我有一个数据矩阵（见下文），我试图将其转换为密度等值线图（Can1和Can2变量），可能使用ks或sm包。 我的问题是如何在数据集中选择那些位于（例如）密度图的80％轮廓上的点？
由于
ID Can1 Can2
4 -12.3235137 -1.0788867664
1 -12.2949912 -0.9321009837
5 -12.2835123 -1.0164225574
2 -12.2571822 -0.7094457036
3 -12.2713779 -0.9908419863
10 -12.9870438 -1.0936405526
6 -12.7167605 -1.4620772026
7 -12.8193776 -1.0911349785
8 -12.9781963 -1.1762698594
9 -12.7983478 -1.3453369581
13 -14.0389948 0.2855210115
11 -14.0015922 0.1467552738
15 -14.0723604 0.0244576488
14 -14.0743560 0.1417245145
12 -13.9898266 0.0005437008
20 -6.5881994 0.5124980991
17 -6.1812321 0.6789584579
16 -6.4704200 0.5942317307
18 -6.6960456 0.5720874622
19 -6.1159788 0.5960966790
22 -2.4794887 2.5493267897
24 -2.4918040 2.7823374576
21 -2.5145044 2.5877290160
23 -2.5048371 2.4916280770
25 -2.5018765 2.8536302559
29 -0.1781852 2.0805229401
26 -0.1581308 2.0151355747
28 -0.2118605 1.9658284615
27 -0.4184119 2.0540218901
30 -0.2994573 2.0205573385
35 2.6254869 1.3858705991
31 2.3146430 1.3510499304
33 2.5346138 1.2524229847
34 2.3741699 1.3842499455
32 2.6008389 1.3446707509
37 3.0920503 1.5807032840
38 3.1559727 1.4924092104
36 3.1593556 1.5803284343
39 3.0801444 1.6031732981
40 3.2562384 1.5810975265
43 4.8414364 2.1539254215
41 4.7938193 2.1613978258
44 4.7919209 2.2151527426
42 4.9830802 2.2374622446
45 4.7629268 2.4217335005
46 5.5631728 0.9986762598
50 5.5250403 1.0549399894
48 5.5833619 1.1368625963
47 5.5660312 1.1881215490
49 5.6224256 1.1634998303
53 5.5536366 0.2513665533
54 5.5276808 0.2685455911
51 5.7103045 0.2193839293
52 5.6014729 0.2353172964
55 5.5959034 0.2447836618
56 5.1542133 0.6070006863
59 5.0043394 0.4518710615
58 5.2314146 0.5656457888
60 5.1318728 0.4771275341
57 5.3599822 0.4918185651
61 7.0235173 -0.2669136870
63 7.0216315 -0.0097862523
64 7.0521253 -0.2457722410
62 7.0150637 -0.1456269078
65 7.0729018 -0.3573952321
69 5.8115406 -1.4652084167
67 5.7624475 -1.4147564126
68 5.8692888 -1.4695783153
70 5.9088094 -1.4927034632
66 5.8400205 -1.4817447808
71 4.8586107 -1.3111515744
73 4.7198564 -1.2891991780
72 4.9153659 -1.4499710448
74 4.7653488 -1.2839433419
75 4.7754971 -1.4655359108
77 3.8955675 -7.0922887151
78 3.8338151 -7.1595858283
80 3.7255063 -7.2147373050
79 3.7367055 -7.3468877516
76 4.0166957 -7.1952570639
答案 0 :(得分：1)
计算80％的分数。一种方式：y<- x[x > 0.8 * max(x)]（我假设你想要80％的最高等级，而不是80％）。
然后绘制y。
答案 1 :(得分：0)
经过一些搜索后，我认为可以使用MASS包中的kde2d函数来实现。