R中电流for回路有哪些优化方法？

我有下面的代码，我想优化，但我目前不知道如何才能做到这一点。首先，让我给你介绍一下这个问题

test.data

包含大约200000行，这使得下面的实现在R中非常缓慢。我尝试做的第一件事是优化函数并尽可能多地删除测试（

if

语句），但是在下面的代码中有两个实例我无法做到这一点

library(data.table)
test.data <- data.table(person = c("A", "B", "C"),
                        duration = c(120,50,30),
                        time = c(159, 231, 312),
                        savings = c(140000, 200000, 300000),
                        ren = c(0.0037, 0.0011, 0.0015),
                        res = c(55, 10, 30))

set.seed(35)

# Deduction series, note that in this example, they are arbitrary.
# They do not follow a pattern. I believe, this is the core of the problem.
# Which makes it extremely difficult to vectorise, since this would result in
# no closed solution.
c_a <- round(runif(max(test.data$duration)), 2) / 10

# Put in as a constant, but it can vary arbitrary.
c_b <- rep(round((8.5 / 12)/100, digits = 4), max(test.data$duration))
rnew <- 0.25
result <- matrix(0, nrow = 6, ncol = 120)

for(j in 1:nrow(test.data)){
  savings <- test.data$savings[j]
  duration <- test.data$duration[j]
  time <- test.data$time[j]
  res <- test.data$res[j]
  m <- matrix(nrow = 6, ncol = duration)

for(i in 1:duration){
  m[1,i] <- ifelse(i == 1, savings, m[6, i-1])

  m[2,i] <- -m[1,i] * c_a[i]

  m[3,i] <- -(m[1,i] + m[2,i]) * c_b[i]

  m[4,i] <- ifelse(i == duration, -(m[1,i] + m[2,i] + m[3,i]), -(m[1,i] + m[2,i]) / (time + 1 - i))

  if(i == res & res < time){
    m[5, i] <- -(m[1,i] + m[2,i]) * (1 - rnew)
  } else {
    m[5, i] <- 0
  }

  m[6, i] <- m[1,i] + m[2,i] + m[3,i] + m[4,i] + m[5,i]
}

  m <- cbind(m, matrix(0, ncol = ncol(result) - ncol(m), nrow = nrow(result)))

  result <- matrix(mapply(sum, result, m, MoreArgs=list(na.rm=T)),ncol=ncol(result))
}

库（data.table）
test.data我唯一能建议的是只初始化
m
一次，其维度与
结果
相同，
并更换外环路的最后2条线路，如下所示。
这将避免重新分配*
m
和使用
mapply
完成的元素方向的
sum

result <- matrix(0, nrow = 6, ncol = 120)
m <- result

for (j in 1:nrow(test.data)) {
  savings <- test.data$savings[j]
  duration <- test.data$duration[j]
  time <- test.data$time[j]
  res <- test.data$res[j]

  for (i in 1:duration) {
    m[1,i] <- ifelse(i == 1, savings, m[6, i-1])

    m[2,i] <- -m[1,i] * c_a[i]

    m[3,i] <- -(m[1,i] + m[2,i]) * c_b[i]

    m[4,i] <- ifelse(i == duration, -(m[1,i] + m[2,i] + m[3,i]), -(m[1,i] + m[2,i]) / (time + 1 - i))

    if (i == res & res < time) {
      m[5, i] <- -(m[1,i] + m[2,i]) * (1 - rnew)
    } else {
      m[5, i] <- 0
    }

    m[6, i] <- m[1,i] + m[2,i] + m[3,i] + m[4,i] + m[5,i]
  }

  result[, 1:duration] <- result[, 1:duration] + m[, 1:duration]
}

result计算未付金额总和的可能方法（即结果中的OPs第1行）。如果需要，可以轻松计算所有中间值（
m[2，j]
，
m[3，j]
，
m[4，j]
，
m[5，j]
）。注意：我没有用实际的dim计时

library(data.table)

calcAmor <- function(ca, cb, rnew, dur, S0, tau, res) {
    amortize <- function(S, ca.t) S - ca.t[1L]*S - (1-ca.t[1L])*cb*S - (S - ca.t[1L]*S) / (tau + 1 - ca.t[2L])

    ans <- Reduce(amortize,
        split(cbind(ca, seq_along(ca)), seq_along(ca)),
        init=S0,
        accumulate=TRUE)[-(dur+1L)]

    ix <- min(res+1L, dur):dur
    tmp <- Reduce(amortize,
        split(cbind(ca[ix], ix), seq_along(ix)),
        init=amortize(ans[res], c(ca[res], res)) - (ans[res] - ans[res]*ca[res])*(1-rnew),
        accumulate=TRUE)
    ans[ix] <- tmp[-length(tmp)]    

    ans
}

set.seed(35)
test.data <- data.table(person = c("A", "B", "C"),
    duration = c(120,50,30),
    time = c(159, 231, 312),
    savings = c(140000, 200000, 300000),
    res = c(55, 10, 30))
maxd <- test.data[, max(duration)]
c_a <- round(runif(maxd), 2) / 10
rnew <- 0.25
cb <- round((8.5 / 12)/100, digits = 4)

test.data[, .(
        dur=seq_len(duration),
        S=calcAmor(ca=c_a[seq_len(duration)], cb, rnew, dur=duration, S0=savings, tau=time, res=res)),
    by=.(person)][, sum(S), by=.(dur)]

另一种可能是将test.data转换为矩阵（删除不使用的第一列）。矩阵在循环中的处理速度更快，但您必须重写代码以指定列号（或名称，例如test.mat[，“duration”]）。
     dur           V1
  1:   1 6.400000e+05
  2:   2 5.783318e+05
  3:   3 5.711966e+05
  4:   4 5.336450e+05
  5:   5 4.774502e+05
 ---                 
116: 116 7.075169e+00
117: 117 6.788631e+00
118: 118 6.339002e+00
119: 119 5.639335e+00
120: 120 5.297898e+00