(sigmoid) curve fitting glm in r

廉价感情. 提交于 2021-01-28 08:32:59

问题


I wish to visualize the relationship between my response variable, detection probability (P.det) and predictor variable (distance) for two categories (transmitter), show error bars and draw a (sigmoidal) curve through the averaged data points.

The dataset is like this:

df <- structure(list(distance = c(50L, 100L, 200L, 300L, 400L, 50L, 
100L, 200L, 300L, 400L), Transmitter = structure(c(1L, 1L, 1L, 
1L, 1L, 2L, 2L, 2L, 2L, 2L), .Label = c("CT", "PT"), class = "factor"), 
    P.det = c(0.918209097, 0.88375438, 0.709288774, 0.534977488, 
    0.341724516, 0.828123952, 0.822201191, 0.543289433, 0.352886247, 
    0.10082457), st.error = c(0.01261614, 0.014990469, 0.024136478, 
    0.027311169, 0.026941438, 0.018663591, 0.019420587, 0.02754911, 
    0.026809247, 0.017041264), ly = c(0.905592958, 0.868763911, 
    0.685152295, 0.50766632, 0.314783078, 0.809460361, 0.802780604, 
    0.515740323, 0.326077, 0.083783306), uy = c(0.930825237, 
    0.898744849, 0.733425252, 0.562288657, 0.368665955, 0.846787544, 
    0.841621778, 0.570838544, 0.379695494, 0.117865833), Valid.detections = c(18, 
    12.5472973, 8.608108108, 4.287162162, 2.158783784, 12.46959459, 
    7.956081081, 4.550675676, 1.682432432, 0.39527027), False.detections = c(0.388513514, 
    0.550675676, 0.368243243, 0.263513514, 0.131756757, 0.533783784, 
    0.385135135, 0.277027027, 0.182432432, 0.14527027)), .Names = c("distance", 
"Transmitter", "P.det", "st.error", "ly", "uy", "Valid.detections", 
"False.detections"), class = "data.frame", row.names = c(NA, 
-10L))

I managed to get the first 2 parts done, but am stuck at the last part. The code to draw the graph with error bars:

library(lattice)
    prepanel.ci <- function(x, y, ly, uy, subscripts, ...)
    {
      x <- as.numeric(x)
      ly <- as.numeric(ly[subscripts])
      uy <- as.numeric(uy[subscripts])
      list(ylim = range(y, uy, ly, finite = TRUE))
    }

    panel.ci <- function(x, y, ly, uy, subscripts, pch = 16, ...)
    {
      x <- as.numeric(x)
      y <- as.numeric(y)
      ly <- as.numeric(ly[subscripts])
      uy <- as.numeric(uy[subscripts])
      panel.arrows(x, ly, x, uy, col = "black",
                   length = 0.25, unit = "native",
                   angle = 90, code = 3)
      panel.xyplot(x, y, pch = pch, ...)
    }

xyplot(P.det~distance, type=c("p","g"),
       ylim=c(0,1),
       ylab="Detection probability", xlab="Distance (m)", 
       group=Transmitter,
       data=df,
       ly = df$ly,
       uy = df$uy,
       prepanel = prepanel.ci,
       panel = panel.superpose,
       panel.groups = panel.ci,
       col=c(1,1),
       layout=c(1,1),
       between=list(x=2),
       scales=list(x=list(alternating=c(1,1), tck=c(1,0)),y=list(alternating=c(1,1), tck=c(1,0))), # ticks inside = tck=c(-1,0)
       aspect=1,
       main="Detection probability vs distance per transmitter type",
)

The reason why I state "glm" in the title is because the data analysis was carried out using a binomial glm() using the lme4 package.

I noticed another thread which is similar to mine: find the intersection of abline with fitted curve, however the difference is that while my graph is also based on 1 y per 1 x, my glm is based on a multitude of y's per x. So following the same codes in this thread returns an error stating that the lengths are not of equal length. It also doesn't seem to work for an "xyplot".

Thanks


回答1:


This is fairly straightforward using ggplot:

library(ggplot2)
ggplot(data = df, aes(x = distance, y = P.det, colour = Transmitter)) +
  geom_pointrange(aes(ymin = P.det - st.error, ymax = P.det + st.error)) +
  geom_smooth(method = "glm", family = binomial, se = FALSE)

enter image description here

Regarding the glmwarning message, see e.g. here.




回答2:


In Lattice, you might add the smoothing into your custom panel function. You can change it to include

panel.ci <- function(x, y, ly, uy, subscripts, type="p", pch = 16, ...)
{
  x <- as.numeric(x)
  y <- as.numeric(y)
  ly <- as.numeric(ly[subscripts])
  uy <- as.numeric(uy[subscripts])
  panel.arrows(x, ly, x, uy, col = "black",
               length = 0.25, unit = "native",
               angle = 90, code = 3)
  panel.xyplot(x, y, pch = pch, type=type, ...)

  #calculate smooth curve
  gg <- glm(y~x, family="binomial")
  gx <- seq(min(x), max(x), length.out=100)
  panel.xyplot(gx, predict(gg, data.frame(x=gx), type="response"),
      pch = pch, type="l", ...)
}

Here we do the glm ourselves and draw the response on the plot.

enter image description here



来源:https://stackoverflow.com/questions/24656498/sigmoid-curve-fitting-glm-in-r

标签
易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!