自己看返回值吧;你是打算提取其概要中信息?
x <- c(0.10,0.11,0.12,0.13,0.14,0.15,0.16,0.17,0.18,0.20,0.21,0.23)
y <- c(42.0,43.5,45.0,45.5,45.0,47.5,49.0,53.0,50.0,55.0,55.0,60.0)
lm.sol <- lm(y~1+x)
summary(lm.sol)
lm.sol$coefficients
summary(lm.sol$coefficients)
A <-summary(lm.sol$coefficients)
A[,1]
Error in A[, 1] : incorrect number of dimensions
如果你打算是要调summary中的Min值的话,应该首先确认他在哪里,比如这样。
str(A)
Classes 'summaryDefault', 'table' Named num [1:6] 28.5 54.1 79.7 79.7 105.2 ...
..- attr(*, "names")= chr [1:6] "Min." "1st Qu." "Median" "Mean" ...
A[1]
Min.
28.49
class(A[1])
[1] "numeric"
但是,如果意图把其数值代入计算程序需要考虑是否需要特定的数值类型转换。 :plain:
而且我发现你自定义函数里面的这一句的返回值存在错误,你需要考虑一下。对应语句:
A[1]-A[2]*qt(1-alpha/2,df)
qt(1-0.05/2,lm.sol$residual)
1 2 3 4 5 6 7
437.728523 60.848902 22.707065 NaN NaN NaN NaN
8 9 10 11 12
3.854130 NaN 2176.742366 NaN 6.548663
Warning message:
In qt(1 - 0.05/2, lm.sol$residual) : 产生了NaNs