(1-previous_record)*current_record 的累积积

Question

数据框包含两个变量 ( timeand rate) 和 10 个观测值

time <- seq(1:10) 
rate <- 1-(0.99^time)
dat <- data.frame(time, rate)

我需要添加一个新列（称为new_rate）。

new_rate 定义如下

注意：new_rate_1是new列new_rate等的第一次观察。

new_rate_1 = rate_1
new_rate_2 = (1-rate_1)*rate_2
new_rate_3 = (1-rate_1)*(1-rate_2)*rate_3
new_rate_4 = (1-rate_1)*(1-rate_2)*(1-rate_3)*rate_4
...
new_rate_10 = (1-rate_1)*(1-rate_2)*(1-rate_3)*(1-rate_4)*(1-rate_5)*(1-rate_6)*(1-rate_7)*(1-rate_8)*(1-rate_9)*rate_10

如何在基础 R或dplyr?

Answer 1

cumprod救援（给@Cole 的帽子提示以简化代码）：

dat$rate * c(1, cumprod(1 - head(dat$rate, -1)))

逻辑是你本质上是在做一个 的cumulative product 1 - dat$rate，乘以当前步骤。
第一步，您可以只保留现有值，但随后您需要偏移两个向量，以便乘法得到所需的结果。

证明：

out <- c(
dat$rate[1],
(1-dat$rate[1])*dat$rate[2],
(1-dat$rate[1])*(1-dat$rate[2])*dat$rate[3],
(1-dat$rate[1])*(1-dat$rate[2])*(1-dat$rate[3])*dat$rate[4],
(1-dat$rate[1])*(1-dat$rate[2])*(1-dat$rate[3])*(1-dat$rate[4])*dat$rate[5],
(1-dat$rate[1])*(1-dat$rate[2])*(1-dat$rate[3])*(1-dat$rate[4])*(1-dat$rate[5])*dat$rate[6],
(1-dat$rate[1])*(1-dat$rate[2])*(1-dat$rate[3])*(1-dat$rate[4])*(1-dat$rate[5])*(1-dat$rate[6])*dat$rate[7],
(1-dat$rate[1])*(1-dat$rate[2])*(1-dat$rate[3])*(1-dat$rate[4])*(1-dat$rate[5])*(1-dat$rate[6])*(1-dat$rate[7])*dat$rate[8],
(1-dat$rate[1])*(1-dat$rate[2])*(1-dat$rate[3])*(1-dat$rate[4])*(1-dat$rate[5])*(1-dat$rate[6])*(1-dat$rate[7])*(1-dat$rate[8])*dat$rate[9],
(1-dat$rate[1])*(1-dat$rate[2])*(1-dat$rate[3])*(1-dat$rate[4])*(1-dat$rate[5])*(1-dat$rate[6])*(1-dat$rate[7])*(1-dat$rate[8])*(1-dat$rate[9])*dat$rate[10]
)

all.equal(
  dat$rate * c(1, cumprod(1 - head(dat$rate, -1))),
  out
)
#[1] TRUE