尽管我更喜欢使用直SQL而不是DSL，但在某些情况下需要使用DSL：当我们要对分组的行执行自定义聚合时。
我不确定的是如何将这种简单的计算转换为 100.0 * (count(*)-sum(dep_delayed))/count(*)
：
DSL UDF中的等效项是什么？是否有必要创建一个AggUDF（/ // list of cities
var liste = [
{ name: "New York", distance: 12, weight: 5},
{ name: "Atlanta", distance: 4, weight: 4},
{ name: "Dallas", distance: 2, weight: 2},
{ name: "Los Angeles", distance: 1, weight: 1},
];;
var repeatTimes = 4;
var choose = [];
choose = [liste.map(x=>x.name), liste.map(x=>x.distance), liste.map(x=>x.weight)];
// randomaly pick a city based on its weight
var rand = function(min, max) {
return Math.random() * (max - min) + min;
};
var getRandomItem = function(choose, weight) {
var total_weight = weight.reduce(function (prev, cur, i, arr) {
return prev + cur;
});
var random_num = rand(0, total_weight);
var weight_sum = 0;
//console.log(random_num)
for (var i = 0; i < choose.length; i++) {
weight_sum += weight[i];
weight_sum = +weight_sum.toFixed(2);
if (random_num <= weight_sum) {
return choose[i];
}
}
// end of function, modify the weights of list again
};
// for the first time pick a city randomaly
var random_item = getRandomItem(choose[0], choose[2]);
console.log(random_item);
newWeights();
// after the first time of picking cities let's modify the weights of list
function newWeights(){
var baseDistance = liste.find(l => l.name == random_item).distance;
console.log(choose[0].indexOf(random_item));
var list = liste.map(c => {
var newWeight = Math.abs(baseDistance - c.distance);
return {
name: c.name,
distance: c.distance,
weight: newWeight
};
});
liste = list.filter(function(value){
return value.weight != 0;
});
choose = [liste.map(x=>x.name), liste.map(x=>x.distance), liste.map(x=>x.weight)];
console.log(liste);
}
// use the modified list to randomaly picking another city
for (var i = 1; i < repeatTimes; i++) {
var random_item = getRandomItem(choose[0], choose[2]);
console.log(random_item);
newWeights();
}
//），如果是，该怎么做？
答案 0 :(得分：0)
我今天碰到了答案：使用
expr("some sql clause")
因此代码可以实现为
.agg(expr("100.0 * (count(*)-sum(dep_delayed))/count(*)"))