如何将梯形积分方法调整为自定义零点？

Question

I want to calculate the integral of a sequence vector. Since there's no function available, I use the trapezoidal method ¹ .

iglTzm <- function(x, y) sum(diff(x) * (head(y, -1) + tail(y, -1))) / 2

The first element of the sequence should be the zero point, so the principle is: if the values of the sequence are predominantly below the first value, the integral should be negative, otherwise positive, or 0.

Consider matrix m1 :

     [,1] [,2] [,3] [,4] [,5] [,6] [,7]
[1,]    6    7    8    8    6    8   10
[2,]    9    9    8    9    9    8    9
[3,]    9   10   10    9    9    9    9
[4,]    9    8    8    8    6    8    9
[5,]   10   10   10    9   10    8    0
[6,]    9    8    9   10    9    9    9

Integration with these raw values will most likely lead to inconsistent values:

> setNames(apply(m1, 1, iglTzm, 0:6), 1:6)
  1   2   3   4   5   6 
 15   2  -2   7 -52   0 

So I adjust the sequences (rows) on their first value (column 1), in order to set the right signs, and get matrix m2 :

     [,1] [,2] [,3] [,4] [,5] [,6] [,7]
[1,]    0    1    2    2    0    2    4
[2,]    0    0   -1    0    0   -1    0
[3,]    0    1    1    0    0    0    0
[4,]    0   -1   -1   -1   -3   -1    0
[5,]    0    0    0   -1    0   -2  -10
[6,]    0   -1    0    1    0    0    0

Logically that doesn't change anything about the values iglTzm() throws, because the diff() is the same:

> setNames(apply(m2, 1, iglTzm, 0:6), 1:6)
  1   2   3   4   5   6 
 15   2  -2   7 -52   0 

Anyway, because I can't simply scale or invert it, I haven't had a brilliant idea yet how to adapt the function to get the right signs, which are assumingly:

#  1   2   3   4   5   6 
# 15  -2   2  -7 -52   0

Does anyone know how to adapt iglTzm() to get the integrals with the correct sign?

The plot of m2 should illustrate the principle a bit more:

---

data

m1 <- matrix(c(6, 7, 8, 8, 6, 8, 10,
                9, 9, 8, 9, 9, 8, 9,
                9, 10, 10, 9, 9, 9, 9,
                9, 8, 8, 8, 6, 8, 9, 
                10, 10, 10, 9, 10, 8, 0, 
                9, 8, 9, 10, 9, 9, 9), 6, byrow=TRUE)

m2 <- t(apply(m1, 1, function(x) scale(x, center=x[1], scale=FALSE)))

# plot
par(mfrow=c(2, 3))
lapply(1:nrow(m2), function(x) {
  plot(m2[x, ], type="l", main=x)
  abline(h=m2[x, 1], col="red", lty=2)
})

Answer 1

Firstly, there is another small but more important issue, although after fixing it your question would still remain valid. What I mean is that the order of x and y as arguments of your function should be reversed due to how you use the function in apply .

But that is not enough and now we come back to your question. For that, let us recall the usual integration: ʃf(x)dx (with limits from a to b) would integrate the area below f, which is what your function already successfully does. Now what you want is to adjust its level. But if we integrate from a to b, that is the same as ʃ(f(x)-f(a))dx = ʃf(x)dx - (ba)f(a), which leads to

iglTzm <- function(y, x) sum(diff(x) * (head(y, -1) + tail(y, -1))) / 2 - y[1] * (max(x) - min(x))
setNames(apply(m1, 1, iglTzm, 0:6), 1:6)
#  1  2  3  4  5  6 
#  9 -2  2 -7 -8  0 

It happens that only two absolute values are different from the version where x and y are reversed. Bet let's look at the first function: should it be 9 or 15? We have that 2*2/2 + 1*2 + 1*2/2 + 2*4/2 = 9, so indeed we want to reverse x and y .

Another way to write the function would be

iglTzm <- function(y, x) sum(diff(x) * (head(y - y[1], -1) + tail(y - y[1], -1))) / 2
setNames(apply(m1, 1, iglTzm, 0:6), 1:6)
#  1  2  3  4  5  6 
#  9 -2  2 -7 -8  0 

Edit : by reversing I only meant the order in the function definition or how you use it in apply ; the function itself in terms of y (function values) and x (grid values) is fine.