查找嵌套数组的最大长度

Question

我有一个数组，其中的节点和边在边内，每个节点都有源、目标、军官数组，我想创建一个军官数组长度的比例。

  var officersScale = d3.scale.linear()
                                        .domain([1,  d3.max(dataset.edges.officers.length, function(d) { return d[0]; })])
                                        .range([1,10]);

数组的样子：

var dataset = {"nodes":[{"name":"Anglo American plc"},{"name":"Associated British Foods plc"},{"name":"Admiral Group plc"},{"name":"ARM Holdings plc"},{"name":"Aviva plc"},{"name":"BAE Systems plc"},{"name":"Barclays PLC"},{"name":"British American Tobacco p.l.c."},{"name":"The Berkeley Group Holdings plc"},{"name":"BHP Billiton plc"},{"name":"BP p.l.c."},{"name":"BT Group plc"},{"name":"Centrica plc"},{"name":"Compass Group PLC"},{"name":"Direct Line Insurance Group PLC"},{"name":"easyJet plc"},{"name":"International Consolidated Airlines Group, S.A."},{"name":"3i Group plc"},{"name":"Intertek Group plc"},{"name":"Lloyds Banking Group plc"},{"name":"Merlin Entertainments plc"},{"name":"Pearson plc"},{"name":"Reckitt Benckiser Group plc"},{"name":"Royal Dutch Shell plc"},{"name":"Rexam PLC"},{"name":"Rio Tinto plc"},{"name":"SABMiller plc"},{"name":"Schroders plc"},{"name":"Standard Life plc"},{"name":"SSE plc"},{"name":"Standard Chartered PLC"},{"name":"Tesco PLC"},{"name":"Whitbread PLC"}],"edges":[{"source":0,"target":29,"officers":["MEDORI, Rene"]},{"source":0,"target":30,"officers":["GROTE, Byron Elmer"]},{"source":0,"target":31,"officers":["GROTE, Byron Elmer"]},{"source":1,"target":13,"officers":["BASON, John George"]}]};

Answer 1

首先，您的数据无效。存在无效令牌，一些内部对象缺少{ 等。

但是，您可以遍历所有官员数组，然后找到最大长度。我希望你知道如何创造规模，这是给你的。

var edges = [{"source":0,"target":29,"officers":["MEDORI, Rene"]},
           {"source":6,"target":28,"officers":["GILLIES, Crawford Scott","GRIMSTONE, Gerald Edgar, Sir"]},
           {"source":9,"target":10,"officers":["ANDERSON, Paul Milton"]},
           {"source":9,"target":19,"officers":["FREW, Anita Margaret"]},
           {"source":9,"target":23,"officers":["BRINDED, Malcolm Arthur"]},
           {"source":15,"target":32,"officers":["KENNEDY, Christopher John"]},
           {"source":24,"target":31,"officers":["CHAMBERS, Stuart John"]},
           {"source":28,"target":29,"officers":["GILLIES, Crawford Scott"]},
           "source":30,"target":31,"officers":["GROTE, Byron Elmer"]}];

使用d3.max

var maxlength = d3.max(edges, function(d) { return d.officers.length; })

使用简单的Array.prototype.map 和Math.max

var lengths = edges.map(function(edge){
     return edge.officers.length;
});

console.log(lengths);
var maxLength = Math.max(...lengths);

console.log(maxLength);

单步，ES6

maxLength = Math.max(...edges.map(edge=> edge.officers.length));