ggplot2中的交叉点

Question

我需要找到我在ggplot2中绘制的两条曲线之间的交点。

这是我的代码：

ggplot(Table1, aes(x=10^H4,y=10^H5, colour = "blabla")) + 
  geom_line(size=1)  + coord_cartesian(xlim=c(100,1000000000))+
 coord_cartesian(ylim=c(0.000000000001,1)) +  xlab("blabla")+
 ylab("blabla")  + ggtitle("blabla")+
scale_y_log10(breaks=c(0.000000000001, 0.000000001, 0.000001, 0.001)
, labels = trans_format("log10", math_format(10^.x)))+
 scale_x_log10(breaks=c(100,1000,10000,100000,1000000,10000000,100000000,1000000000)   
 , labels = trans_format("log10", math_format(10^.x))) + 
  geom_line(data=Table1,aes(10^H4,10^H6, colour = "blabla1"))+ 
  scale_color_manual("", values =c("blabla"="blue", "blabla1" = "red")
, labels=c("blabla","blabla1"))

我尝试过使用locator()这是有用的，但并不像我预期的那样精确：

Desired point

期望的点是24600。

我也尝试使用intercept(x,y)：

a <- 10^A5
b <- 10^A6
intercept(a,b)
3.689776e-07 1.963360e-07 6.622165e-07

事实并非如此，我认为可能没有考虑到这是一个对数日志的事实。

我的数据：

structure(list(H4 = c(2, 2.1, 2.2, 2.3, 2.4, 2.5, 2.6, 2.7, 2.8, 
2.9, 3, 3.1, 3.2, 3.3, 3.4, 3.5, 3.6, 3.7, 3.8, 3.9, 4, 4.1, 
4.2, 4.3, 4.4, 4.5, 4.6, 4.7, 4.8, 4.9, 5, 5.1, 5.2, 5.3, 5.4, 
5.5, 5.6, 5.7, 5.8, 5.9, 6, 6.1, 6.2, 6.3, 6.4, 6.5, 6.6, 6.7, 
6.8, 6.9, 7, 7.1, 7.2, 7.3, 7.4, 7.5, 7.6, 7.7, 7.8, 7.9, 8, 
8.1, 8.2, 8.3, 8.4, 8.5, 8.6, 8.7, 8.8, 8.9, 9), H5 = c(-8.979, 
-8.927, -8.877, -8.829, -8.782, -8.736, -8.691, -8.648, -8.606, 
-8.565, -8.525, -8.485, -8.445, -8.405, -8.364, -8.323, -8.282, 
-8.241, -8.2, -8.159, -8.116, -8.064, -7.982, -7.826, -7.592, 
-7.333, -7.101, -6.91, -6.759, -6.64, -6.543, -6.461, -6.387, 
-6.321, -6.264, -6.218, -6.182, -6.155, -6.132, -6.117, -6.111, 
-6.12, -6.23, -6.433, -6.574, -6.664, -6.712, -6.726, -6.722, 
-6.707, -6.704, -6.748, -6.82, -6.864, -6.872, -6.859, -6.83, 
-6.796, -6.757, -6.717, -6.678, -6.636, -6.594, -6.549, -6.502, 
-6.454, -6.402, -6.349, -6.295, -6.238, -6.179), H6 = c(-5.116, 
-5.31, -5.495, -5.669, -5.823, -5.958, -6.075, -6.179, -6.271, 
-6.355, -6.433, -6.506, -6.575, -6.642, -6.707, -6.769, -6.829, 
-6.886, -6.941, -6.993, -7.044, -7.095, -7.144, -7.192, -7.237, 
-7.28, -7.321, -7.36, -7.398, -7.435, -7.47, -7.504, -7.536, 
-7.569, -7.602, -7.64, -7.684, -7.735, -7.789, -7.848, -7.917, 
-8.003, -8.131, -8.312, -8.494, -8.668, -8.823, -8.963, -9.095, 
-9.225, -9.365, -9.531, -9.711, -9.859, -9.965, -10.041, -10.098, 
-10.141, -10.175, -10.203, -10.236, -10.263, -10.285, -10.301, 
-10.314, -10.323, -10.33, -10.335, -10.339, -10.342, -10.344)), .Names = c("H4", 
"H5", "H6"), row.names = c(NA, -71L), class = "data.frame")

Answer 1

我们可以使用approxfun()函数，它执行线性插值。我们使用两条曲线之间的差异（或在数据点集之间），然后找到x轴上的近似值，使y中的差值或值等于0.

