【发布时间】:2020-03-12 16:10:43
【问题描述】:
我在 csv 文件中有以下数据。
y,x1,x2,x3,x4,x5,x6,x7,x8,x9
10,2113,1985,38.9,64.7,4,868,59.7,2205,1917
11,2003,2855,38.8,61.3,3,615,55,2096,1575
11,2957,1737,40.1,60,14,914,65.6,1847,2175
13,2285,2905,41.6,45.3,-4,957,61.4,1903,2476
10,2971,1666,39.2,53.8,15,836,66.1,1457,1866
11,2309,2927,39.7,74.1,8,786,61,1848,2339
10,2528,2341,38.1,65.4,12,754,66.1,1564,2092
11,2147,2737,37,78.3,-1,761,58,1821,1909
4,1689,1414,42.1,47.6,-3,714,57,2577,2001
2,2566,1838,42.3,54.2,-1,797,58.9,2476,2254
7,2363,1480,37.3,48,19,984,67.5,1984,2217
example = data.frame(x1,x2,x3,x4,y)
如何使用scatter3D(x,y,z) 绘制变量x1、x2、x3?
我试过了:
library("plot3D")
with(example,scatter3D(y ~ x1 + x2 + x3))
但我得到错误:
min(x,na.rm) 中的错误:参数的“类型”(列表)无效
【问题讨论】:
-
您尝试过什么,收到了什么错误消息?您是否阅读过函数的手册页或软件包的小插图?
-
是的,但是获取错误的参数类型无效
-
您想要从这些 3D 数据中得到回归平面还是曲面?
-
我可以尝试将所有数据 x1、x2、x3 转换为数组
xx = as.array(cbind(x1,x2,x3))。我想要的图表:plot points,但是 "z" 它需要什么值? -
这是 3d 而不是 4d。试试
with(example, scatter3D(y, x1, x2))。
标签: r linear-regression scatter3d plot3d