如何估计 R 中沿图形线的点的坐标？

Question

Assume I have data:

x <- c(1900,1930,1944,1950,1970,1980,1983,1984)
y <- c(100,300,500,1500,2500,3500,4330,6703)

I then plot this data and add a line graph between my known x and y coordinates:

plot(x,y)
lines(x,y)

Is there a way to predict coordinates of unknown points along the graphed line?

Answer 1

You can use approxfun .

f <- approxfun(x, y=y)

f(seq(1900, 2000, length.out = 10))
# [1]  100.0000  174.0741  248.1481  347.6190  574.0741 1777.7778 2333.3333
# [8] 3277.7778        NA        NA

Note the NA, when the sequence is outside the range of interpolated points (there are left and right options to approxfun).

Answer 2

You can manually calculate the slopes of each line.

The equation of a line is given by

y − y1 = grad*(x − x1)

where grad is calculated by the change in y divided by the change in x

We can produce equations for each line by using two points from each line in the plot.

f2 <- function(xnew, X=x, Y=y) {
             id0 <- findInterval(xnew, X, rightmost=T)
             id1 <- id0 + 1
             grad <- (Y[id1] - Y[id0]) / (X[id1] - X[id0]) 
             Y[id0] + grad* (xnew - X[id0])
 }

 f2(x)
#[1]  100  300  500 1500 2500 3500 4330 6703

f <- approxfun(x, y=y) # bunks
f(seq(1900, 2000, length.out = 10))
# [1]  100.0000  174.0741  248.1481  347.6190  574.0741 1777.7778 2333.3333 3277.7778        NA        NA
f2(seq(1900, 2000, length.out = 10))
# [1]  100.0000  174.0741  248.1481  347.6190  574.0741 1777.7778 2333.3333 3277.7778        NA        NA

If you wanted to extrapolate using the final slope, you can do this by adding the all.in=TRUE argument to findInterval .

With that said, approxfun does it better & easier!

Answer 3

Or, if you want to approximate points along the lines in regular intervals, you can use approx instead of approxfun . For that purpose, this maybe saves you a tiny bit of coding.

x <- c(1900,1930,1944,1950,1970,1980,1983,1984)
y <- c(100,300,500,1500,2500,3500,4330,6703)

new_points <- approx(x, y)
lapply(new_points, head)
#> $x
#> [1] 1900.000 1901.714 1903.429 1905.143 1906.857 1908.571
#> 
#> $y
#> [1] 100.0000 111.4286 122.8571 134.2857 145.7143 157.1429
plot(new_points)
lines(x,y)

^{Created on 2022-11-20 with reprex v2.0.2}