我有一条曲线,来自经验数据,我可以得到一个合理的模型。我需要识别一个点(x,y),其中曲线与已知中心和半径的圆相交。以下代码说明了这个问题。
x <- c(0.05, 0.20, 0.35, 0.50, 0.65, 0.80, 0.95,
1.10, 1.25, 1.40, 1.55, 1.70, 1.85, 2.00,
2.15, 2.30, 2.45, 2.60, 2.75, 2.90, 3.05)
y <- c(1.52, 1.44, 1.38, 1.31, 1.23, 1.15, 1.06,
0.96, 0.86, 0.76, 0.68, 0.61, 0.54, 0.47,
0.41, 0.36, 0.32, 0.29, 0.27, 0.26, 0.26)
fit <- loess(y ~ x, control = loess.control(surface = "direct"))
newx <- data.frame(x = seq(0, 3, 0.01))
fitline <- predict(fit, newdata = newx)
est <- data.frame(newx, fitline)
plot(x, y, type = "o",lwd = 2)
lines(est, col = "blue", lwd = 2)
library(plotrix)
draw.circle(x = 3, y = 0, radius = 2, nv = 1000, lty = 1, lwd = 1)
答案 0 :(得分:10)
为了获得交点,我们可以使用r中的optim函数来执行此操作:
circle=function(x){
if(4<(x-3)^2) return(NA)# Ensure it is limited within the radius
sqrt(4-(x-3)^2)
}
fun=function(x)predict(fit,data.frame(x=x))
g=function(x)(circle(x)-fun(x))# We need to set this to zero. Ie solve this
sol1=optimise(function(x)abs(g(x)),c(1,5))$min
[1] 1.208466
因此,这两个函数应该在x=1.208466
..
为了更加精确,您可以使用optim
功能:
sol2= optim(1,function(x)abs(g(x)),g,method="Brent",upper=5,lower=1)$par
[1] 1.208473
现在你可以评估:
circle(sol1)
[1] 0.889047
fun(sol1)
1
0.8890654
circle(sol2)
[1] 0.889061
fun(sol2)
1
0.889061
从上面可以看出,解决方案2非常接近..
在图表上绘制这一点将具有挑战性,因为draw.circle
函数与zxes成比例绘制圆圈。因此每次都会根据绘图区域的大小进行更改。
如果您要编写自己的圆圈功能:
circleplot=function(x,y,r){
theta=seq(0,2*pi,length.out = 150)
cbind(x+r*cos(theta),y+r*sin(theta))
}
然后你可以这样做:
plot(x, y, type = "o",lwd = 2)
lines(est, col = "blue", lwd = 2)
lines(circleplot(3,0,2))
abline(v=sol2,col=2)
points(sol2,fun(sol2),col=2,pch=16)
答案 1 :(得分:8)
使用sf
包中的函数可以直接找到交集。
计算圆值(受this answer和as done by @Onyambu启发)
circ <- function(xc = 0, yc = 0, r = 1, n = 100){
v <- seq(0, 2 * pi, len = n)
cbind(x = xc + r * cos(v),
y = yc + r * sin(v))
}
m <- circ(xc = 3, yc = 0, r = 2)
将预测值和圆值转换为&#34;简单特征&#34; (LINESTRING
),找到他们的交集(POINT
):
library(sf)
int <- st_intersection(st_linestring(as.matrix(est)),
st_linestring(m))
int
# POINT (1.2091 0.8886608)
将交集添加到地块中:
plot(x, y, type = "o", lwd = 2)
lines(est, col = "blue", lwd = 2)
lines(m)
points(int[1], int[2], col = "red", pch = 19)