结合两个csv(d3)

Question

我正在使用以下代码加载两个csv文件

    d3.csv("sqrt100train.csv", function(error, data2) {
    d3.csv("sqrt100test.csv", function(error, data) {

sqrt100train看起来像这样:

    Trees   Train
    1   0.059286
    2   0.057857
    3   0.032857
    4   0.03619
    5   0.022619

和sqrt100test看起来像这样:

    Trees   Test
    1   0.072222
    2   0.072778
    3   0.044444
    4   0.049444
    5   0.042778

如何组合这些数组来实现此目的:

    Trees   Train   Test
    1   0.059286    0.072222
    2   0.057857    0.072778
    3   0.032857    0.044444
    4   0.036190    0.049444
    5   0.022619    0.042778

Answer 1

问题:长度或X值不同

serie1
X,  Y
1,  11
3,  13
5,  15
6,  16

serie2
X,  Y
1,  21
2,  22
4,  24
6,  26
7,  27

BothSeries
X,  Y1, Y2
1,  11, 21
2,    , 22
3,  13, 
4,    , 24
5,  15, 
6,  16, 26
7,    , 27

这段代码解决了它.

var serie1 = [];
var serie2 = [];
var BothSeries = [[null,null,null]];


function merge() {
    BothSeries = [];
    var i, j, aux;
    //copy serie1 a BothSeries
    if(serie1.length!=0){
        for (var i = 0; i < serie1.length; i++) {
            //null represent the serie2 value
            BothSeries[i] = [serie1[i][0], serie1[i][1],null];
        }
    }
    //set the value Y of serie2 in the BothSeries when their X are equals.
    if(serie2.length!=0){
        for (var i = 0; i < serie2.length; i++) {
            for (var j = 0; j < BothSeries.length; j++) {
                if(BothSeries[j][0] == serie2[i][0]){
                    BothSeries[j][2] = serie2[i][1];
                }
            }
        }
    }
    //set the value X and Y of serie2 in the BothSeries when their X aren't equals or doesn't exist.
    var cont=0;
    if(serie2.length!=0){
        for (var i = 0; i < serie2.length; i++) {
            for (var j = 0; j < BothSeries.length; j++) {
                if(BothSeries[j][0] != serie2[i][0]){
                    cont++;
                }
            }
            if(cont==BothSeries.length)
            {
                BothSeries[BothSeries.length] = [serie2[i][0],null,serie2[i][1]];
            }
            cont=0;
        }
    }
    //order with bubble method by X
    if(BothSeries.length!=0){
        for(i=0;i<BothSeries.length-1;i++){
            for(j=0;j<BothSeries.length-i-1;j++){
                if(BothSeries[j+1][0] < BothSeries[j][0])
                {
                    aux=BothSeries[j+1];
                    BothSeries[j+1]=BothSeries[j];
                    BothSeries[j]=aux;
                }
            }
        }
    }
}