如果我使用predict()函数计算特定x值的y值,则得到的值与我可以使用显式拟合方程计算的值不同。
我使用nls(MyEquation)拟合了以下数据,并获得了m1,m2,...参数。 然后,我想使用predict(m)函数或我用于拟合的显式方程(输入所需的x值)来反向计算特定x值的y值。 对于相同的x值,我获得了不同的y值。哪个是正确的?
> df
pH activity
1 3.0 0.88
2 4.0 1.90
3 5.0 19.30
4 6.0 70.32
5 7.0 100.40
6 7.5 100.00
7 8.0 79.80
8 9.0 7.75
9 10.0 1.21
x <- df$pH
y <- df$activity
m<-nls(y~(m1*(10^(-x))+m2*10^(-m3))/(10^(-m3)+10^(-x)) - (m5*(10^(-x))+1*10^(-i))/(10^(-i)+10^(-x)), start = list(m1=1,m2=100,m3=7,m5=1))
> m
Nonlinear regression model
model: y ~ (m1 * (10^(-x)) + m2 * 10^(-m3))/(10^(-m3) + 10^(-x)) - (m5 * (10^(-x)) + 1 * 10^(-i))/(10^(-i) + 10^(-x))
data: parent.frame()
m1 m2 m3 m5
-176.032 13.042 6.282 -180.704
residual sum-of-squares: 1522
Number of iterations to convergence: 14
Achieved convergence tolerance: 5.805e-06
list2env(as.list(coef(m)), .GlobalEnv)
#calculate y based on fitting parameters
# choose the 7th x value (i.e. x[7]) that corresponds to pH = 8
# (using predict)
> x_pH8 <- x[7]
> predict(m)[7]
[1] 52.14299
# (using the explicit fitting equation with the fitted parameters
> x1 <- x_pH8
> (m1*(10^(-x1))+m2*10^(-m3))/(10^(-m3)+10^(-x1)) - (m5*(10^(-x1))+1*10^(-8.3))/(10^(-8.3)+10^(-x1))
[1] 129.5284
如您所见: 预测(m)[7]得出y = 52.14299(对于x = 8)
同时
(m1 *(10 ^(-x1))+ m2 * 10 ^(-m3))/(10 ^(-m3)+10 ^(-x1))-(m5 *(10 ^(-x1) ))+ 1 * 10 ^(-8.3))/(10 ^(-8.3)+10 ^(-x1))得出y = 129.5284(对于x = 8)
答案 0 :(得分:0)
您在手动计算中使用的package hr.atos.praksa.josipmaricevic.zadatak15;
import java.util.Scanner;
import java.util.regex.Pattern;
public class Functions {
static Scanner input = new Scanner(System.in);
final static private Pattern patternNotBlank = Pattern.compile("\\S+");
final static private Pattern patternOIB = Pattern.compile("^[0-9]{11}$");
final static private Pattern patternString = Pattern.compile("[A-Z][a-z]*");
final static private Pattern patternInt = Pattern.compile("[0-9]+$");
final static private Pattern patternFloat = Pattern.compile("^[0-9]+([.]?[0-9]+|[0-9]*)");
public static String inputCheck(String print, String _pattern, boolean andStatement, boolean orStatement ) {
String string = "";
Pattern pattern = patternFloat;
switch(_pattern) {
case "int":
pattern = patternInt;
break;
case "float":
pattern = patternFloat;
break;
case "string":
pattern = patternString;
break;
case "oib":
pattern = patternOIB;
break;
case "notblank":
pattern = patternNotBlank;
break;
}
while((andStatement && pattern.matcher(string).matches()==false) || orStatement){
System.out.print(print);
string = input.nextLine();
}
return string;
}
}
的值可能与模型拟合中使用的值不同。我没有任何差异:
i