情景:
我有d3呈现多个网络。
它由单个节点,多个连接的节点网络组成。
我正在尝试附加d3-force(模拟),以便在拖动时节点不重叠,整个网络保持相对定位。
使用了很多人的力量。
问题:
如果有人能指出我正确的解决方案,我将不胜感激。
这是一个小小提琴:https://jsfiddle.net/pdubey84/or0tn49k/2/
var nodes = [
{ id: "mammal", group: 0, label: "Mammals", level: 1 },
{ id: "dog" , group: 0, label: "Dogs" , level: 2 },
{ id: "cat" , group: 0, label: "Cats" , level: 2 },
{ id: "fox" , group: 0, label: "Foxes" , level: 2 },
{ id: "elk" , group: 0, label: "Elk" , level: 2 },
{ id: "insect", group: 1, label: "Insects", level: 1 },
{ id: "ant" , group: 1, label: "Ants" , level: 2 },
{ id: "bee" , group: 1, label: "Bees" , level: 2 },
{ id: "fish" , group: 2, label: "Fish" , level: 1 },
{ id: "carp" , group: 2, label: "Carp" , level: 2 },
{ id: "pike" , group: 2, label: "Pikes" , level: 2 },
{ id: "pike" , group: 3, label: "Pikes" , level: 1 },
{ id: "pike" , group: 4, label: "Pikes" , level: 1 },
{ id: "pike" , group: 4, label: "Pikes" , level: 1 },
{ id: "foo" , group: 4, label: "foo" , level: 1 },
{ id: "bar" , group: 4, label: "bar" , level: 1 }
]
var links = [
{ target: "mammal", source: "dog" , strength: 0.7 },
{ target: "mammal", source: "cat" , strength: 0.7 },
{ target: "mammal", source: "fox" , strength: 0.7 },
{ target: "mammal", source: "elk" , strength: 0.7 },
{ target: "insect", source: "ant" , strength: 0.7 },
{ target: "insect", source: "bee" , strength: 0.7 },
{ target: "fish" , source: "carp", strength: 0.7 },
{ target: "fish" , source: "pike", strength: 0.7 },
{ target: "cat" , source: "elk" , strength: 0.1 },
{ target: "carp" , source: "ant" , strength: 0.1 },
{ target: "elk" , source: "bee" , strength: 0.1 },
{ target: "dog" , source: "cat" , strength: 0.1 },
{ target: "fox" , source: "ant" , strength: 0.1 },
{ target: "pike" , source: "cat" , strength: 0.1 },
{ target: "foo" , source: "bar" , strength: 0.1 }
]
function getNodeColor(node, neighbors) {
return 'red'
}
function getLinkColor(node, link) {
return '000'
}
var width = window.innerWidth
var height = window.innerHeight
var svg = d3.select('svg')
svg.attr('width', width).attr('height', height)
// simulation setup with all forces
var linkForce = d3
.forceLink()
.id(function (link) { return link.id })
.strength(function (link) { return link.strength })
var simulation = d3
.forceSimulation()
.force('link', linkForce)
.force('charge', d3.forceManyBody().strength(-500).distanceMax(50))
.force('center', d3.forceCenter(width / 2, height / 2))
var dragDrop = d3.drag().on('start', function (node) {
node.fx = node.x
node.fy = node.y
}).on('drag', function (node) {
simulation.alphaTarget(0.7).restart()
node.fx = d3.event.x
node.fy = d3.event.y
}).on('end', function (node) {
if (!d3.event.active) {
simulation.alphaTarget(0)
}
node.fx = null
node.fy = null
})
var linkElements = svg.append("g")
.attr("class", "links")
.selectAll("line")
.data(links)
.enter().append("line")
.attr("stroke-width", 1)
.attr("stroke", "rgba(50, 50, 50, 0.2)")
var nodeElements = svg.append("g")
.attr("class", "nodes")
.selectAll("circle")
.data(nodes)
.enter().append("circle")
.attr("r", 10)
.attr("fill", getNodeColor)
.call(dragDrop)
simulation.nodes(nodes).on('tick', () => {
nodeElements
.attr('cx', function (node) { return node.x })
.attr('cy', function (node) { return node.y })
linkElements
.attr('x1', function (link) { return link.source.x })
.attr('y1', function (link) { return link.source.y })
.attr('x2', function (link) { return link.target.x })
.attr('y2', function (link) { return link.target.y })
})
simulation.force("link").links(links)
答案 0 :(得分:0)
这两种行为都是预期的。这就是你可以为这些问题做的事情:
网络分散:我想你正在谈论拖动功能,因为加载时模拟只运行300个滴答。如果是这种情况,请使用simulation.alpha
代替simulation.alphaTarget
。 simulation.alpha
是控制熵的正确方法:
simulation.alpha(0.7).restart();
当您拖动一个网络时,所有网络都会移动:这是因为您使用的是d3.forceCenter()
。根据{{3}}:
定心力均匀地转换节点,使得所有节点的平均位置(如果所有节点具有相同的权重,则为质心)位于给定位置。 (强调我的)
你可以做的是在第一次拖动时取消此力:
simulation.force("center", null)
以下是这两个变化的小提琴:API