Dear Axel,
Assuming that you're not wedded to using mutate():
> D1 <- 1 - as.matrix(sim_data_wide[, 2:11])
> D2 <- matrix(0, 10, 10)
> colnames(D2) <- paste0("PC_", 1:10)
> for (i in 1:10) D2[, i] <- 1 - apply(D1[, 1:i, drop=FALSE], 1, prod)
> all.equal(D2, as.matrix(sim_data_wide[, 22:31]))
[1] TRUE
I hope this helps,
John
> On May 9, 2020, at 7:45 PM, Axel Urbiz <axel.urbiz at gmail.com>
wrote:
>
> Hello,
>
> Is there a less verbose approach to obtaining the PC_i variables inside the
mutate?
>
> library(tidyverse)
> sim_data <- data.frame(borrower_id = sort(rep(1:10, 20)),
> quarter = rep(1:20, 10),
> pd = runif(length(rep(1:20, 10)))) # conditional
probs
>
> sim_data_wide <- tidyr::spread(sim_data, quarter, pd)
> colnames(sim_data_wide)[-1] <- paste0("P_",
colnames(sim_data_wide)[-1])
>
> # Compute cumulative probs
> sim_data_wide <- sim_data_wide %>%
> mutate(PC_1 = P_1,
> PC_2 = 1-(1-P_1)*(1-P_2),
> PC_3 = 1-(1-P_1)*(1-P_2)*(1-P_3),
> PC_4 = 1-(1-P_1)*(1-P_2)*(1-P_3)*(1-P_4),
> PC_5 = 1-(1-P_1)*(1-P_2)*(1-P_3)*(1-P_4)*(1-P_5),
> PC_6 =
1-(1-P_1)*(1-P_2)*(1-P_3)*(1-P_4)*(1-P_5)*(1-P_6),
> PC_7 =
1-(1-P_1)*(1-P_2)*(1-P_3)*(1-P_4)*(1-P_5)*(1-P_6)*(1-P_7),
> PC_8 =
1-(1-P_1)*(1-P_2)*(1-P_3)*(1-P_4)*(1-P_5)*(1-P_6)*(1-P_7)*(1-P_8),
> PC_9 =
1-(1-P_1)*(1-P_2)*(1-P_3)*(1-P_4)*(1-P_5)*(1-P_6)*(1-P_7)*(1-P_8)*(1-P_9),
> PC_10 =
1-(1-P_1)*(1-P_2)*(1-P_3)*(1-P_4)*(1-P_5)*(1-P_6)*(1-P_7)*(1-P_8)*(1-P_9)*(1-P_10)
> )
>
>
> Thanks,
> Axel.
> [[alternative HTML version deleted]]
>
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