折线图系列的对象数组?(+ 表示交叉过滤器解决方案)

Question

在JavaScript中，如何将"长表格式"转换为适合折线图的数据结构：

var data = [
  {"type":"A"," year":2000," value":50},
  {"type":"A"," year":2001," value":51},
  {"type":"A"," year":2002," value":52},
  {"type":"B"," year":2000," value":60},
  {"type":"B"," year":2001," value":55},
  {"type":"B"," year":2002," value":57}
]
=>
var series = [ 
  {type: "A", values : [{x: 2000, y: 50}, {x: 2001, y: 52},] },
  {type: "B", values : [{x: 2000, y: 60}, {x: 2001, y: 55},] },
]

一个普通的JavaScript解决方案以及一个带有交叉过滤器库的解决方案 - 可以在数据的任何维度上工作 - 将是有价值的：https://github.com/square/crossfilter/wiki/API-Reference

Answer 1

好问题，但不是前行的;

看看 https://github.com/NickQiZhu/dc.js/pull/91

你会有类似的东西：

cr = crossfilter(data);
typeDim=cr.dimension(function(d) {return d.type});
typeGroup=typeDim.group();
yearDim=cr.dimension(function(d) {return d.year});
valDim=cr.dimension(function(d) {return d.value});
yearTypeGroup = dateDimension.group();
yearTypeGroup.reduce( 
    // add
    // pivot[0].idx()[i] returns the index of statusGroup
    // the reduce function hence construct a key-value object, keys being indexes of the pivot group. 
    function(p, v, i, pivot) { 
         ++p[pivot[0].idx()[i]].count;
        p[pivot[0].idx()[i]].value += +v.value;
        return p;
    },
    //remove
    function(p, v,i,pivot) {
        --p[pivot[0].idx()[i]].count;
        p[pivot[0].idx()[i]].value -= +v.value;
        return p;
    },
    //init
    function(pivot) {
        var l, i, obj ={};
        if(l = pivot[0].all().length){
            for(i=0; i<l; ++i) {
                obj[i] = {count:0, value:0}
            }
        }
        return obj
    }
);
yearTypeGroup.pivot(typeGroup)

祝你好运 ; ）三.