I have a dataframe
Date repair
<date> <dbl>
2018-07-01 4420
2018-07-02 NA
2018-07-03 NA
2018-07-04 NA
2018-07-05 NA
Where 4420 is time in minutes. I'm trying to get this:
Date repair
<date> <dbl>
2018-07-01 1440
2018-07-02 1440
2018-07-03 1440
2018-07-04 100
2018-07-05 NA
Where 1440 - minutes in one day and 100 what is left. I made it with loop. Can this be achieved in a more elegant way?
With dplyr
:
library(dplyr)
df %>%
mutate(
repair = c(rep(1440, floor(repair[1] / 1440)),
repair[1] %% 1440,
rep(NA, n() - length(c(rep(1440, floor(repair[1] / 1440)), repair[1] %% 1440))))
)
Output:
Date repair
1 2018-07-01 1440
2 2018-07-02 1440
3 2018-07-03 1440
4 2018-07-04 100
5 2018-07-05 NA
You could write a little function for that task
f <- function(x, y, length_out) {
remainder <- x %% y
if(remainder == 0) {
`length<-`(rep(y, x %/% y), length_out)
} else {
`length<-`(c(rep(y, x %/% y), remainder), length_out)
}
}
Input
x <- 4420
y <- 24 * 60
Result
f(x, y, length_out = 10)
# [1] 1440 1440 1440 100 NA NA NA NA NA NA
length_out
should probably be equal to nrow(your_data)
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