使用拟合的 2 次多项式模型从 Y 值预测 X 值

Question

我有一个格式如下的数据集：

dataset1 = data.frame(
caliber = c("5000", "2500", "1250", "625", "312.5", "156", "80", "40", "20", "0"),
var1 = c(NA, NA, NA, 30458, 13740,11261, 9729, 5039, 3343, 367),
var2 = c(463000, 271903, 154611,87204, 47228, 28082, 14842, 8474, 5121, 1308),
var3 = c(308385, 184863, 89719, 48986, 27968, 18557, 9191, 5248, 3210, 703), 
var4 = c(290159, 149061, 64045, 36864, 19092, 12515, 6805, 3933, 2339, 574), 
var5 = c(270801, 163657, 51642, 48197, 23582, 14544, 7877, 4389, 2663, 482), 
var6 = c(NA, NA, NA, 37316, 21305, 11823, 5692, 3070, 1781, 363))

描述口径与其他变量之间关系的最佳方式是通过 2 次多项式方程：var = poly(caliber, 2, raw=T)

我的问题是如何使用一组新的变量来识别 calibre 变量的值。正如您在下面看到的，我已经有了每个变量的结果，但我需要确定口径的值。

dataset2 = data.frame(
caliber = c(NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA),
var1 = c(1120, 1296, 1132, 1280, 1096, 1124, 1004, 8384, 1072, 1104, 1568, 1044, 1108, 1012),
var2 = c(5044, 4924, 5088, 4804, 4824, 4844, 4964, 4788, 4804, 4964, 4824, 4788, 4844, 4944),
var3 = c(2836, 2744, 2744, 2668, 2688, 2940, 2756, 2720, 2668, 2892, 2636, 2700, 2836, 2668),
var4 = c(8872, 61580, 3036, 4468, 12132, 3000, 7920, 6868, 6896, 9392, 4728, 6896, 21076, 3228),
var5 = c(2312, 4236, 1928, 4448, 2388, 2108, 3644, 3060, 2168, 1912, 1812, 3528, 4100, 2176),
var6 = c(1156, 1228, 1224, 1364, 1128, 1176, 1184, 1640, 1188, 1300, 1332, 1176, 1176, 1152))

我知道之前有几个关于这个话题的话题，比如

但是 none 有所帮助。主要问题是：

formula <- lm(var2~poly(caliber,2,raw=T), dataset1)
approx(x = formula$fitted, y = formula$caliber, xout = 0)$y

公式 $caliber 的 NA 值

mod<-lm(var2~poly(caliber, 2, raw=T), data=dataset1); summary(mod)
newdata=data.frame("var2"=dataset2[1:24,c("var2")])
pred<-predict(mod,newdata, type = 'response')

Error in poly(caliber, 2, coefs = list(alpha = c(998.35, 3691.21383929929 :object 'caliber' not found

无法将预测传递给另一个数据集

具有不同行的数据集

X 和 Y 之间的插值给出了错误的值

Answer 1

根据讨论，我所了解的，我为您提供以下解决方案

dataset1 = data.frame(
  caliber = c(5000, 2500, 1250, 625, 312.5, 156, 80, 40, 20, 0),
  var1 = c(NA, NA, NA, 30458, 13740,11261, 9729, 5039, 3343, 367),
  var2 = c(463000, 271903, 154611,87204, 47228, 28082, 14842, 8474, 5121, 1308),
  var3 = c(308385, 184863, 89719, 48986, 27968, 18557, 9191, 5248, 3210, 703), 
  var4 = c(290159, 149061, 64045, 36864, 19092, 12515, 6805, 3933, 2339, 574), 
  var5 = c(270801, 163657, 51642, 48197, 23582, 14544, 7877, 4389, 2663, 482), 
  var6 = c(NA, NA, NA, 37316, 21305, 11823, 5692, 3070, 1781, 363))

formula <- lm(caliber ~ poly(var2, degree = 2, raw=T), dataset1)

dataset2 = data.frame(
  caliber = c(NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA),
  var1 = c(1120, 1296, 1132, 1280, 1096, 1124, 1004, 8384, 1072, 1104, 1568, 1044, 1108, 1012),
  var2 = c(5044, 4924, 5088, 4804, 4824, 4844, 4964, 4788, 4804, 4964, 4824, 4788, 4844, 4944),
  var3 = c(2836, 2744, 2744, 2668, 2688, 2940, 2756, 2720, 2668, 2892, 2636, 2700, 2836, 2668),
  var4 = c(8872, 61580, 3036, 4468, 12132, 3000, 7920, 6868, 6896, 9392, 4728, 6896, 21076, 3228),
  var5 = c(2312, 4236, 1928, 4448, 2388, 2108, 3644, 3060, 2168, 1912, 1812, 3528, 4100, 2176),
  var6 = c(1156, 1228, 1224, 1364, 1128, 1176, 1184, 1640, 1188, 1300, 1332, 1176, 1176, 1152))

predict(formula, dataset2, type = 'response')

predict 函数的输出将为您提供数据集 2 中口径的值。

我已经更正了你的数据集1。如果将值放在双引号内，它就会变成字符。所以，我从 caliber 变量中删除了双引号。

使用拟合的 2 次多项式模型从 Y 值预测 X 值

Predict X value from Y value with a fitted 2-degree polynomial model

regression

r

prediction

predict