The total amount with conditions in R

Question

Say I have

a <- c(0, 22, 0, 2, 0, 0, 20, 20, 20, 0, 0)

I want to make a cumulative amount in which I minus 5for each value in a, and then add the previous value.

However, I also have the condition that if ait gets smaller 0, I want cumsum to be 0, and if ait gets bigger 40, for cumsum to 40.

So I want to get

(0, 17, 12, 9, 4, 0, 15, 30, 40, 35, 30)

Can anyone help? I’ve tried it for hours now for several hours!

@Holger, this method does not always work. Therefore, if I add a few extra zeros, it will not have the right solution

a <- c(0, 22, 0, 2, 0, 0, 0, 0, 20, 20, 20, 0, 0)

gives

 0 17 12  9  4  0  1  7 22 37 52 47 42

Answer 1

Here are a few alternatives:

1) Loop :

b <- a; for(i in seq_along(b)[-1]) b[i] <- min(40, max(0, a[i] - 5 + b[i-1]))
b
## [1]  0 17 12  9  4  0 15 30 40 35 30

2)

f <- function(b, a) min(40, max(0, a - 5 + b))
Reduce(f, a, acc = TRUE)
## [1]  0 17 12  9  4  0 15 30 40 35 30

3) . , .

rec <- function(a) {
   n <- length(a)
   if (n <= 1) a
   else {
     rec.hd <- Recall(a[-n])
     c(rec.hd, min(40, max(0, rec.hd[n-1] + a[n] - 5)))
   }
}
rec(a)
## [1]  0 17 12  9  4  0 15 30 40 35 30

Answer 2

Try

cumsum_up_low <- function(a, d=5, up=40, low=0 ){
  out = rep(0, length(a))
  out[1] = a[1]*(a[1]>=0 && a[1]<=40) + 0*(a[1]<0) + 40*(a[1] > 40)
  for(i in 2:length(a)){
    if(out[i-1] + a[i] - d > low && out[i-1] + a[i] - d < up){
      out[i] = out[i-1] + a[i] - d
    } else if(out[i-1] + a[i] - d <= low){
      out[i] = 0          
    } else out[i] = 40
  }
  out
}

cumsum_up_low(a, d=5, up=40, low=0)
# [1]  0 17 12  9  4  0 15 30 40 35 30

a <- sample(a, 1e6, TRUE)
system.time(cumsum_up_low(a))
#   user  system elapsed 
#   3.59    0.00    3.59 
library(compiler)
cumsum_up_low_compiled <- cmpfun(cumsum_up_low)
system.time(cumsum_up_low_compiled(a))
#   user  system elapsed 
#   0.28    0.00    0.28 

library(Rcpp)
cppFunction('
NumericVector cumsum_up_low_cpp(NumericVector a, double d, double up, double low) {
  NumericVector out(a.size());
  out[0] = a[0];
  for(int i=1; i<a.size(); i++){
    if(out[i-1] + a[i] - d > low & out[i-1] + a[i] - d < up){
      out[i] = out[i-1] + a[i] - d;
    } else if(out[i-1] + a[i] - d <= low){
      out[i] = 0;          
    } else out[i] = 40;
  }
  return out;
}')

a <- sample(a, 5e6, replace = TRUE)
system.time(cumsum_up_low_compiled(a, d=5, up=40, low=0))
#   user  system elapsed 
#   1.45    0.00    1.46 
system.time(cumsum_up_low_cpp(a, d=5, up=40, low=0))
#   user  system elapsed 
#   0.04    0.02    0.05 

Answer 3

, :

a <- c(0, 22, 0, 2, 0, 0, 20, 20, 20, 0, 0)

## Initialize another vector just like a
c <- a


## Do it easy-to-understand'ly in a for loop:
for (i in seq_along(a)){
  b <- a[i]
  if (i>1) {
    b <- b+c[i-1]
    b <- b-5
  } 
  if (b<0) b <- 0
  if (b>40) b <- 40
  c[i] <- b
  print(c[i])
}

, , !

Answer 4

Reduce max min pmin pmax, .

, 0 40 . .

:

Reduce(function(x, y) min(max(x + y - 5, 0), 40), a, 0, accumulate=TRUE)
[1]  0  0 17 12  9  4  0 15 30 40 35 30

pmin(pmax(Reduce(function(x, y) x + y - 5, a, 0, accumulate=TRUE), 0), 40)
[1]  0  0 12  7  4  0  0  9 24 39 34 29