如何使用 JMESPATH 和 javascript 获取具有特定名称值的最后 3 个元素？

Question

我有以下 json 数据数组：

[{"name":"Button1","date":"1596959802144"},{"name":"Button2","date":"1596959804238"},{"name":"Button3","date":"1596959809334"},{"name":"Button1","date":"1597000878135"},{"name":"Button2","date":"1597000896335"},{"name":"Button3","date":"1597000901536"},{"name":"Button2","date":"1597000904437"},{"name":"Button3","date":"1597000909535"},{"name":"Button1","date":"1597000912250"},{"name":"Button2","date":"1597000939937"},{"name":"Button3","date":"1597000957940"},{"name":"Button2","date":"1597000964640"},{"name":"Button1","date":"1597001005141"},{"name":"Button2","date":"1597001010240"},{"name":"Button3","date":"1597001014845"},{"name":"Button2","date":"1597001021644"},{"name":"Button1","date":"1597001025738"},{"name":"Button2","date":"1597001049030"},{"name":"Button3","date":"1597001054139"},{"name":"Button1","date":"1597001057741"},{"name":"Button2","date":"1597001060340"},{"name":"Button3","date":"1597001062445"},{"name":"Button1","date":"1597002599045"},{"name":"Button1","date":"1597002604128"},{"name":"Button1","date":"1597002609546"},{"name":"Button1","date":"1597002613435"},{"name":"Button1","date":"1597002681736"},{"name":"Button1","date":"1597002690843"},{"name":"Button1","date":"1597002694136"},{"name":"Button1","date":"1597002696349"},{"name":"Button1","date":"1597002699243"}]

我想使用 JMESPath javascript 库来获取每个不同名称值的最后 3 个条目。 例如：

[{"name":"Button3","date":"1597001014845"},{"name":"Button2","date":"1597001021644"},{"name":"Button2","date":"1597001049030"},{"name":"Button3","date":"1597001054139"},{"name":"Button2","date":"1597001060340"},{"name":"Button3","date":"1597001062445"},{"name":"Button1","date":"1597002694136"},{"name":"Button1","date":"1597002696349"},{"name":"Button1","date":"1597002699243"}]

所以每个名字的最后 3 次出现 = Button*

检查stackOverflow，我看到使用JQ可以使用这个function： map_values(delpaths(keys_unsorted[:-2] | map([.])))

获取 JSON object 的每个项目的最后 N 个元素

有什么办法吗？ 还是使用其他 javascript 模块？

Answer 1

如果您不关心结果数组中的顺序，这将是一种纯粹的 JavaScript 方法：

 const getLastNForEveryName = (arr, n) => { const lastNOfEach = arr.reduce((acc, curr) => { if(acc[curr.name] == null) { // If the key doesnt exist yet, create it with the current item in the array acc[curr.name] = [curr]; } else { if(acc[curr.name].length >= n) // If the array is as big as the desired size alread, remove the first added one acc[curr.name].shift(); acc[curr.name].push(curr); // push the current item in the array } return acc; }, {}) return Object.values(lastNOfEach).flatMap(l => l); // Just get the values of the object and flatMap it, so that we dont have arrays of arrays } // Testing const values = [{"name":"Button1","date":"1596959802144"},{"name":"Button2","date":"1596959804238"},{"name":"Button3","date":"1596959809334"},{"name":"Button1","date":"1597000878135"},{"name":"Button2","date":"1597000896335"},{"name":"Button3","date":"1597000901536"},{"name":"Button2","date":"1597000904437"},{"name":"Button3","date":"1597000909535"},{"name":"Button1","date":"1597000912250"},{"name":"Button2","date":"1597000939937"},{"name":"Button3","date":"1597000957940"},{"name":"Button2","date":"1597000964640"},{"name":"Button1","date":"1597001005141"},{"name":"Button2","date":"1597001010240"},{"name":"Button3","date":"1597001014845"},{"name":"Button2","date":"1597001021644"},{"name":"Button1","date":"1597001025738"},{"name":"Button2","date":"1597001049030"},{"name":"Button3","date":"1597001054139"},{"name":"Button1","date":"1597001057741"},{"name":"Button2","date":"1597001060340"},{"name":"Button3","date":"1597001062445"},{"name":"Button1","date":"1597002599045"},{"name":"Button1","date":"1597002604128"},{"name":"Button1","date":"1597002609546"},{"name":"Button1","date":"1597002613435"},{"name":"Button1","date":"1597002681736"},{"name":"Button1","date":"1597002690843"},{"name":"Button1","date":"1597002694136"},{"name":"Button1","date":"1597002696349"},{"name":"Button1","date":"1597002699243"}]; console.log(getLastNForEveryName(values, 3));