将 sample_n output 从 tibble 转换为变量

Question

I am trying to create a random set and one of the variables I need is a randomly selected date. The problem that I have is that the output of the sample_n is a tibble itself. This becomes problematic when I am trying to bind by row multiple sets.

Please see the example below.

library(tidyverse)
dts <- 
  tibble(date=seq(as.Date("2020-07-01"),as.Date("2020-07-31"),by="1 days")) %>% 
  mutate(wday=weekdays(date)) %>% 
  filter(wday!="Saturday" & wday!="Sunday")

dt1 <- tibble(
  date=sample_n(dts[1], 5),
  x=runif(5, 10, 20)) 
glimpse(dt1)

dt2 <- tibble(
  date=sample_n(dts[1], 4),
  x=runif(4, 10, 20)) 
glimpse(dt2)

From this you can see that the variable date in dt1 and dt2 is a tibble. I was expecting this to be just a date variable.

This becomes problematic when I am trying to combine these two sets.

dt <- dt1 %>% bind_rows(dt2)

When I run the last line I get the following error message:

Error: Argument 1 can't be a list containing data frames

Thanks

Answer 1

You can just pull the result to convert it into a vector:

dts %>% sample_n(5) %>% pull(date)
[1] "2020-07-22" "2020-07-08" "2020-07-03" "2020-07-15" "2020-07-17"

Answer 2

Why not just use sample() in base ? dplyr::sample_n() is to sample rows of a dataset. In your case you only need to sample a vector, ie dts$date , so dplyr::sample_n() is a detour.

dt1 <- tibble(
  date = sample(dts$date, 5),
  x = runif(5, 10, 20)
)

dt1

# # A tibble: 5 x 2
#   date           x
#   <date>     <dbl>
# 1 2020-07-09  13.0
# 2 2020-07-17  18.0
# 3 2020-07-13  16.7
# 4 2020-07-28  16.5
# 5 2020-07-03  14.6