R Corrgram 显示丰度为零的频率对 'Pie Method'
R Corrgram showing frequency pairs that have zero abundance 'Pie Method'
我正在尝试使用 Zuur et al (2010) 可重现的 R 代码(下方)重现一个 corrgram(下图;图 1),显示水鸟物种对的丰度均为零的频率。圆圈的颜色和填充量对应于具有双零的观测值的比例。从左下角到右上角的对角线 运行ning 表示变量为零的观测值百分比。
我已经为我的数据调整了这段代码,但是在 运行 为两个数据集设置代码后我遇到了同样的问题。当我 运行 代码时,corrgram 内的圆圈没有填充,并且保持为空(下图;图 2)。
但是我很困惑为什么我会遇到这个问题。如果有人能解决为什么会发生这种情况,那么我将非常感谢您的帮助。
数据:Zuur 等人 (2010)
数据太大,无法包含在这个 post 中,但可以在名为 ElphickBirdData.txt [=16] 的 supporting materials section 中找到=]
R 代码:Zuur 等人 (2010)
RiceField <- read.table(file="ElphickBirdData.txt", header = TRUE)
AllS <- c(
"TUSW", "GWFG", "WHGO", "CAGO", "MALL",
"GADW", "GWTE", "CITE", "UNTE", "AMWI", "NOPI",
"NOSH", "RIDU", "CANV", "BUFF", "WODU", "RUDU",
"EUWI", "UNDU", "PBGB", "SORA", "COOT", "COMO",
"AMBI", "BCNH", "GBHE", "SNEG", "GREG", "WFIB",
"SACR", "AMAV", "BNST", "BBPL", "KILL", "LBCU",
"GRYE", "LEYE", "LBDO", "SNIP", "DUNL", "WESA",
"LESA", "PEEP", "RUFF", "UNSH", "RBGU", "HEGU",
"CAGU", "GUSP")
#Determine species richness
Richness <- colSums(RiceField[,AllS] > 0, na.rm = TRUE)
#Remove all covariates
Birds <- RiceField[,AllS]
#To reduce the of variables in the figure, we only used the
#20 species that occured at more than 40 sites.
#As a result, N = 20. Else it becomes a mess.
Birds2 <- Birds[, Richness > 40]
N <- ncol(Birds2)
AllNames <- names(Birds2)
A <- matrix(nrow = N, ncol = N)
for (i in 1:N){
for (j in 1:N){
A[i,j] <- sum(RiceField[,AllS[i]]==0 & RiceField[,AllS[j]]==0, na.rm=TRUE)
}}
A1 <- A/2035
print(A1, digits = 2)
rownames(A1) <- AllNames
colnames(A1) <- AllNames
library(lattice)
library(RColorBrewer)
panel.corrgram.2 <- function(x, y, z, subscripts, at = pretty(z), scale = 0.8, ...)
{
require("grid", quietly = TRUE)
x <- as.numeric(x)[subscripts]
y <- as.numeric(y)[subscripts]
z <- as.numeric(z)[subscripts]
zcol <- level.colors(z, at = at, ...)
for (i in seq(along = z))
{
lims <- range(0, z[i])
tval <- 2 * base::pi *
seq(from = lims[1], to = lims[2], by = 0.01)
grid.polygon(x = x[i] + .5 * scale * c(0, sin(tval)),
y = y[i] + .5 * scale * c(0, cos(tval)),
default.units = "native",
gp = gpar(fill = zcol[i]))
grid.circle(x = x[i], y = y[i], r = .5 * scale,
default.units = "native")
}
}
levelplot(A1,xlab=NULL,ylab=NULL,
at=do.breaks(c(0.5,1.01),101),
panel=panel.corrgram.2,
scales=list(x=list(rot=90)),
colorkey=list(space="top"),
col.regions=colorRampPalette(c("red","white","blue")))
#Grey colours
levelplot(A1.bats,xlab=NULL,ylab=NULL,
at=do.breaks(c(0.5,1.01),101),
panel=panel.corrgram.2,
scales=list(x=list(rot=90)),
colorkey=list(space="top"),
col.regions=colorRampPalette(c(grey(0.8),grey(0.5),grey(0.2))))
图 1.
图2
你的问题的原因是grid.circles
用白色涂抹grid.polygon
。您可以通过更改 grid.circle
和 grid.polygon
的顺序来解决它(或将 gp = gpar(fill=NA)
添加到 grid.circle()
)。
panel.corrgram.2.2 <- function(x, y, z, subscripts, at = pretty(z), scale = 0.8, ...)
{
require("grid", quietly = TRUE)
x <- as.numeric(x)[subscripts]
y <- as.numeric(y)[subscripts]
z <- as.numeric(z)[subscripts]
zcol <- level.colors(z, at = at, ...)
