求出拟合曲线的斜线的交点 [英] find the intersection of abline with fitted curve
问题描述
我使用以下代码绘制了一条拟合的逻辑曲线:
I plotted a logistic curve with its fit using the following codes:
数据:L50
data:L50
str(L50)
"data.frame":10磅.的3个变量:
'data.frame': 10 obs. of 3 variables:
$ Length.Class:int 50 60 70 80 90 100 110 120 130 140
$ Length.Class: int 50 60 70 80 90 100 110 120 130 140
$ Total.Ind:int 9 20 18 8 4 4 1 0 1 2
$ Total.Ind : int 9 20 18 8 4 4 1 0 1 2
$ Mature.Ind:int 0 0 6 5 3 2 1 0 1 2
$ Mature.Ind : int 0 0 6 5 3 2 1 0 1 2
plot(L50$Mature.Ind/L50$Total.Ind ~ L50$Length.Class, data=L50,pch=20,xlab="Length class(cm)",ylab="Proportion of mature individuals")
glm.out<-glm(cbind(L50$Mature.Ind, L50$Total.Ind-L50$Mature.Ind) ~ L50$Length.Class,family=binomial(logit), data=L50)
glm.out
呼叫:glm(公式= cbind(L50 $ Mature.Ind，L50 $ Total.Ind-L50 $ Mature.Ind)〜
L50 $ Length.Class，家庭=二项式(logit)，数据= L50)
glm.out
Call: glm(formula = cbind(L50$Mature.Ind, L50$Total.Ind - L50$Mature.Ind) ~
L50$Length.Class, family = binomial(logit), data = L50)
系数:
(拦截)L50 $ Length.Class
-8.6200 0.1053
Coefficients:
(Intercept) L50$Length.Class
-8.6200 0.1053
自由度:总计8(即无效)； 7残留 空偏差:38.14 残余偏差:9.924 AIC:23.4
Degrees of Freedom: 8 Total (i.e. Null); 7 Residual Null Deviance: 38.14 Residual Deviance: 9.924 AIC: 23.4
lines(L50$Length.Class, glm.out$fitted,type="l", col="red",lwd=2)
abline(h=0.5,col="black",lty=2,lwd=2)
我得到以下曲线:
问题是我需要在拟合曲线上找到与Y = 0.5对应的点，并通过它在x轴上的值绘制一条线段....对此有任何帮助吗? 谢谢
The question is that i need to find the point that corresponds to Y=0.5 on the fitted curve and draw a line segment through it with its value on the x-axis....Any help with that? Thank you
这是你的要求
dput(L50)
structure(list(Length.Class = c(50L, 60L, 70L, 80L, 90L, 100L,
110L, 120L, 130L, 140L), Total.Ind = c(9L, 20L, 18L, 8L, 4L,
4L, 1L, 0L, 1L, 2L), Mature.Ind = c(0L, 0L, 6L, 5L, 3L, 2L, 1L,
0L, 1L, 2L), MatF = c(0L, 0L, 1L, 4L, 1L, 2L, 0L, 0L, 1L, 2L), MatM = c(0L, 0L, 5L, 1L, 2L, 0L, 1L, 0L, 0L, 0L)), .Names = c("Length.Class",
"Total.Ind", "Mature.Ind", "MatF", "MatM"), class = "data.frame", row.names = c(NA,-10L))
structure(list(Length.Class = c(50L, 60L, 70L, 80L, 90L, 100L,
110L, 120L, 130L, 140L), Total.Ind = c(9L, 20L, 18L, 8L, 4L,
4L, 1L, 0L, 1L, 2L), Mature.Ind = c(0L, 0L, 6L, 5L, 3L, 2L, 1L,
0L, 1L, 2L), MatF = c(0L, 0L, 1L, 4L, 1L, 2L, 0L, 0L, 1L, 2L), MatM = c(0L, 0L, 5L, 1L, 2L, 0L, 1L, 0L, 0L, 0L)), .Names = c("Length.Class",
"Total.Ind", "Mature.Ind", "MatF", "MatM"), class = "data.frame", row.names = c(NA,-10L))
推荐答案
您的系数表示y = -8.62 + 0.1053x，所以x = (glm.out$family$linkfun(.5)+8.62)/ 0.1053.话虽如此，您可能希望使用文档完善的功能，例如MASS软件包中的dose.p(myFit, 0.5)，这样您还可以获得标准错误等.
Your coefficients say that y = -8.62 + 0.1053x, so x = (glm.out$family$linkfun(.5)+8.62)/ 0.1053. Having said that, you'll probably want to use a well documented function, such as dose.p(myFit, 0.5) from the MASS package, so that you also get standard errors etc.
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