如何创建带对应编号的重复字符串向量：生成user_1至user_33适配purrr函数.y参数

Got it, let's get this sorted for you! You're aiming to generate a vector of strings like user_1, user_2, ..., user_33 to pass as the .y parameter in a purrr function—your initial rep(user_, [1:33]) approach isn't working because rep() repeats existing elements, but you need to combine the base string with sequential numbers instead.

Here are the simplest ways to make this vector: #

1. Base R with paste0()

This is the most straightforward method using base R, no extra packages needed:

user_vec <- paste0("user_", 1:33)

paste0() concatenates strings without any separator, so it perfectly pairs "user_" with each number from 1 to 33.

2. Tidyverse-friendly with stringr::str_c()

If you're already using purrr (part of the tidyverse), str_c() from the stringr package is a great alternative—it works the same way but fits better with tidyverse workflows:

library(stringr)
user_vec <- str_c("user_", 1:33)

Using this vector with purrr #

Once you have user_vec, you can pass it directly to purrr functions. For example, if you want to map a function over each user ID:

library(purrr)

# Example function to process a user ID
process_user <- function(user_id) {
  message("Working on: ", user_id)
}

# Pass user_vec as .y (or as the first argument to map, which defaults to .x)
map(.y = user_vec, .f = process_user)
# Or the more common shorthand:
map(user_vec, process_user)

Why your initial rep() approach didn't work #

The rep(user_, 1:33) syntax would repeat the string user_ 1 time, then 2 times, up to 33 times—resulting in a vector like c("user_", "user_", "user_", ...) (total of 1+2+...+33 = 561 elements) instead of the numbered sequence you need.

内容的提问来源于stack exchange，提问作者Stackstudent_09