在构建分数多项式函数时避免使用eval(parse())

Question

我的目标是在R中编写一个接受分数多项式(FP)系数的函数,并返回一个矢量化函数,该函数评估给定输入数的指定FP.FP定义有两个重要规则:

• x ^ 0定义为log(x)
• 幂可以有多个系数,其中幂p的第二个系数将log(x)的因子加到加法项(x ^ p*log(x)),第三个加上log(x)^ 2(x)因子^ p*log(x)^ 2),依此类推

我下面的当前解决方案将FP函数构建为字符串,解析字符串并返回一个计算表达式的函数.我的问题是,是否有更好/更快的方法可以避免eval(parse())- 可能使用一些substitute()魔法.

该函数必须处理每个电源的系数数量事先未知,但在被调用时指定.最终的FP评估需要很快,因为它经常被调用.

这将是很好不限于标准功率-2,-1,-0.5,0,0.5%,1,2,3,理想情况下,所需的功能会做两个步骤在一次:接受FP-系数以及数字向量并返回输入的FP值,同时仍然很快.

getFP <- function(p_2, p_1, p_0.5, p0, p0.5, p1, p2, p3, ...) {
    p <- as.list(match.call(expand.dots=TRUE)[-1])         # all args
    names(p) <- sub("^p", "", names(p))     # strip "p" from arg names
    names(p) <- sub("_", "-", names(p))     # replace _ by - in arg names

    ## for one power and the i-th coefficient: build string
    getCoefStr <- function(i, pow, coefs) {
        powBT  <- ifelse(as.numeric(pow), paste0("x^(", pow, ")"), "log(x)")
        logFac <- ifelse(i-1,             paste0("*log(x)^", i-1), "")
        paste0("(", coefs[i], ")*", powBT, logFac)
    }

    onePwrStr <- function(pow, p) { # for one power: build string for all coefs
        coefs  <- eval(p[[pow]])
        pwrStr <- sapply(seq(along=coefs), getCoefStr, pow, coefs)
        paste(pwrStr, collapse=" + ")
    }

    allPwrs <- sapply(names(p), onePwrStr, p)  # for each power: build string
    fpExpr  <- parse(text=paste(allPwrs, collapse=" + "))
    function(x) { eval(fpExpr) }
}

一个例子是-1.5*x^(-1) - 14*log(x) - 13*x^(0.5) + 6*x^0.5*log(x) + 1*x^3指定了具有系数(-1.5,-14,-13,6,1)的幂(-1,0,0.5,0.5,3).

> fp <- getFP(p_1=-1.5, p0=-14, p0.5=c(-13, 6), p3=1)
> fp(1:3)
[1] -13.50000000 -14.95728798   0.01988127

Answer 1

首先,我们创建一个将在序列中生成单个术语的函数

one <- function(p, c = 1, repeated = 1) {
  if (p == 0) {
    lhs <- substitute(c * log(x), list(c = c))
  } else {
    lhs <- substitute(c * x ^ p, list(c = c, p = p))
  }

  if (repeated == 1) return(lhs)
  substitute(lhs * log(x) ^ pow, list(lhs = lhs, pow = repeated - 1))
}
one(0)
# 1 * log(x)
one(2)
# 1 * x^2

one(2, 2)
# 2 * x^2

one(2, r = 2)
# 1 * x ^ 2 * log(x)^1
one(2, r = 3)
# 1 * x ^ 2 * log(x)^2

这里的关键工具substitute()是这里解释的.

接下来我们编写一个将两个术语加在一起的函数.再次使用替代品:

add_expr_1 <- function(x, y) {
  substitute(x + y, list(x = x, y = y))
}

add_expr_1(one(0, 1), one(2, 1))

我们可以使用它来创建一个函数来添加任意数量的术语:

add_expr <- function(x) Reduce(add_expr_1, x)
add_expr(list(one(0, 1), one(1, 1), one(2, 3)))

有了这些部分,最终的功能很简单 - 我们计算出代表的数量,然后用来为每个组合Map()调用one()一次powers,coefs并且reps:

fp <- function(powers, coefs) {
  # Determine number of times each power is repeated. This is too
  # clever approach but I think it works
  reps <- ave(powers, powers, FUN = seq_along)

  # Now generate a list of expressions using one
  components <- Map(one, powers, coefs, reps)

  # And combine them together with plus
  add_expr(components)
}

powers <- c(-1, 0, 0.5, 0.5, 3)
coefs <-  c(-1.5, -14, -13, 6, 1)
fp(powers, coefs)
# -1.5 * x^-1 + -14 * log(x) + -13 * x^0.5 + 6 * x^0.5 * log(x)^1 + 
#   1 * x^3