得到R中的表面图

本文介绍了得到R中的表面图的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我正在尝试从数据框AAA获取表面图:

I'm trying to obtain a surface plot from data frame AAA:

j a m p o f 13929 0.86739583 19 165.83 0.1588727 13.24444 13930 0.63166667 19 178.19 0.6105804 12.68333 13932 0.90212963 17 157.77 0.3345627 12.52222 13933 0.80152778 68 146.19 0.1219885 12.35000 13934 0.75784722 62 134.88 0.1531627 12.36667 13935 0.57763889 66 123.80 0.4093869 12.47500 13936 0.56201389 88 112.87 0.9095722 12.45833 13937 0.51680556 26 102.03 0.8494420 12.37500 13938 0.46093333 28 91.20 0.9153419 12.21111 13939 0.16645833 24 80.30 0.8309784 12.04444 13940 0.15451389 36 69.23 2.2847927 12.15556 13941 0.51347222 134 57.92 2.9551087 12.42500 13942 0.33763889 128 46.31 3.5784096 12.53333 13943 0.12937500 38 34.33 3.7371723 12.47778 13944 0.42760870 63 22.00 4.7831677 12.46667 13945 0.09962121 8 9.36 4.8281897 12.30000 13950 0.97901515 18 57.70 0.0000000 12.15833 13951 0.85333333 14 71.07 0.0000000 12.48333 13952 0.92811594 14 84.28 10.0444672 12.49167 13953 0.84812500 42 97.29 7.8020987 12.51667

我的代码:

require(fields) fitx <- Tps( AAA[, 4:6], AAA$a) out.p <- predict.surface(fitx, xy = c(4,5)) plot.surface(out.p, type="p")

但是，它并没有通过.显然，网格不足以表示数据，并且无法获取predict.surface.

However, it doesn't run through. Apparently, the grid is insufficient to represent the data and it's not able to get the predict.surface.

推荐答案

在Tps函数中，您的x矩阵为AAA[, 4:6]，因此具有三列.

In the Tps function, your x matrix is AAA[, 4:6] and hence has three columns.

但是在predict.surface函数中，您指定了xy = c(4,5).传递给xy参数的值是相对于fitx对象中的矩阵的.由于用于使用predict.surface函数创建fitx的矩阵有三列，因此您不能引用第4列和第5列.相反，原始数据的第4列和第5列.帧AAA对应于fitx的第1列和第2列.

But in the predict.surface function, you specified xy = c(4,5). The values passed to the xyparamter are relative to the matrix in your fitx object. Since the matrix used for creating fitx with the predict.surface function has three columns, you can't refer to the 4th and 5th column. Instead, the columns 4 and 5 of your original data.frame AAA correspond to columns 1 and 2 in fitx.

您不妨尝试:

library(fields) fitx <- Tps(AAA[, 4:6], AAA$a) out.p <- predict.surface(fitx, xy = c(1,2)) # Note the different argument passed to `xy` plot.surface(out.p, type="p")