遇到0时重置的累计和

Question

我想对字段进行累积求和,但只要遇到0就重置聚合值.

这是我想要的一个例子:

data.frame(campaign = letters[1:4] , 
       date=c("jan","feb","march","april"),
       b = c(1,0,1,1) ,
       whatiwant = c(1,0,1,2)
       )

 campaign  date b whatiwant
1        a   jan 1         1
2        b   feb 0         0
3        c march 1         1
4        d april 1         2

Answer 1

另一个基础就是

with(df, ave(b, cumsum(b == 0), FUN = cumsum))
## [1] 1 0 1 2

这将b根据0外观将列分组,并计算b每个组的累积总和

使用最新data.table版本的另一种解决方案(v 1.9.6+)

library(data.table) ## v 1.9.6+
setDT(df)[, whatiwant := cumsum(b), by = rleid(b == 0L)]
#    campaign  date b whatiwant
# 1:        a   jan 1         1
# 2:        b   feb 0         0
# 3:        c march 1         1
# 4:        d april 1         2

每条评论都有一些基准

set.seed(123)
x <- sample(0:1e3, 1e7, replace = TRUE)
system.time(res1 <- ave(x, cumsum(x == 0), FUN = cumsum))
# user  system elapsed 
# 1.54    0.24    1.81 
system.time(res2 <- Reduce(function(x, y) if (y == 0) 0 else x+y, x, accumulate=TRUE))
# user  system elapsed 
# 33.94    0.39   34.85 
library(data.table)
system.time(res3 <- data.table(x)[, whatiwant := cumsum(x), by = rleid(x == 0L)])
# user  system elapsed 
# 0.20    0.00    0.21 

identical(res1, as.integer(res2))
## [1] TRUE
identical(res1, res3$whatiwant)
## [1] TRUE

Answer 2

另一个迟到的想法

ff = function(x)
{
    cs = cumsum(x)
    cs - cummax((x == 0) * cs)
}
ff(c(0, 1, 3, 0, 0, 5, 2))
#[1] 0 1 4 0 0 5 7

并比较:

library(data.table)
ffdt = function(x) 
    data.table(x)[, whatiwant := cumsum(x), by = rleid(x == 0L)]$whatiwant

x = as.numeric(x) ##because 'cumsum' causes integer overflow
identical(ff(x), ffdt(x))
#[1] TRUE
microbenchmark::microbenchmark(ff(x), ffdt(x), times = 25)
#Unit: milliseconds
#    expr      min       lq   median       uq      max neval
#   ff(x) 315.8010 362.1089 372.1273 386.3892 405.5218    25
# ffdt(x) 374.6315 407.2754 417.6675 447.8305 534.8153    25

Answer 3

您可以将该Reduce函数与自定义函数一起使用,当遇到的新值为0时返回0,否则将新值添加到累计值:

Reduce(function(x, y) if (y == 0) 0 else x+y, c(1, 0, 1, 1), accumulate=TRUE)
# [1] 1 0 1 2