我是R的新手，我正在尝试以下面的格式构建非线性相关。我在R中尝试了一个脚本，但它无法工作并返回错误消息“初始参数估计时的奇异梯度矩阵”。有人可以帮助我使用正确的脚本输入R，以便根据新的数据集估计更新的相关系数吗？数据集由3个变量Z，X和Y组成。我想估计Z = f（x，y）。 谢谢
等式
z=a+bx+cy+dx^2+ey^2+fxy+gx^3+hy^3+ixy^2+jx^2y
a 0.065119008
b -0.002506607
c 0.004586821
d 3.73635E-05
e 8.41116E-07
f -1.7902E-05
g -1.28967E-07
h -1.04123E-10
i -2.40641E-09
j 4.42138E-08
X | Y | Z
_______ | _______ | _______
60 | 100 | 0.41994
60 | 200 | 0.79807
60 | 300 | 1.18778
60 | 400 | 1.58945
60 | 500 | 2.00336
60 | 600 | 2.42971
60 | 700 | 2.86858
60 | 800 | 3.31989
60 | 900 | 3.78335
60 | 1000 | 4.25842
60 | 1100 | 4.74429
60 | 1200 | 5.23983
60 | 1300 | 5.74359
60 | 1400 | 6.25381
60 | 1500 | 6.76844
60 | 1600 | 7.28523
60 | 1700 | 7.80179
60 | 1800 | 8.31574
60 | 1900 | 8.82475
60 | 2000 | 9.32668
80 | 100 | 0.40357
80 | 200 | 0.76552
80 | 300 | 1.13711
80 | 400 | 1.5185
80 | 500 | 1.90979
80 | 600 | 2.311
80 | 700 | 2.72205
80 | 800 | 3.14274
80 | 900 | 3.57269
80 | 1000 | 4.01141
80 | 1100 | 4.45817
80 | 1200 | 4.91207
80 | 1300 | 5.37202
80 | 1400 | 5.83674
80 | 1500 | 6.30477
80 | 1600 | 6.77453
80 | 1700 | 7.24438
80 | 1800 | 7.71262
80 | 1900 | 8.17761
80 | 2000 | 8.63777
100 | 100 | 0.38847
100 | 200 | 0.73573
100 | 300 | 1.09104
100 | 400 | 1.45447
100 | 500 | 1.82598
100 | 600 | 2.20551
100 | 700 | 2.59287
100 | 800 | 2.9878
100 | 900 | 3.38993
100 | 1000 | 3.79877
100 | 1100 | 4.21372
100 | 1200 | 4.63401
100 | 1300 | 5.0588
100 | 1400 | 5.48709
100 | 1500 | 5.91781
100 | 1600 | 6.3498
100 | 1700 | 6.78184
100 | 1800 | 7.21271
100 | 1900 | 7.64119
100 | 2000 | 8.06612
120 | 100 | 0.37451
120 | 200 | 0.70832
120 | 300 | 1.04892
120 | 400 | 1.39627
120 | 500 | 1.7503
120 | 600 | 2.11085
120 | 700 | 2.47771
120 | 800 | 2.85059
120 | 900 | 3.22913
120 | 1000 | 3.61287
120 | 1100 | 4.00129
120 | 1200 | 4.39376
120 | 1300 | 4.78958
120 | 1400 | 5.18797
120 | 1500 | 5.58809
120 | 1600 | 5.98905
120 | 1700 | 6.38994
120 | 1800 | 6.78981
120 | 1900 | 7.18777
120 | 2000 | 7.58291
140 | 100 | 0.36155
140 | 200 | 0.683
140 | 300 | 1.01021
140 | 400 | 1.34307
140 | 500 | 1.68148
140 | 600 | 2.02523
140 | 700 | 2.37411
140 | 800 | 2.72783
140 | 900 | 3.08602
140 | 1000 | 3.4483
140 | 1100 | 3.81418
140 | 1200 | 4.18314
140 | 1300 | 4.55459
140 | 1400 | 4.9279
140 | 1500 | 5.3024
140 | 1600 | 5.67739
140 | 1700 | 6.05216
140 | 1800 | 6.42596
140 | 1900 | 6.7981
140 | 2000 | 7.16787
160 | 100 | 0.34948
160 | 200 | 0.65953
160 | 300 | 0.97447
160 | 400 | 1.29419
160 | 500 | 1.61852
160 | 600 | 1.94728
160 | 700 | 2.28022
160 | 800 | 2.61706
160 | 900 | 2.95748
160 | 1000 | 3.3011
160 | 1100 | 3.64752
160 | 1200 | 3.99628
160 | 1300 | 4.34688
160 | 1400 | 4.6988
160 | 1500 | 5.05149
160 | 1600 | 5.40438
160 | 1700 | 5.7569
160 | 1800 | 6.10847
160 | 1900 | 6.4585
160 | 2000 | 6.80647
180 | 100 | 0.33822
180 | 200 | 0.6377
180 | 300 | 0.94137
180 | 400 | 1.24907
180 | 500 | 1.56064
180 | 600 | 1.87588
180 | 700 | 2.19455
180 | 800 | 2.51639
180 | 900 | 2.84109
180 | 1000 | 3.16833
180 | 1100 | 3.49772
180 | 1200 | 3.82888
180 | 1300 | 4.16138
180 | 1400 | 4.49478
180 | 1500 | 4.82863
180 | 1600 | 5.16245
180 | 1700 | 5.49577
180 | 1800 | 5.82812
180 | 1900 | 6.15903
180 | 2000 | 6.48806
200 | 100 | 0.32767
200 | 200 | 0.61734
200 | 300 | 0.91058
200 | 400 | 1.20725
200 | 500 | 1.50717
200 | 600 | 1.81015
200 | 700 | 2.11596
200 | 800 | 2.42434
200 | 900 | 2.73502
200 | 1000 | 3.04768
200 | 1100 | 3.36202
200 | 1200 | 3.67767
200 | 1300 | 3.99427
200 | 1400 | 4.31145
200 | 1500 | 4.62882
200 | 1600 | 4.94597
200 | 1700 | 5.26253
200 | 1800 | 5.57809
200 | 1900 | 5.89227
200 | 2000 | 6.2047
答案 0 :(得分：0)
我不完全确定你想做什么，或者为什么谷歌不能令人满意，但也许这些方面的东西会给你一个想法：
x <- rep(c(60,80,100,160,200), each = 10)
y <- c(seq(from = 100, to = 2000, length.out = 25),seq(1800, 200, length.out = 25))
z <- rnorm(50, 6)
df <- data.frame(x,y,z)
mod <- lm(z ~ 1 + x + y + I(x^2) + I(y^2) + I(x*y) + I(x^3) + I(y^3) + I(x*y^2) + I(x*y^3), data =df)
summary(mod)
summary(mod)$adj