[0]) > iterable(value)){
return arr_size-1;
}
return binarySearch(array, 0, arr_size-1, value, iterable);
}
if(iterable(array[pivot]) < iterable(value)){
return pivot+1;
} else if(iterable(array[pivot+1]) > iterable(value)){
return pivot;
}
if (iterable(array[pivot]) == iterable(value))
return pivot;
if (iterable(array[0]) <= iterable(value))
return binarySearch(array, 0, pivot-1, value, iterable);
else
return binarySearch(array, pivot+1, arr_size-1, value, iterable);
}
以下是测试用例:http://jsbin.com/ratufewa/1/edit
希望这是正确的方向。
迭代解决方案http://jsbin.com/ratufewa/3/edit:
function findPivot(arr, low, high, iterable)
{
while(true){
// base cases
if (high < low) return -1;
if (high == low) return low;
var mid = (low + high) >>> 1;
if (mid < high && iterable(arr[mid]) > iterable(arr[mid + 1]))
return mid;
if (mid > low && iterable(arr[mid]) < iterable(arr[mid - 1]))
return (mid-1);
if (iterable(arr[low]) >= iterable(arr[mid]))
high = mid-1;
else
low = mid + 1;
}
}
function binarySearch(arr, low, high, val, iterable)
{
while(true){
if (high < low)
return low;
var mid = (low + high) >>> 1;
if (iterable(val) == iterable(arr[mid]))
return mid;
if (iterable(val) > iterable(arr[mid]))
low = mid + 1;
else
high = mid -1;
}
}
function sortedIndex(array, value, iterable) {
var arr_size = array.length;
var pivot = findPivot(array, 0, arr_size-1, iterable);
if (pivot == -1) {
if(iterable(array[arr_size-1]) < iterable(value)){
return 0;
} else if(iterable(array[0]) > iterable(value)){
return arr_size-1;
}
return binarySearch(array, 0, arr_size-1, value, iterable);
}
if(iterable(array[pivot]) < iterable(value)){
return pivot+1;
} else if(iterable(array[pivot+1]) > iterable(value)){
return pivot;
}
if (iterable(array[pivot]) == iterable(value))
return pivot;
if (iterable(array[0]) <= iterable(value))
return binarySearch(array, 0, pivot-1, value, iterable);
else
return binarySearch(array, pivot+1, arr_size-1, value, iterable);
}
答案 1 :(得分:1)
通常,在circular buffer中,您会存储实际占用位置的开头和结尾。有了这些信息,它应该是相当微不足道的,因为我们可以区分三种情况:
function sortedIndex(buffer, item, getValue) {
if (buffer.start < buffer.end)
// do standard binary search between start and end indices
else if (getValue(buffer[0]) <= getValue(item))
// do standard binary search between 0 and end index
else // getValue(buffer[0] > getValue(item)
// do standard binary search between start and buffer.length
}