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我想为我的数据绘制一个回归平面：

结构（列表（L = C（96.4155，76.803，71.5615，68.193，65.6975，74.627，67.82，67.82，64.26，62.06，62.06，60.35，68.35，68.284 58.27, 56.42, 54.91, 53.65), a = c(-0.8375, -20.47, -22.1875, -22.6125, -22.5845, -2.2415, -12.91, -16.16, -17.9, -18.77, -2.4, -10.28, - 13.64, -15.33, -16.37, -2.4495, -8.66, -11.74, -13.46, -14.55, -2.4645, -7.58, -10.38, -12.06, -13.09), b = c(1.437, -24.5915, -29.3275) , -31.892, -33.524, -5.151, -15.22, -19.61, -22.65, -24.8, -6.239, -13.24, -17.11, -19.64, -21.55, -6.7845, -12.18, -15.45, -17.68, - 19.47, -7.124, -11.53, -14.38, -16.41, -17.89)), .Names = c("L", "a", "b"), row.names = c(NA, 25L), class = “数据帧”）

我从这里尝试了一个例子。这是我的尝试：

L<-X[1:25,2]
a<-X[1:25,3]
b<-X[1:25,4]
mod2 = lm(L~a*b)
open3d()
plot3d(x=a, y=b, z=L, type="s", col="red", size=1)

grd <- expand.grid(x1=a, x2=b )
grd$pred <-predict(mod2, newdata=grd)
persp3d(x=unique(grd[,1]), y=unique(grd[,2]), 
    z=matrix(grd[,3]), add=TRUE, alpha=0.5)

3d-Plot 看起来不错，但对于网格，我收到此错误：

Error in persp3d.default(x = grd[, 1], y = grd[, 2], z = matrix(grd[,  : 
increasing 'x' and 'y' values expected

我的代码中的问题在哪里，是否有另一种方法可以解决我的问题？谢谢你的帮助！

更新：

我已经用 plane3d() 试过了，但得到了这个错误：

Error in segments(x, z1, x + y.max * yx.f, z2 + yz.f * y.max, lty = ltya,  : 
cannot mix zero-length and non-zero-length coordinates

但是，在寻找其他解决方案后，我发现我的意图是错误的。我正在寻找这样的东西......

数据
意图

...但是不知道如何实现它，或者根据我对 R 的实际了解是否有可能。有人知道吗？

更新2：

我已经使用以下代码尝试了 scatter3d() 函数： 

scatter3d(a,b,L,fit="smooth")

但收到此错误： 

Error in smooth.construct.tp.smooth.spec(object, dk$data, dk$knots) : 
A term has fewer unique covariate combinations than specified maximum degrees of freedom

我已经发现，我必须使用 choose.k 来调整自由度，但它总是针对 gam 函数进行解释，我不知道如何在 scatter3d 函数中使用它。任何想法？

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1 回答 1

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I guess I've solved my problem with a very nasty solution: 

data <- structure(list(L = c(96.4155, 76.803, 71.5615, 68.193, 65.6975, 74.627, 67.82, 64.26,    
62.06, 60.35, 68.284, 63.7, 61.04, 59.05, 57.56, 64.2695, 60.69, 58.47, 56.78, 55.42, 
61.3715, 58.27, 56.42, 54.91, 53.65), a = c(-0.8375, -20.47, -22.1875, -22.6125, 
-22.5845, -2.2415, -12.91, -16.16, -17.9, -18.77, -2.4, -10.28, -13.64, -15.33, -16.37, 
-2.4495, -8.66, -11.74, -13.46, -14.55, -2.4645, -7.58, -10.38, -12.06, -13.09), b = 
c(1.437, -24.5915, -29.3275, -31.892, -33.524, -5.151, -15.22, -19.61, -22.65, -24.8, 
-6.239, -13.24, -17.11, -19.64, -21.55, -6.7845, -12.18, -15.45, -17.68, -19.47, -7.124, 
-11.53, -14.38, -16.41, -17.89)), .Names = c("L", "a", "b"), row.names = c(NA, 25L),
class = "data.frame")

L<-data[1:25,1]
a<-data[1:25,2]
b<-data[1:25,3]
plot3d(a,b,L, col="red", size="3", xlab="a*", ylab="b*", zlab="L*", main=NULL)

#Generate order for quadruplets

a[c(1:64)] <- a[c(1,2,7,6,2,3,8,7,3,4,9,8,4,5,10,9,
                  6,7,12,11,7,8,13,12,8,9,14,13,9,10,15,14,
                  11,12,17,16,12,13,18,17,13,14,19,18,14,15,20,19,
                  16,17,22,21,17,18,23,22,18,19,24,23,19,20,25,24)] 
b[c(1:64)] <- b[c(1,2,7,6,2,3,8,7,3,4,9,8,4,5,10,9,
                  6,7,12,11,7,8,13,12,8,9,14,13,9,10,15,14,
                  11,12,17,16,12,13,18,17,13,14,19,18,14,15,20,19,
                  16,17,22,21,17,18,23,22,18,19,24,23,19,20,25,24)] 
L[c(1:64)] <- L[c(1,2,7,6,2,3,8,7,3,4,9,8,4,5,10,9,
                  6,7,12,11,7,8,13,12,8,9,14,13,9,10,15,14,
                  11,12,17,16,12,13,18,17,13,14,19,18,14,15,20,19,
                  16,17,22,21,17,18,23,22,18,19,24,23,19,20,25,24)] 

quads3d(a[1:64],b[1:64],L[1:64],col="red",alpha=0.5)

Is there any function who does exactly the same?

于 2014-04-05T21:30:35.107 回答