R Create new column and shift values 1 row - r

Question

I want to create a new column and then I want to have the values of the following day in the following row. In my example-dataframe I have 3 columns: date, price and the return. Now I want to detect overreactions. If the returns are higher than the mean + 1 standard deviation, then the return is an overreaction. If not, then the value is 'NA'.
library(tidyverse)
library(quantmod)
df <- tibble(
date = lubridate::today() +0:9,
price = c(1,2.5,2,3,5,6.5,4,9,3,4))
df <- mutate(df, return = Delt(price))
df <- df %>% mutate(overreaction=
ifelse(return > mean(df$return, na.rm = TRUE) + sd(df$return, na.rm = TRUE),
yes = return, no = NA
)
)
Now I'm creating a new column, that gives me the return of the following day, if an overreaction took place at the previous day.
df <- df %>% mutate(following_day =
ifelse(overreaction != "NA",
yes= return%>% data.table::shift(n=1L, fill=NA, type=c("lead")),
no=NA)
)
print(df)
# A tibble: 10 x 5
date price return overreaction following_day
<date> <dbl> <dbl> <dbl> <dbl>
1 2019-02-04 1 NA NA NA
2 2019-02-05 2.5 1.5 1.5 -0.200
3 2019-02-06 2 -0.200 NA NA
4 2019-02-07 3 0.5 NA NA
5 2019-02-08 5 0.667 NA NA
6 2019-02-09 6.5 0.3 NA NA
7 2019-02-10 4 -0.385 NA NA
8 2019-02-11 9 1.25 1.25 -0.667
9 2019-02-12 3 -0.667 NA NA
10 2019-02-13 4 0.333 NA NA
And it works except for one problem:
I want that the values in the following_day-column are shiftetd by 1 row, so that they are in the original position.
This is how the dataframe should look like:
# A tibble: 10 x 5
date price return overreaction following_day
<date> <dbl> <dbl> <dbl> <dbl>
1 2019-02-04 1 NA NA NA
2 2019-02-05 2.5 1.5 1.5 NA
3 2019-02-06 2 -0.200 NA -0.200
4 2019-02-07 3 0.5 NA NA
5 2019-02-08 5 0.667 NA NA
6 2019-02-09 6.5 0.3 NA NA
7 2019-02-10 4 -0.385 NA NA
8 2019-02-11 9 1.25 1.25 NA
9 2019-02-12 3 -0.667 NA -0.667
10 2019-02-13 4 0.333 NA NA
Can someone help me?

Answer 1

Enclose df$following_day in dplyr::lag:
library(tidyverse)
library(quantmod)
df <- tibble(
date = lubridate::today() +0:9,
price = c(1,2.5,2,3,5,6.5,4,9,3,4)) %>%
mutate(return= Delt(price))
df <- mutate(df, overreaction =
ifelse( return > mean(df$return, na.rm = TRUE) + sd(df$return, na.rm = TRUE),
return, NA))
df <- mutate(df, following_day = ifelse(!is.na(overreaction),
data.table::shift(df$return, type = "lead"),
NA))
df$following_day <- dplyr::lag(df$following_day) # or data.table::shift
Output:
> df
# A tibble: 10 x 5
date price return overreaction following_day
<date> <dbl> <dbl> <dbl> <dbl>
1 2019-02-04 1 NA NA NA
2 2019-02-05 2.5 1.5 1.5 NA
3 2019-02-06 2 -0.200 NA -0.200
4 2019-02-07 3 0.5 NA NA
5 2019-02-08 5 0.667 NA NA
6 2019-02-09 6.5 0.3 NA NA
7 2019-02-10 4 -0.385 NA NA
8 2019-02-11 9 1.25 1.25 NA
9 2019-02-12 3 -0.667 NA -0.667
10 2019-02-13 4 0.333 NA NA
The same can be achieved by subbing dplyr::lag with data.table::shift(df$following_day, type = "lag")