我可以重现您的错误。 nls
函数缺少其中的参数data
。
m<-nls(y ~ b/(x^2)+a, start=list(a=a_start, b=b_start))
# Error in nls(y ~ b/(x^2) + a, start = list(a = a_start, b = b_start)) :
# parameters without starting value in 'data': y, x
现在数据df
被创建并传递给nls函数。请确保,I()
中的绝缘表达是预期的。
df <- data.frame(x = c(71.33, 74.98 , 80 , 85.35 , 90.03),
y = c(119.17, 107.73 , 99.72 , 75, 54.59))
a_start <- -39.5
b_start <- 800000
m <- nls(y ~ I(b/(x^2+a)), data = df, start=list(a=a_start, b=b_start))
summary(m)
# Formula: y ~ I(b/(x^2 + a))
#
# Parameters:
# Estimate Std. Error t value Pr(>|t|)
# a -1743.2 872.5 -1.998 0.1396
# b 412486.2 89981.4 4.584 0.0195 *
# ---
# Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
#
# Residual standard error: 9.103 on 3 degrees of freedom
#
# Number of iterations to convergence: 6
# Achieved convergence tolerance: 4.371e-06
阅读公式手册页以了解绝缘表达式。
?formula
的formula
手册页说,
The^operator indicates crossing to the specified degree. For example (a+b+c)^2 is identical to (a+b+c)*(a+b+c) which in turn expands to a formula containing the main effects for a, b and c together with their second-order interactions
此外,它建议使用I()
防止公式运算符和算术运算符之间的模糊。
这里为来自公式手册页
avoid this confusion, the function I() can be used to bracket those portions of a model formula where the operators are used in their arithmetic sense. For example, in the formula y ~ a + I(b+c), the term b+c is to be interpreted as the sum of b and c.
的另一句名言也是这人页是值得一读
?AsIs
In function formula. There it is used to inhibit the interpretation of operators such as "+", "-", "*" and "^" as formula operators, so they are used as arithmetical operators. This is interpreted as a symbol by terms.formula.
Sathish所在,谢谢! – aSportsguy
是的,完美的,并且改变工作!我不是贸易程序员,所以我发现一些R文档很难阅读,或者在这里寻找某些语法解决方案。 。 。 – aSportsguy