动物的平均家庭范围

Question

这几天一直在为此苦苦挣扎。我在这里使用我的数据的一个子集，3 个人，每人 7 次搬迁，跟踪 3 个季节。并非所有的鱼都存在于所有季节，并且它们的位置因季节而异。我需要估计每个季节鱼类的平均栖息地面积并绘制它。

我认为 Anthony Fischbach 在下面的帖子中的回复是我需要做的，我只是中途停留（主要是因为我对空间分析和 adehabitatHR 非常陌生）。

http://r-sig-geo.2731867.n2.nabble.com/averaging-home-range-contours-td7499301.html#a7503972

df <- structure(list(Season = structure(c(2L, 2L, 2L, 2L, 2L, 2L, 2L, 
3L, 3L, 3L, 3L, 3L, 3L, 3L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L), .Label = c("Spring", "Summer", "Fall"), class = "factor"), 
Easting = c(425952.746437786, 425930.124103401, 425904.838430173, 
425897.396103185, 425859.77203381, 425769.155466566, 425698.934422453, 
425172.162553713, 425201.54784946, 425265.188960297, 425293.907156855, 
425334.519543017, 425351.98115297, 425341.657291998, 426499.521534761, 
426491.617459401, 426489.21804795, 426458.425998741, 426444.572705444, 
426435.13014907, 426428.521796826, 425206.363932968, 425214.055516556, 
425214.686198032, 425219.462846981, 425224.610973811, 425225.509004604, 
425231.130704745, 425295.438309455, 425295.033258023, 425298.385156621, 
425286.605300153, 425292.736058183, 425296.563442322, 425289.068691518, 
425262.047903812, 425245.264536305, 425236.068286685, 425246.961026478, 
425244.291558159, 425234.858476333, 425220.329721481), Northing = c(5630038.14127036, 
5630042.68938361, 5630053.17010795, 5630050.05067675, 5630050.03021924, 
5630044.42581987, 5630080.68394433, 5631660.65042482, 5631622.41326505, 
5631580.90371775, 5631564.80939618, 5631533.30748204, 5631498.46768363, 
5631476.0382073, 5629521.34564057, 5629530.46592294, 5629545.1806574, 
5629555.1802463, 5629557.15541335, 5629572.08088964, 5629598.08820948, 
5630913.54144105, 5630904.64457731, 5630904.41307655, 5630898.56103167, 
5630889.14462961, 5630887.90831366, 5630886.93779994, 5630645.78035996, 
5630647.00958702, 5630649.85318956, 5630643.90518234, 5630648.93328633, 
5630650.5466496, 5630648.65226268, 5631661.36063509, 5631635.46509059, 
5631632.70549606, 5631635.88559859, 5631631.47516021, 5631626.93945528, 
5631610.91014489), ID = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 
3L, 3L, 3L, 3L, 3L, 3L), .Label = c("01", "04", "06"), class = "factor")), .Names =         
c("Season", 
"Easting", "Northing", "ID"), row.names = c(NA, 42L), class = "data.frame")

## following the advice in the link above:
library(adehabitatHR)
tagList <- unique(df$ID)

kud1 <- list()
kud_spdf <- list()
vd <- list()

## running a loop across animals, estimating 95% 
for(i in tagList){
thing <- SpatialPoints(subset(df[df$ID == i,], select = c(Easting, Northing)),   
proj4string = CRS("+proj=utm +zone=11 +ellps=WGS84"))
thing <- SpatialPointsDataFrame(coords = thing, data = subset(df[df$ID == i,], select = c(Season)))
thing$Season <- as.character(thing$Season)
kud1[[i]] <- kernelUD(thing, grid = 120, extent = 0.2, same4all=TRUE)
kud_spdf[[i]] <- estUDm2spixdf(kud1[[i]]) 
    fullgrid(kud_spdf[[i]]) <- TRUE
vd[[i]] <- kud_spdf[[i]]@data 
     }

然后我被卡住了，因为 vd 中的每条鱼都有不同数量的点（因为据我了解，网格是为每条鱼单独计算的）。如何继续对鱼进行平均、按像素求和和标准化数据等？

我对绘图部分没问题，只要我可以在 kernelUD 对象上使用getverticeshr，然后运行fortify，并使用 ggmap 和 ggplot 逐组绘图。

任何帮助将不胜感激......

Answer 1

我不确定我是否理解你的问题，但你为什么不能将物种和季节一起工作？

sp_df <- subset(df, select = c(IDS, Easting, Northing))
coordinates(sp_df) <- ~Easting+Northing
#plot(sp_df, axes = T, col = )
kud <- kernelUD(sp_df[ ,1], grid = 200, same4all=TRUE)
hr <- kernel.area(kud, percent = 95)

    X01Fall X01Summer X04Spring X04Summer X06Spring X06Summer
95 8.638694  8.118291  1.353048    2581.2    2581.2 0.4163226

并从这里开始按个人汇总数据

这里有很多选择...

hr1 <- as.data.frame(t(hr))
hr1$ID <- factor(rep(c(1,4,6), each = 2))
names(hr1)[1] <- 'p95'
# Using base functions
aggregate(p95~ ., hr1, mean)

# Fast and simple
library(dplyr)
hr1 %>%
  group_by(ID)%>%
  summarise(mean_hr = mean(p95))
    Source: local data frame [3 x 2]

  ID   mean_hr
1  1 8.4037848
2  4 0.8040197
3  6 0.2109327

编辑

我不明白当个人表现出明显的空间位移时获得平均家庭范围的相关性，例如在这种情况下。

kareas <- getverticeshr(kud, 95)
library(ggplot2)
kdareas <- fortify(kareas)
ggplot(kdareas) +
  geom_polygon(aes(x=long, y=lat, group = group, fill = id, colour = id),
               alpha = .4) +
  theme_bw() +
  coord_equal()

我不明白你为什么应该这样做，但那就是我......