I'm trying to implement a lag function but it seems i need an existing x column for it to work
lets say i have this data frames
df <- data.frame(AgeGroup=c("0-4", "5-39", "40-59","60-69","70 "),
px=c(.99, .97, .95, .96,.94))
i want a column Ix that is lag(Ix)*lag(px) starting from 1000.
The data i want is
df2 <- data.frame(AgeGroup=c("0-4", "5-39", "40-59","60-69","70 "),
px=c(.99, .97, .95, .96, .94),
Ix=c(1000, 990, 960.3, 912.285, 875.7936))
I've tried
library(dplyr)
df2<-mutate(df,Ix = lag(Ix, default = 1000)*lag(px))
ifelse statements don't work after the creation of a reference
df$Ix2=NA
df[1,3]=1000
df$Ix<-ifelse(df[,3]==1000,1000,
lag(df$Ix, default = 1000)*lag(px,default =1))
and have been playing around with creating separate Ix column with Ix=1000 then run the above but it doesn't seem to work. Does anyone have any ideas how i can create the x 1 column?
CodePudding user response:
You could use cumprod()
combined with dplyr::lag()
for this:
> df$Ix <- 1000*dplyr::lag(cumprod(df$px), default = 1)
> df
AgeGroup px Ix
1 0-4 0.99 1000.0000
2 5-39 0.97 990.0000
3 40-59 0.95 960.3000
4 60-69 0.96 912.2850
5 70 0.94 875.7936
CodePudding user response:
You can also use accumulate
from purrr
. Using head(px, -1)
includes all values in px
except the last one, and the initial Ix
is set to 1000.
library(tidyverse)
df %>%
mutate(Ix = accumulate(head(px, -1), prod, .init = 1000))
Output
AgeGroup px Ix
1 0-4 0.99 1000.0000
2 5-39 0.97 990.0000
3 40-59 0.95 960.3000
4 60-69 0.96 912.2850
5 70 0.94 875.7936