Question

我有两个向量：

x <- c(-2.0,-1.75,-1.50,-1.25,-1.00,-0.75,-0.50,-0.25,0.00,0.25,0.50,0.75,1.00,1.25,1.50,1.75,2.00,2.25,2.50,2.75)
y <- c(37.0000,24.1602,15.06250,8.91016,5.00000,2.72266,1.56250,1.09766,1.00000,1.03516,1.06250,1.03516,1.00000,1.09766,1.56250,2.72266,5.00000,8.91016,15.06250,24.16016)

我正在尝试创建一个从向量x给出数字的函数，它返回相应的y值（相同的索引）。例如，func(-2.0)应返回37.0000。

目前我有这个超级丑陋的功能，我认为这不是我应该做的：

func1 <- function(x) {
  if (x==-2.0) {return (37.0000)}
  else if (x==-1.75){return (24.1602)}
  else if (x==-1.50){return (15.06250)}
  else if (x==-1.25){return (8.91016)}
  else if (x==-1.00){return (5.00000)}
  else if (x==-0.75){return (2.72266)}
  else if (x==-0.50){return (1.56250)}
  else if (x==-0.25){return (1.09766)}
  else if (x==0.00){return (1.00000)}
  else if (x==0.25){return (1.03516)}
  else if (x==0.50){return (1.06250)}
  else if (x==0.75){return (1.03516)}
  else if (x==1.00){return (1.00000)}
  else if (x==1.25){return (1.09766)}
  else if (x==1.50){return (1.56250)}
  else if (x==1.75){return (2.72266)}
  else if (x==2.00){return (5.00000)}
  else if (x==2.25){return (8.91016)}
  else if (x==2.50){return (15.06250)}
  else if (x==2.75){return (24.16016)}
  else {return (Inf)}
}

Answer 1

For exact matching:

foo = function(u) {res=y[pmatch(u,x)];ifelse(is.na(res), Inf, res)}
#> foo(-2)
#[1] 37
#> foo(-1.8)
#[1] Inf
#> foo(-4)
#[1] Inf

Not sure about what you need but beware you can use linear interpolation (you can put method as constant instead of linear):

foo = approxfun(x,y, yleft=Inf, yright=Inf)
#> foo(-2)
#[1] 37
#> foo(-1.8)
#[1] 26.72816
#> foo(-4)
#[1] Inf

In this last case, the value is not Inf in the boundary domain defined by x.

Answer 2

You seem to be doing an interpolation. The R function for interpolation is approx().

approx(x, y, xout = -2)

$x
[1] -2

$y
[1] 37

In practise it's easier to create an interpolation function using approxfun(). Try this:

foo <- approxfun(x, y)
foo(-2)
[1] 37

You should probably avoid using an exact matching strategy using == or match(). The reason is simple - if you use calculated values to find the index position, you could find that the match is inexact.

Compare:

y[ which(x == -2.0) ]
[1] 37

y[ which(x == -2.00000000001) ]
numeric(0)

Similarly:

y[match(-2.0, x)]
[1] 37

y[match(-2.0000000000001, x)]
[1] NA

Answer 3

Since x and y are the same length, put x as names of y

names(y)<-x
y

      -2    -1.75     -1.5    -1.25       -1    -0.75     -0.5    -0.25        0     0.25      0.5     0.75        1     1.25      1.5     1.75        2 
37.00000 24.16020 15.06250  8.91016  5.00000  2.72266  1.56250  1.09766  1.00000  1.03516  1.06250  1.03516  1.00000  1.09766  1.56250  2.72266  5.00000 
    2.25      2.5     2.75 
 8.91016 15.06250 24.16016

This way, you can call by the names, e.g.

y["-2"]

-2 
37

y["-1.75"]

-1.75 
24.1602

Answer 4

You don't really need a function for this, I suggest you just use:

y[x == -1.75]

x == -1.75 returns a boolean vector so it will pick the right value for y.

If you really want a function:

f <- function(x,y,xi){
return(y[x == xi])
}

创建从向量到另一个向量的函数

4 个答案: