CODEWITHSUNDEEP
rplotcoordinates
Victor Hanson
Victor Hanson

Reputation: 13

Coordinates of x axis where line cuts the curve

I have a graph as shown below. How do I find the coordinates on x-axis where the horizontal grey line cuts the curve?

enter image description here

below is the toy dataset that I used to generate x, y values 

df <- structure(list(x = c(0, -1, -2, -3, -4, -5, -6, -7, -8), 
    y = c(-5.22672371336103, -2.04798328990208, 
    -0.998312848674327, -1.13656559451279, -1.80175393429754, 
    -2.67597356058193, -3.62933726371666, -4.61213085819315, 
    -5.60579419730348)), .Names = c("x", "y"
), row.names = c(NA, -9L), class = "data.frame")

plot(df$x, df$y, asp = 1)
abline(h=-1.92,  col = "gray60")
lines(df$x, df$y)

Upvotes: 1

Views: 303

Answers (2)

Neoromanzer
Neoromanzer

Reputation: 434

Here comes another solution.

Let me define the x and y vectors separately first. 

x= c(0, -1, -2, -3, -4, -5, -6, -7, -8)
y= c(-5.22672371336103, -2.04798328990208, 
     -0.998312848674327, -1.13656559451279, -1.80175393429754, 
     -2.67597356058193, -3.62933726371666, -4.61213085819315, 
     -5.60579419730348)

Instead of focusing on the intersection between the two curves, to simplify, what I´m going to do is to displace/move your curve h units up. 

y<-y+1.92

Now my problem is much simpler: Calculating the roots of the curve.

I´ll fit a 4th degree polynomial (this was a bit random, I have to admit). 

fit4 <- lm(y~poly(x,4,raw=TRUE))
summary(fit4)

Coefficients:
                          Estimate Std. Error t value Pr(>|t|)    
(Intercept)             -3.2879360  0.0525516  -62.57 3.91e-07 ***
poly(x, 4, raw = TRUE)1 -4.4218115  0.1044875  -42.32 1.86e-06 ***
poly(x, 4, raw = TRUE)2 -1.4833140  0.0583804  -25.41 1.42e-05 ***
poly(x, 4, raw = TRUE)3 -0.1799201  0.0112986  -15.92 9.09e-05 ***
poly(x, 4, raw = TRUE)4 -0.0080516  0.0007005  -11.49 0.000327 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.05373 on 4 degrees of freedom
Multiple R-squared:  0.9995,    Adjusted R-squared:  0.9991 
F-statistic:  2130 on 4 and 4 DF,  p-value: 6.605e-07

As you can see, my R-squared is quite good... so enough.

Now, I get the coefficients and get the roots of my polynomial. 

coef<-fit4$coefficients
polyroot(coef)

Which are -1.094 and -4.136.

Upvotes: 2

AkselA
AkselA

Reputation: 8846

I'm sure there are more clever ways of doing this, but here's a method using splines and brute force.

spl <- splinefun(df)
s <- seq(min(df$x), max(df$x), by=5e-3)
est <- spl(s)

xs <- s[diff(sign(diff(c(0, abs(-1.92 - est))))) > 0]

plot(df$x, df$y, asp=1)
abline(h=-1.92,  col = "gray60")
lines(s, est)
abline(v=xs, col="blue")

enter image description here

Upvotes: 2

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