I can't find out how should I put up this ques so I used this method.
我不知道如何提出这个问题所以我使用了这种方法。
I have a latitude-longitude dataset. The image posted below is what I want to produce.. This is my dataset:
我有一个纬度 - 经度数据集。下面张贴的图片是我想要制作的图片。这是我的数据集:
Latitude Longitude
21.06941667 71.07952778
21.06941667 71.07952778
21.06666667 71.08158333
21.07186111 71.08688889
21.08625 71.07083333
21.08719444 71.07286111
21.08580556 71.07686111
21.07894444 71.08225
....
I have used geom_path() to find the path. Now, As shown in fig. I have highlighted the variance with white color around the path which I want to do. This is how I calculated variance:
我使用geom_path()来查找路径。现在,如图所示。我突出了我想要做的路径周围的白色差异。这就是我计算方差的方法:
var.latitude <- var(Data$Latitude)
var.longitude <- var(Data$Longitude)
I have marked the variance over the points using geom_errorbar():
我使用geom_errorbar()标记了点的方差:
geom_errorbar(aes(x=Latitude,y=Longitude, ymin=Longitude-var.longitude, ymax=Longitude+var.longitude),width=0.001)+
geom_errorbarh(aes(xmin=Latitude-var.latitude,xmax=Latitude+var.latitude),height=0.001)
Can anyone tell me how should I highlight the white area?
谁能告诉我如何突出白色区域?
1 个解决方案
#1
4
I'm approaching this with the polygon feature of ggplot, see the documentation
我正在接近ggplot的面要素,请参阅文档
library(ggplot2)
data = rbind.data.frame(c(21.06941667, 71.07952778),
c(21.06666667, 71.08158333 ),
c(21.07186111, 71.08688889 ),
c(21.08625 , 71.07083333 ),
c(21.08719444, 71.07286111 ),
c(21.08580556, 71.07686111 ),
c(21.07894444, 71.08225 ))
names(data) = c("Latitude", "Longitude")
Your variance is quite small, I multiplied by 10 for it to be visible in the graph. Note that in the graph in your question you draw the area from the fins of the errorbars, which is almost certainly not what you want.
你的方差非常小,我乘以10使其在图中可见。请注意,在您的问题的图表中,您从错误栏的鳍中绘制区域,这几乎肯定不是您想要的。
var.latitude <- var(data$Latitude)*10
var.longitude <- var(data$Longitude)*10
Calculating this area as one is a menial task as also noted in the comments above. I found the easiest way to do this is overlapping two polygons for each path plus a polygon for each point. There sure may be more elegant ways, but hey, it works.
将该区域计算为一个区域是一项琐碎的任务,如上述评论中所述。我发现最简单的方法是为每个路径重叠两个多边形,并为每个点重叠一个多边形。肯定会有更优雅的方式,但嘿,它的确有效。
pos.poly = data.frame(id = paste0("c", as.character(1)),
x = c(data$Latitude[1]-var.latitude, data$Latitude[1], data$Latitude[1]+var.latitude, data$Latitude[1]),
y = c(data$Longitude[1], data$Longitude[1]+var.longitude, data$Longitude[1], data$Longitude[1]-var.longitude))
for(i in 2:dim(data)[1]){
loc.pos1 = data.frame(id = paste0("a", as.character(i)),
x = c(data$Latitude[i-1]-var.latitude, data$Latitude[i]-var.latitude,
data$Latitude[i]+var.latitude, data$Latitude[i-1]+var.latitude),
y = c(data$Longitude[i-1], data$Longitude[i], data$Longitude[i], data$Longitude[i-1]))
pos.poly = rbind(pos.poly, loc.pos1)
loc.pos2 = data.frame(id = paste0("b", as.character(i)),
x = c(data$Latitude[i-1], data$Latitude[i], data$Latitude[i], data$Latitude[i-1]),
y = c(data$Longitude[i-1]+var.longitude, data$Longitude[i]+var.longitude,
data$Longitude[i]-var.longitude, data$Longitude[i-1]-var.longitude))
pos.poly = rbind(pos.poly, loc.pos2)
loc.pos3 = data.frame(id = paste0("c", as.character(i)),
x = c(data$Latitude[i]-var.latitude, data$Latitude[i], data$Latitude[i]+var.latitude, data$Latitude[i]),
y = c(data$Longitude[i], data$Longitude[i]+var.longitude, data$Longitude[i], data$Longitude[i]-var.longitude))
pos.poly = rbind(pos.poly, loc.pos3)
}
This is plotted from two datasets so we need to specify data
and the aes
a couple more times.
这是从两个数据集绘制的,因此我们需要多次指定数据和aes。
plot1 = ggplot(pos.poly, aes(x=x, y=y)) + geom_polygon(aes(group=id), fill="white") + geom_path(data = data, aes(x=Latitude, y=Longitude))
plot1 = plot1 + xlab("Latitude") + ylab("Longitude") +
geom_errorbar(data = data, aes(x=Latitude,y=Longitude, ymin=Longitude-var.longitude, ymax=Longitude+var.longitude)) +
geom_errorbarh(data = data, aes(xmin=Latitude-var.latitude,xmax=Latitude+var.latitude, x=Latitude, y=Longitude))
print(plot1)
#1
4
I'm approaching this with the polygon feature of ggplot, see the documentation
我正在接近ggplot的面要素,请参阅文档
library(ggplot2)
data = rbind.data.frame(c(21.06941667, 71.07952778),
c(21.06666667, 71.08158333 ),
c(21.07186111, 71.08688889 ),
c(21.08625 , 71.07083333 ),
c(21.08719444, 71.07286111 ),
c(21.08580556, 71.07686111 ),
c(21.07894444, 71.08225 ))
names(data) = c("Latitude", "Longitude")
Your variance is quite small, I multiplied by 10 for it to be visible in the graph. Note that in the graph in your question you draw the area from the fins of the errorbars, which is almost certainly not what you want.
你的方差非常小,我乘以10使其在图中可见。请注意,在您的问题的图表中,您从错误栏的鳍中绘制区域,这几乎肯定不是您想要的。
var.latitude <- var(data$Latitude)*10
var.longitude <- var(data$Longitude)*10
Calculating this area as one is a menial task as also noted in the comments above. I found the easiest way to do this is overlapping two polygons for each path plus a polygon for each point. There sure may be more elegant ways, but hey, it works.
将该区域计算为一个区域是一项琐碎的任务,如上述评论中所述。我发现最简单的方法是为每个路径重叠两个多边形,并为每个点重叠一个多边形。肯定会有更优雅的方式,但嘿,它的确有效。
pos.poly = data.frame(id = paste0("c", as.character(1)),
x = c(data$Latitude[1]-var.latitude, data$Latitude[1], data$Latitude[1]+var.latitude, data$Latitude[1]),
y = c(data$Longitude[1], data$Longitude[1]+var.longitude, data$Longitude[1], data$Longitude[1]-var.longitude))
for(i in 2:dim(data)[1]){
loc.pos1 = data.frame(id = paste0("a", as.character(i)),
x = c(data$Latitude[i-1]-var.latitude, data$Latitude[i]-var.latitude,
data$Latitude[i]+var.latitude, data$Latitude[i-1]+var.latitude),
y = c(data$Longitude[i-1], data$Longitude[i], data$Longitude[i], data$Longitude[i-1]))
pos.poly = rbind(pos.poly, loc.pos1)
loc.pos2 = data.frame(id = paste0("b", as.character(i)),
x = c(data$Latitude[i-1], data$Latitude[i], data$Latitude[i], data$Latitude[i-1]),
y = c(data$Longitude[i-1]+var.longitude, data$Longitude[i]+var.longitude,
data$Longitude[i]-var.longitude, data$Longitude[i-1]-var.longitude))
pos.poly = rbind(pos.poly, loc.pos2)
loc.pos3 = data.frame(id = paste0("c", as.character(i)),
x = c(data$Latitude[i]-var.latitude, data$Latitude[i], data$Latitude[i]+var.latitude, data$Latitude[i]),
y = c(data$Longitude[i], data$Longitude[i]+var.longitude, data$Longitude[i], data$Longitude[i]-var.longitude))
pos.poly = rbind(pos.poly, loc.pos3)
}
This is plotted from two datasets so we need to specify data
and the aes
a couple more times.
这是从两个数据集绘制的,因此我们需要多次指定数据和aes。
plot1 = ggplot(pos.poly, aes(x=x, y=y)) + geom_polygon(aes(group=id), fill="white") + geom_path(data = data, aes(x=Latitude, y=Longitude))
plot1 = plot1 + xlab("Latitude") + ylab("Longitude") +
geom_errorbar(data = data, aes(x=Latitude,y=Longitude, ymin=Longitude-var.longitude, ymax=Longitude+var.longitude)) +
geom_errorbarh(data = data, aes(xmin=Latitude-var.latitude,xmax=Latitude+var.latitude, x=Latitude, y=Longitude))
print(plot1)