Mutate using ifelse with a condition in R

user_16249516 0
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I want to create data frame for temperature

Time frame : 2020-01-01 ~ 2020-12-31
Temperature :
2020-01-01 ~ 2020-02-29 and 2020-10-01 ~ 2020-12-31 : random integer between 10 ~ 24
2020-03-01 ~ 2020-05-31 : random integer between 5 ~ 17
2020-06-01 ~ 2020-09-31 : random integer between 1 ~ 8

my code is

library(tidyverse)
library(lubridate)
library(summarytools)
library(dplyr)

start <- as.POSIXct("2020-01-01", "%Y-%m-%d", tz = "UTC")
end   <- as.POSIXct("2020-12-31", "%Y-%m-%d", tz = "UTC")

remT <- seq(start, end, by = "1 day") 
date_df <- as.data.frame(remT)
date_df <- setNames(date_df, c("Date")) 
date_df <- date_df %>% arrange(Date)

cond_1 <- for(i in date_df$Date){(i >= as.Date('2020-01-01') && i <= as.Date('2020-02-29')) || (i >= as.Date('2020-10-01') && i <= as.Date('2020-12-31'))}
cond_2 <- for(j in date_df$Date)(j >= as.Date('2020-06-01') && j <= as.Date('2020-09-30'))

x <- sample(10:24, 152, replace=TRUE)
y <- sample(1:8, 122, replace=TRUE)
z <- sample(5:17, 92, replace=TRUE)


date_df <- date_df %>%
  mutate(Test = 
           ifelse(cond_1, x,
                     ifelse(cond_2, y , z)
                  )
         )

However, this code returned the error message unfortunately

Error: Problem with mutate() column Test. ℹ Test = ifelse(cond_1, x, ifelse(cond_2, y, z)). ℹ Test must be size 366 or 1, not 0.

How to solve this error?

dplyr case-when if-statement r
2 Answers

mutate needs either 1 or a complete set of values. case_when or ifelse or if_else will take care where to replace values according to given condition. So sample 366 values for all conditions.

Best way is to use n() in place of any value. Moreover, you may do this in one single pipe

library(tidyverse)
library(lubridate, warn.conflicts = F)

set.seed(123)
seq.Date(as.Date('2020-01-01'), as.Date('2020-12-31'), by = 1) %>%
  as.data.frame() %>% setNames('Date') %>%
  mutate(temp = case_when(month(Date) %in% c(1:2, 10:12)~ sample(10:24, n(), T),
                          month(Date) %in% c(3:5) ~ sample(5:17, n(), T),
                          TRUE ~ sample(1:8, n(), T))) -> df

head(df, 5)
#>         Date temp
#> 1 2020-01-01   24
#> 2 2020-01-02   24
#> 3 2020-01-03   12
#> 4 2020-01-04   23
#> 5 2020-01-05   12

tail(df, 5)
#>           Date temp
#> 362 2020-12-27   15
#> 363 2020-12-28   19
#> 364 2020-12-29   19
#> 365 2020-12-30   15
#> 366 2020-12-31   21

Created on 2021-06-17 by the reprex package (v2.0.0)

user_28848590

You can use sapply with switch and sample:

library(lubridate)
start <- as.POSIXct("2020-01-01", "%Y-%m-%d", tz = "UTC")
end   <- as.POSIXct("2020-12-31", "%Y-%m-%d", tz = "UTC")

remT <- seq(start, end, by = "1 day") 

set.seed(1)
temp <- sapply(as.character(month(remT)), function(m)
  switch(m,
         `1` = , `2` = , `10` = , `11` = , `12` = sample(10:24, 1),
         `3` = , `4` = , `5` = sample(5:17, 1),
         `6` = , `7` = , `8` = , `9` = sample(1:8, 1)))
  

sort(unique(temp[month(remT) %in% c(1:2, 10:12)]))
#R>  [1] 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
sort(unique(temp[month(remT) %in% 3:5]))
#R>  [1]  5  6  7  8  9 10 11 12 13 14 15 16 17
sort(unique(temp[month(remT) %in% 6:9]))
#R> [1] 1 2 3 4 5 6 7 8

A slightly faster solution may be to use vapply, switch and sample.int:

set.seed(1)
temp <- vapply(as.character(month(remT)), function(m)
  switch(m,
         `1` = , `2` = , `10` = , `11` = , `12` = sample.int(15, 1) + 9L,
         `3` = , `4` = , `5` = sample.int(13, 1) + 4L,
         `6` = , `7` = , `8` = , `9` = sample.int(8, 1)), integer(1))
  

sort(unique(temp[month(remT) %in% c(1:2, 10:12)]))
#R>  [1] 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
sort(unique(temp[month(remT) %in% 3:5]))
#R>  [1]  5  6  7  8  9 10 11 12 13 14 15 16 17
sort(unique(temp[month(remT) %in% 6:9]))
#R> [1] 1 2 3 4 5 6 7 8

Otherwise, ifelse needs an equal length of values as the first argument as AnilGoyal points out. Thus, the following does work:

set.seed(1)
temp <- ifelse(
  month(remT) %in% c(1:2, 10:12), sample(10:24, length(remT), TRUE),
  ifelse(month(remT) %in% 3:5, sample(5:17, length(remT), TRUE),
         sample(1:8, length(remT), TRUE)))

sort(unique(temp[month(remT) %in% c(1:2, 10:12)]))
#R>  [1] 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
sort(unique(temp[month(remT) %in% 3:5]))
#R>  [1]  5  6  7  8  9 10 11 12 13 14 15 16 17
sort(unique(temp[month(remT) %in% 6:9]))
#R> [1] 1 2 3 4 5 6 7 8
user_58612440
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