Hello!

```
# I have got this vector:
a <- c("1", "2", "3", "4", "5", "6", "7")
#[1] "1" "2" "3" "4" "5" "6" "7"
# and I would like to have this vector:
# [1] 1_A1 1_A2 1_A7 2_A1 2_A2 2_A7 3_A1 3_A2 3_A7 4_A1 4_A2 4_A7 5_A1 5_A2 5_A7 6_A1 6_A2 6_A7 7_A1 7_A2 7_A7
# My solution started with:
r1 <- rep(a, each=3)
# [1] "1" "1" "1" "2" "2" "2" "3" "3" "3" "4" "4" "4" "5" "5" "5" "6" "6" "6" "7" "7" "7"
# Then I created this vector:
b <- c("_A1","_A2","_A7")
# [1] "_A1" "_A2" "_A7"
# And then I replicated the previous vector 7 times
r2 <- rep(b,times=7)
# [1] "_A1" "_A2" "_A7" "_A1" "_A2" "_A7" "_A1" "_A2" "_A7" "_A1" "_A2" "_A7" "_A1" "_A2" "_A7" "_A1" "_A2" "_A7" "_A1" "_A2" "_A7"
```

Now I would like to combine vector r1 and vector r2 to achieve my desired vector. Any help?

Or is my solution quite cumbersome and there is a better solution? Maybe with a for-loop?

Thank you!

I don't know if it is that much quicker or easier, but you could use `expand_grid()`

, then `unite()`

.

```
library(tidyverse)
a <- c("1", "2", "3", "4", "5", "6", "7")
b <- c("_A1","_A2","_A7")
expand_grid(a, b) %>%
unite("c", a:b, sep = "") %>%
pull()
[1] "1_A1" "1_A2" "1_A7" "2_A1" "2_A2" "2_A7" "3_A1" "3_A2" "3_A7" "4_A1" "4_A2" "4_A7" "5_A1" "5_A2" "5_A7" "6_A1" "6_A2" "6_A7" "7_A1" "7_A2" "7_A7"
```

2 Likes

Yes, thank you! Nice code!

I don't really understand the purpose of "c" in your code. It seems I can write anything instead of "c"...

But it works perfectly for my data.

1 Like

Yeah, you can write anything instead of "c". See comments below:

```
expand_grid(a, b) %>% # creates a grid with all possibilities of a and b
unite("c", a:b, sep = "") %>% # creates a new column 'c' from a and b, with no separator, removing the original cols.
pull() # pulls all of it in to a vector.
```

1 Like

Here's an answer using just `paste`

and `rep`

:

```
# method 1
X <- paste0(
rep(x = 1:7, each = 3),
"_A",
rep(x = c(1, 2, 7), times = 7)
)
X
#> [1] "1_A1" "1_A2" "1_A7" "2_A1" "2_A2" "2_A7" "3_A1" "3_A2" "3_A7" "4_A1"
#> [11] "4_A2" "4_A7" "5_A1" "5_A2" "5_A7" "6_A1" "6_A2" "6_A7" "7_A1" "7_A2"
#> [21] "7_A7"
# method 2
a <- rep(x = as.character(x = 1:7), each = 3)
b <- c("_A1", "_A2", "_A7")
Y <- paste0(a, b)
Y
#> [1] "1_A1" "1_A2" "1_A7" "2_A1" "2_A2" "2_A7" "3_A1" "3_A2" "3_A7" "4_A1"
#> [11] "4_A2" "4_A7" "5_A1" "5_A2" "5_A7" "6_A1" "6_A2" "6_A7" "7_A1" "7_A2"
#> [21] "7_A7"
# check
all.equal(target = X, current = Y)
#> [1] TRUE
```

1 Like

Also a nice solution. paste() was the function I didn't think of.

Thank you Yarnabrina and Williaml; now I've learned something.

1 Like

system
Closed
April 29, 2021, 5:57pm
7
This topic was automatically closed 7 days after the last reply. New replies are no longer allowed. If you have a query related to it or one of the replies, start a new topic and refer back with a link.