I'm very new to R, and I'm having trouble figuring out a 3d surface plot of the data. I typically have something like this: 0.1 0.2 0.3 0.4 0.5 0.001 40960.16 40960.16 40960.16 40960.16 40960.16 0.0025 40960.16 40960.16 40960.16 40960.16 40960.16 0.0039 40960.16 40960.16 40960.16 40960.16 40960.16 0.0061 40960.16 40960.16 40960.16 40960.16 40960.16 0.0095 40960.16 40960.16 40960.16 40960.16 40960.16 0.0147 40960.16 40960.16 40960.16 33756.49 25979.93 0.023 40960.16 40960.16 28130.81 19838.51 14891.95 0.0358 40960.16 26877.74 16258.07 11004.66 7898.252 0.0558 35941.34 15461.96 8752.924 5487.987 3601.922 0.087 21231.18 8254.491 4115.487 2189.69 1152.315 0.1357 11883.95 3815.37 1425.492 464.9616 84.35834 0.2115 6029.579 1264.774 166.2719 0 0 0.3296 2500.958 115.8725 0 0 0 0.5138 605.4039 0 0 0 0 0.801 0 0 0 0 0 0.9999 0 0 0 0 0 a first row corresponding to the values in y and a first column corresponding to the values in x,the matrix of residual values are in z as the following:> x1 <- t(x) #transpose x[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] V1 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1> y1 <- t(y)[,1] V1 0.0010 V2 0.0010 V3 0.0039 V4 0.0061 V5 0.0095 V6 0.0147 V7 0.0230 V8 0.0358 V9 0.0558 V10 0.0870 V11 0.1357 V12 0.2115 V13 0.3296 V14 0.5138 V15 0.8010 V16 0.9999> z_t <- t(z)[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] V1 40960.16 40960.16 40960.16 40960.16 40960.16 40960.16 40960.165 40960.165 V2 40960.16 40960.16 40960.16 40960.16 40960.16 40960.16 40960.165 26877.741 V3 40960.16 40960.16 40960.16 40960.16 40960.16 40960.16 28130.814 16258.071 V4 40960.16 40960.16 40960.16 40960.16 40960.16 33756.49 19838.507 11004.663 V5 40960.16 40960.16 40960.16 40960.16 40960.16 25979.93 14891.946 7898.252 V6 40960.16 40960.16 40960.16 40960.16 35282.13 20810.88 11618.426 5865.759 V7 40960.16 40960.16 40960.16 40960.16 29505.17 17131.90 9301.074 4447.296 V8 40960.16 40960.16 40960.16 40960.16 25179.81 14384.26 7581.456 3412.924 V9 40960.16 40960.16 40960.16 36868.08 21822.20 12257.56 6260.442 2634.903 V10 40960.16 40960.16 40960.16 32663.23 19142.03 10565.56 5218.556 2036.595 [,9] [,10] [,11] [,12] [,13] [,14] [,15] [,16] V1 35941.3351 21231.183500 11883.95464 6029.5794 2500.9576 605.4039 0 0 V2 15461.9567 8254.491283 3815.37041 1264.7739 115.8725 0.0000 0 0 V3 8752.9239 4115.487198 1425.49197 166.2719 0.0000 0.0000 0 0 V4 5487.9866 2189.690272 464.96165 0.0000 0.0000 0.0000 0 0 V5 3601.9222 1152.314526 84.35834 0.0000 0.0000 0.0000 0 0 V6 2406.3472 561.860696 0.00000 0.0000 0.0000 0.0000 0 0 V7 1606.2609 229.200968 0.00000 0.0000 0.0000 0.0000 0 0 V8 1054.1819 59.850348 0.00000 0.0000 0.0000 0.0000 0 0 V9 668.1935 1.390629 0.00000 0.0000 0.0000 0.0000 0 0 V10 399.1614 0.000000 0.00000 0.0000 0.0000 0.0000 0 0 I tried to use surface3d() and persp3d(), but the results was error.> library(rgl) > surface3d(x1,y1,z_t)Error in rgl.surface(x = c(0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, : bad dimension for rows What should I do? Thanks, [[alternative HTML version deleted]]
On 13-05-20 12:17 AM, Nguyen Hoang wrote:> I'm very new to R, and I'm having trouble figuring out a 3d surface > plot of the data. I typically have something like this: > > > > > > > 0.1 > 0.2 > 0.3 > 0.4 > 0.5 > > > 0.001 > 40960.16 > 40960.16 > 40960.16 > 40960.16 > 40960.16 > > > 0.0025 > 40960.16 > 40960.16 > 40960.16 > 40960.16 > 40960.16 > > > 0.0039 > 40960.16 > 40960.16 > 40960.16 > 40960.16 > 40960.16 > > > 0.0061 > 40960.16 > 40960.16 > 40960.16 > 40960.16 > 40960.16 > > > 0.0095 > 40960.16 > 40960.16 > 40960.16 > 40960.16 > 40960.16 > > > 0.0147 > 40960.16 > 40960.16 > 40960.16 > 33756.49 > 25979.93 > > > 0.023 > 40960.16 > 40960.16 > 28130.81 > 19838.51 > 14891.95 > > > 0.0358 > 40960.16 > 26877.74 > 16258.07 > 11004.66 > 7898.252 > > > 0.0558 > 35941.34 > 15461.96 > 8752.924 > 5487.987 > 3601.922 > > > 0.087 > 21231.18 > 8254.491 > 4115.487 > 2189.69 > 1152.315 > > > 0.1357 > 11883.95 > 3815.37 > 1425.492 > 464.9616 > 84.35834 > > > 0.2115 > 6029.579 > 1264.774 > 166.2719 > 0 > 0 > > > 0.3296 > 2500.958 > 115.8725 > 0 > 0 > 0 > > > 0.5138 > 605.4039 > 0 > 0 > 0 > 0 > > > 0.801 > 0 > 0 > 0 > 0 > 0 > > > 0.9999 > 0 > 0 > 0 > 0 > 0 > > a first row corresponding to the values in y and a first column > corresponding to the values in x,the matrix of residual values are in > z as the following: > > >> x1 <- t(x) #transpose x > [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] > V1 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 >> y1 <- t(y) > [,1] > V1 0.0010 > V2 0.0010 > V3 0.0039 > V4 0.0061 > V5 0.0095 > V6 0.0147 > V7 0.0230 > V8 0.0358 > V9 0.0558 > V10 0.0870 > V11 0.1357 > V12 0.2115 > V13 0.3296 > V14 0.5138 > V15 0.8010 > V16 0.9999 > > >> z_t <- t(z) > [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] > V1 40960.16 40960.16 40960.16 40960.16 40960.16 40960.16 40960.165 40960.165 > V2 40960.16 40960.16 40960.16 40960.16 40960.16 40960.16 40960.165 26877.741 > V3 40960.16 40960.16 40960.16 40960.16 40960.16 40960.16 28130.814 16258.071 > V4 40960.16 40960.16 40960.16 40960.16 40960.16 33756.49 19838.507 11004.663 > V5 40960.16 40960.16 40960.16 40960.16 40960.16 25979.93 14891.946 7898.252 > V6 40960.16 40960.16 40960.16 40960.16 35282.13 20810.88 11618.426 5865.759 > V7 40960.16 40960.16 40960.16 40960.16 29505.17 17131.90 9301.074 4447.296 > V8 40960.16 40960.16 40960.16 40960.16 25179.81 14384.26 7581.456 3412.924 > V9 40960.16 40960.16 40960.16 36868.08 21822.20 12257.56 6260.442 2634.903 > V10 40960.16 40960.16 40960.16 32663.23 19142.03 10565.56 5218.556 2036.595 > [,9] [,10] [,11] [,12] [,13] [,14] [,15] [,16] > V1 35941.3351 21231.183500 11883.95464 6029.5794 2500.9576 605.4039 0 0 > V2 15461.9567 8254.491283 3815.37041 1264.7739 115.8725 0.0000 0 0 > V3 8752.9239 4115.487198 1425.49197 166.2719 0.0000 0.0000 0 0 > V4 5487.9866 2189.690272 464.96165 0.0000 0.0000 0.0000 0 0 > V5 3601.9222 1152.314526 84.35834 0.0000 0.0000 0.0000 0 0 > V6 2406.3472 561.860696 0.00000 0.0000 0.0000 0.0000 0 0 > V7 1606.2609 229.200968 0.00000 0.0000 0.0000 0.0000 0 0 > V8 1054.1819 59.850348 0.00000 0.0000 0.0000 0.0000 0 0 > V9 668.1935 1.390629 0.00000 0.0000 0.0000 0.0000 0 0 > V10 399.1614 0.000000 0.00000 0.0000 0.0000 0.0000 0 0 > > > I tried to use surface3d() and persp3d(), but the results was error. > > >> library(rgl) >> surface3d(x1,y1,z_t) > Error in rgl.surface(x = c(0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, : > bad dimension for rows > > > What should I do?Don't convert x and y into matrices, leave them as vectors. Unlike persp, persp3d allows you to specify x and/or y as matrices, but then they need to be the same shape as z. It was complaining because it thought you had done that, but got the shape wrong. Duncan Murdoch
WIth x, y are vectors, the grid is plotted by surface3d. Thanks for your
answer.
My code is here;
#SurfacePlot is plotted by above surface data
Surface <- read.csv("D:/R/Surface.csv", header=F)
Surface <- as.matrix(Surface)
x <- Surface[2:17,1]
x <- as.numeric(x) # x is vector 1x16
y <- Surface[1,2:11]
y <- t(y)
y <- as.numeric(y) #y is vector 1x10
z <- Surface[2:17,2:11] # z is matrix 16x10
plot3d(ScatterPlot$x1, ScatterPlot$y1, ScatterPlot$z1,
xlab="x",ylab="y",zlab="z", type="s",
col = "blue", size=0.25, lit=FALSE)
# ScatterPlot is showed as below data.
surface3d(x,y,z,alpha=0.4,front="lines",back="lines")
I combined surface3d and plot3d() in graph.
And now, I want to add line segments showing the error between points of
scatter plot and grid of surface3d.
ScatterPlot Data:
X1 Y1 Z1
1 25000.030 0.001218202 0.1915856
2 5118.15450 0.061169029 0.4711610
3 17005.82875 0.007177435 0.3222662
4 5300.22698 0.029688212 0.2770485
...
(3000 points)
I think I should build linear model to predict z1_new values from x1 & y1
based on Surface data, And use spheres3d() to plot points of Scatterplot on
surface grid, and then use segment3d() to add line segments between points
of scatter plot and grid of surface3d coordinate.
But. i don't know to do that. What should I do? my method is true ?
Nguyen.Hoang
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