for (i in seq(along = z))
{
lims <- range(0, z[i])
tval <- 2 * base::pi *
seq(from = lims[1], to = lims[2], by = 0.01)
grid.circle(x = x[i], y = y[i], r = .5 * scale, # change the order
default.units = "native")
grid.polygon(x = x[i] + .5 * scale * c(0, sin(tval)),
y = y[i] + .5 * scale * c(0, cos(tval)),
default.units = "native",
gp = gpar(fill = zcol[i]))
}
}
levelplot(A1,xlab=NULL,ylab=NULL,
at=do.breaks(c(0.5,1.01),101),
panel=panel.corrgram.2.2,
scales=list(x=list(rot=90)),
colorkey=list(space="top"),
col.regions=colorRampPalette(c("red","white","blue")))
我正在尝试使用 Zuur et al (2010) 可重现的 R 代码(下方)重现一个 corrgram(下图;图 1),显示水鸟物种对的丰度均为零的频率。圆圈的颜色和填充量对应于具有双零的观测值的比例。从左下角到右上角的对角线 运行ning 表示变量为零的观测值百分比。
我已经为我的数据调整了这段代码,但是在 运行 为两个数据集设置代码后我遇到了同样的问题。当我 运行 代码时,corrgram 内的圆圈没有填充,并且保持为空(下图;图 2)。
但是我很困惑为什么我会遇到这个问题。如果有人能解决为什么会发生这种情况,那么我将非常感谢您的帮助。
数据:Zuur 等人 (2010)
数据太大,无法包含在这个 post 中,但可以在名为 ElphickBirdData.txt [=16] 的 supporting materials section 中找到=]
R 代码:Zuur 等人 (2010)
RiceField <- read.table(file="ElphickBirdData.txt", header = TRUE)
AllS <- c(
"TUSW", "GWFG", "WHGO", "CAGO", "MALL",
"GADW", "GWTE", "CITE", "UNTE", "AMWI", "NOPI",
"NOSH", "RIDU", "CANV", "BUFF", "WODU", "RUDU",
"EUWI", "UNDU", "PBGB", "SORA", "COOT", "COMO",
"AMBI", "BCNH", "GBHE", "SNEG", "GREG", "WFIB",
"SACR", "AMAV", "BNST", "BBPL", "KILL", "LBCU",
"GRYE", "LEYE", "LBDO", "SNIP", "DUNL", "WESA",
"LESA", "PEEP", "RUFF", "UNSH", "RBGU", "HEGU",
"CAGU", "GUSP")
#Determine species richness
Richness <- colSums(RiceField[,AllS] > 0, na.rm = TRUE)
#Remove all covariates
Birds <- RiceField[,AllS]
#To reduce the of variables in the figure, we only used the
#20 species that occured at more than 40 sites.
#As a result, N = 20. Else it becomes a mess.
Birds2 <- Birds[, Richness > 40]
N <- ncol(Birds2)
AllNames <- names(Birds2)
A <- matrix(nrow = N, ncol = N)
for (i in 1:N){
for (j in 1:N){
A[i,j] <- sum(RiceField[,AllS[i]]==0 & RiceField[,AllS[j]]==0, na.rm=TRUE)
}}
A1 <- A/2035
print(A1, digits = 2)
rownames(A1) <- AllNames
colnames(A1) <- AllNames
library(lattice)
library(RColorBrewer)
panel.corrgram.2 <- function(x, y, z, subscripts, at = pretty(z), scale = 0.8, ...)
{
require("grid", quietly = TRUE)
x <- as.numeric(x)[subscripts]
y <- as.numeric(y)[subscripts]
z <- as.numeric(z)[subscripts]
zcol <- level.colors(z, at = at, ...)
for (i in seq(along = z))
{
lims <- range(0, z[i])
tval <- 2 * base::pi *
seq(from = lims[1], to = lims[2], by = 0.01)
grid.polygon(x = x[i] + .5 * scale * c(0, sin(tval)),
y = y[i] + .5 * scale * c(0, cos(tval)),
default.units = "native",
gp = gpar(fill = zcol[i]))
grid.circle(x = x[i], y = y[i], r = .5 * scale,
default.units = "native")
}
}
levelplot(A1,xlab=NULL,ylab=NULL,
at=do.breaks(c(0.5,1.01),101),
panel=panel.corrgram.2,
scales=list(x=list(rot=90)),
colorkey=list(space="top"),
col.regions=colorRampPalette(c("red","white","blue")))
#Grey colours
levelplot(A1.bats,xlab=NULL,ylab=NULL,
at=do.breaks(c(0.5,1.01),101),
panel=panel.corrgram.2,
scales=list(x=list(rot=90)),
colorkey=list(space="top"),
col.regions=colorRampPalette(c(grey(0.8),grey(0.5),grey(0.2))))
图 1.
你的问题的原因是grid.circles
用白色涂抹grid.polygon
。您可以通过更改 grid.circle
和 grid.polygon
的顺序来解决它(或将 gp = gpar(fill=NA)
添加到 grid.circle()
)。
panel.corrgram.2.2 <- function(x, y, z, subscripts, at = pretty(z), scale = 0.8, ...)
{
require("grid", quietly = TRUE)
x <- as.numeric(x)[subscripts]
y <- as.numeric(y)[subscripts]
z <- as.numeric(z)[subscripts]
zcol <- level.colors(z, at = at, ...)
for (i in seq(along = z))
{
lims <- range(0, z[i])
tval <- 2 * base::pi *
seq(from = lims[1], to = lims[2], by = 0.01)
grid.circle(x = x[i], y = y[i], r = .5 * scale, # change the order
default.units = "native")
grid.polygon(x = x[i] + .5 * scale * c(0, sin(tval)),
y = y[i] + .5 * scale * c(0, cos(tval)),
default.units = "native",
gp = gpar(fill = zcol[i]))
}
}
levelplot(A1,xlab=NULL,ylab=NULL,
at=do.breaks(c(0.5,1.01),101),
panel=panel.corrgram.2.2,
scales=list(x=list(rot=90)),
colorkey=list(space="top"),
col.regions=colorRampPalette(c("red","white","blue")))