在下面的示例中,我试图通过Respond_Id
的唯一出现来求和。例如。在这种情况下,它应该是3
,"Respond_Id"
258,261
和345
。
这是我的数据:
{"Respond_Id":258,"Gender":"Female","Age":"18-21","Answer":424},
{"Respond_Id":258,"Gender":"Female","Age":"18-21","Answer":428},
{"Respond_Id":261,"Gender":"Male","Age":"22-26", "Answer":427},
{"Respond_Id":261,"Gender":"Male","Age":"22-26", "Answer":432},
{"Respond_Id":345,"Gender":"Female","Age":"27-30","Answer":424},
{"Respond_Id":345,"Gender":"Female","Age":"27-30","Answer":425},
{"Respond_Id":345,"Gender":"Female","Age":"27-30","Answer":433},
我知道我应该使用group reduce,所以我试过(改编自一个例子):
var ntotal = answerDim.group().reduce(
function(p, d) {
if(d.Respond_Id in p.Respond_Ids){
p.Respond_Ids[d.Respond_Id]++;
}
else {
p.Respond_Ids[d.Respond_Id] = 1;
p.RespondCount++;
}
return p;
},
function (p, d) {
p.Respond_Ids[d.Respond_Id]--;
if(p.Respond_Ids[d.Respond_Id] === 0){
delete p.Respond_Ids[d.Respond_Id];
p.RespondCount--;
}
return p;
},
function () {
return {
RespondCount: 0,
Respond_Ids: {}
};
}
);
然后:
numberDisplay
.group(ntotal)
.valueAccessor(function(d){ return d.value.RespondCount; });
dc.renderAll();
但似乎没有用。有人知道如何使它工作吗?谢谢
答案 0 :(得分:0)
根据您的JSFiddle,您的设置如下:
var RespondDim = ndx.dimension(function (d) { return d.Respond_Id;});
var ntotal = RespondDim.group().reduce(
function(p, d) {
if(d.Respond_Id in p.Respond_Ids){
p.Respond_Ids[d.Respond_Id]++;
}
else {
p.Respond_Ids[d.Respond_Id] = 1;
p.RespondCount++;
}
return p;
},
function (p, d) {
p.Respond_Ids[d.Respond_Id]--;
if(p.Respond_Ids[d.Respond_Id] === 0){
delete p.Respond_Ids[d.Respond_Id];
p.RespondCount--;
}
return p;
},
function () {
return {
RespondCount: 0,
Respond_Ids: {}
};
});
这里需要注意的重要一点是,默认情况下,您的组密钥与维度密钥相同。因此,每个受访者ID将有一个组。这不是你想要的。
您可以切换到使用专为此用例设计的dimension.groupAll
,但不幸的是dimension.groupAll.reduce
签名略有不同。最简单的解决方法是将维度定义为具有单个值:
var RespondDim = ndx.dimension(function (d) { return true;});
现在您将看到ntotal.all()
将如下所示:
{key: true, value: {RespondCount: 3, Respond_Ids: {258: 2, 261: 2, 345: 3}}}