#Finding the x value in the log10 scale

f1 <- approxfun(10^Table1$H5 - 10^Table1$H6,Table1$H4, rule=2)

f1(0)
[1] 4.516067

x11(); ggplot(Table1, aes(x=10^H4,y=10^H5, colour = "blabla")) + 
  geom_line(size=1)  + coord_cartesian(xlim=c(100,1000000000))+
 coord_cartesian(ylim=c(0.000000000001,1)) +  xlab("blabla")+
 ylab("blabla")  + ggtitle("blabla")+
scale_y_log10(breaks=c(0.000000000001, 0.000000001, 0.000001, 0.001)
, labels = trans_format("log10", math_format(10^.x)))+
 scale_x_log10(breaks=c(100,1000,10000,100000,1000000,10000000,100000000,1000000000)   
 , labels = trans_format("log10", math_format(10^.x))) + 
  geom_line(data=Table1,aes(10^H4,10^H6, colour = "blabla1"))+ 
  scale_color_manual("", values =c("blabla"="blue", "blabla1" = "red")
, labels=c("blabla","blabla1"))+
geom_vline(xintercept=10^ f1(0)) # adding the verical line

数据：

structure(list(H4 = c(2, 2.1, 2.2, 2.3, 2.4, 2.5, 2.6, 2.7, 2.8, 
2.9, 3, 3.1, 3.2, 3.3, 3.4, 3.5, 3.6, 3.7, 3.8, 3.9, 4, 4.1, 
4.2, 4.3, 4.4, 4.5, 4.6, 4.7, 4.8, 4.9, 5, 5.1, 5.2, 5.3, 5.4, 
5.5, 5.6, 5.7, 5.8, 5.9, 6, 6.1, 6.2, 6.3, 6.4, 6.5, 6.6, 6.7, 
6.8, 6.9, 7, 7.1, 7.2, 7.3, 7.4, 7.5, 7.6, 7.7, 7.8, 7.9, 8, 
8.1, 8.2, 8.3, 8.4, 8.5, 8.6, 8.7, 8.8, 8.9, 9), H5 = c(-8.979, 
-8.927, -8.877, -8.829, -8.782, -8.736, -8.691, -8.648, -8.606, 
-8.565, -8.525, -8.485, -8.445, -8.405, -8.364, -8.323, -8.282, 
-8.241, -8.2, -8.159, -8.116, -8.064, -7.982, -7.826, -7.592, 
-7.333, -7.101, -6.91, -6.759, -6.64, -6.543, -6.461, -6.387, 
-6.321, -6.264, -6.218, -6.182, -6.155, -6.132, -6.117, -6.111, 
-6.12, -6.23, -6.433, -6.574, -6.664, -6.712, -6.726, -6.722, 
-6.707, -6.704, -6.748, -6.82, -6.864, -6.872, -6.859, -6.83, 
-6.796, -6.757, -6.717, -6.678, -6.636, -6.594, -6.549, -6.502, 
-6.454, -6.402, -6.349, -6.295, -6.238, -6.179), H6 = c(-5.116, 
-5.31, -5.495, -5.669, -5.823, -5.958, -6.075, -6.179, -6.271, 
-6.355, -6.433, -6.506, -6.575, -6.642, -6.707, -6.769, -6.829, 
-6.886, -6.941, -6.993, -7.044, -7.095, -7.144, -7.192, -7.237, 
-7.28, -7.321, -7.36, -7.398, -7.435, -7.47, -7.504, -7.536, 
-7.569, -7.602, -7.64, -7.684, -7.735, -7.789, -7.848, -7.917, 
-8.003, -8.131, -8.312, -8.494, -8.668, -8.823, -8.963, -9.095, 
-9.225, -9.365, -9.531, -9.711, -9.859, -9.965, -10.041, -10.098, 
-10.141, -10.175, -10.203, -10.236, -10.263, -10.285, -10.301, 
-10.314, -10.323, -10.33, -10.335, -10.339, -10.342, -10.344)), .Names = c("H4", 
"H5", "H6"), row.names = c(NA, -71L), class = "data.frame")