我有一个带有许多边的密集点图和一个与几乎任何其他节点都有连接的节点。
例如：
import {
Pipe,
PipeTransform
} from '@angular/core';
@Pipe({
name: 'orderBy',
pure: false
})
export class OrderByConnectedPipe implements PipeTransform {
value: string[] = [];
private static _orderByComparator(a: any, b: any): number {
if ((a === null || a === undefined) && (b === null || b === undefined)) {
return 0;
}
if (a === null || a === undefined) {
a = (isNaN(parseFloat(b)) || !isFinite(b)) ? '' : 0;
}
if (b === null || b === undefined) {
b = (isNaN(parseFloat(a)) || !isFinite(a)) ? '' : 0;
}
if ((isNaN(parseFloat(a)) || !isFinite(a)) || (isNaN(parseFloat(b)) || !isFinite(b))) {
if (a.toLocaleLowerCase() < b.toLocaleLowerCase()) {
return -1;
}
if (a.toLocaleLowerCase() > b.toLocaleLowerCase()) {
return 1;
}
} else {
if (parseFloat(a) < parseFloat(b)) {
return -1;
}
if (parseFloat(a) > parseFloat(b)) {
return 1;
}
}
return 0;
}
transform(input: any, config: string = '+'): any {
//invalid input given
if (!input) return input;
//make a copy of the input's reference
this.value = [...input];
let value = this.value;
if (!Array.isArray(value)) {
return value;
}
if (!Array.isArray(config) || (Array.isArray(config) && config.length == 1)) {
let propertyToCheck: string = !Array.isArray(config) ? config : config[0];
let desc = propertyToCheck.substr(0, 1) == '-';
//Basic array
if (!propertyToCheck || propertyToCheck == '-' || propertyToCheck == '+') {
return !desc ? value.sort() : value.sort().reverse();
} else {
let property: string = propertyToCheck.substr(0, 1) == '+' || propertyToCheck.substr(0, 1) == '-' ?
propertyToCheck.substr(1) :
propertyToCheck;
return value.sort(function(a: any, b: any) {
let aValue = a[property];
let bValue = b[property];
let propertySplit = property.split('.');
if (typeof aValue === 'undefined' && typeof bValue === 'undefined' && propertySplit.length > 1) {
aValue = a;
bValue = b;
for (let j = 0; j < propertySplit.length; j++) {
aValue = aValue[propertySplit[j]];
bValue = bValue[propertySplit[j]];
}
}
return !desc ?
OrderByConnectedPipe._orderByComparator(aValue, bValue) :
-OrderByConnectedPipe._orderByComparator(aValue, bValue);
});
}
} else {
//Loop over property of the array in order and sort
return value.sort(function(a: any, b: any) {
for (let i: number = 0; i < config.length; i++) {
let desc = config[i].substr(0, 1) == '-';
let property = config[i].substr(0, 1) == '+' || config[i].substr(0, 1) == '-' ?
config[i].substr(1) :
config[i];
let aValue = a[property];
let bValue = b[property];
let propertySplit = property.split('.');
if (typeof aValue === 'undefined' && typeof bValue === 'undefined' && propertySplit.length > 1) {
aValue = a;
bValue = b;
for (let j = 0; j < propertySplit.length; j++) {
aValue = aValue[propertySplit[j]];
bValue = bValue[propertySplit[j]];
}
}
let comparison = !desc ?
OrderByConnectedPipe._orderByComparator(aValue, bValue) :
-OrderByConnectedPipe._orderByComparator(aValue, bValue);
//Don't return 0 yet in case of needing to sort by next property
if (comparison != 0) {
return comparison;
}
}
return 0; //equal each other
});
}
}
}
export let ORDERBY_PROVIDERS = [
OrderByConnectedPipe
];我想隐藏digraph TEST
{
rankdir=LR;
node [shape=plaintext];
graph [compound=true];
A[label=<<table border="0" cellspacing="0" cellborder="1" >
<tr><td >A</td></tr>
<tr><td port='a1'>a1</td></tr>
<tr><td port='a2'>a2</td></tr>
</table>>];
B;
C;
D;
E;
F;
G;
H;
I;
A:a1:e->B:w;
A:a1:e->C:w;
A:a1:e->D:w;
A:a1:e->E:w;
A:a1:e->F:w;
A:a1:e->G:w;
A:a1:e->H:w;
A:a1:e->I:w;
A:a2:e->B:w;
B:e->C:w;
C:e->D:w;
D:e->E:w;
E:e->F:w;
E:e->G:w;
E:e->H:w;
F:e->I:w
G:e->I:w
H:e->I:w
}
个连接并添加某种跳转标签，以便为读者提供信息。
例如：
a1将属性 A:a1:e->{a1_out[label="a1"]};
{a1_Bin[label="a1"]}->B:w;
{a1_Cin[label="a1"]}->C:w;
{a1_Din[label="a1"]}->D:w;
{a1_Ein[label="a1"]}->E:w;
{a1_Fin[label="a1"]}->F:w;
{a1_Gin[label="a1"]}->G:w;
{a1_Hin[label="a1"]}->H:w;
{a1_Iin[label="a1"]}->I:w;
添加到[style=invis] - 连接只是不渲染它们，但保持布局与它们一样。结果是由于空间和其他地方的密集连接，节点和标签的位置看起来很奇怪。
完全删除连接确实会改变图形的语义和图形节点的等级（不是在这个示例中，而是在其他示例中）。
所以我正在寻找一种方法来提供a1信息，以正确计算一侧节点之间的所有依赖关系并建议它，而不是绘制渲染并在另一侧绘制这些连接。
答案 0 :(得分：1)