Can I "automate" uniting columns that are V1:V148? - r

Question

I have a dataset that has 148 columns. I need to combine them in groups of 4.
For example V1,V2,V3,V4 =X1
V1 <-c(0,3)
V2 <-c(F,F)
V3 <-c (2,4)
V4 <-c(A,C)
X1
0F2A
3F4C
I know I can use
```{r}
new_data_4 <-new%>%
unite(V1:V4)%>%
unite(V5:V8)%>%
unite(V9:V12)%>%
unite(V13:V16)
```
with great success but I would like to make this function. My wish is that it can count the number of columns and do it automatically without hardcoding the numbers. I have more files to go over. I have looked around StackOverFlow and found LOTS of examples with specific problems that don't really jive with what I have.
I have tried:
```{r}
unite_columns <-function(x){
united_cols <-tidyr::unite(x, seq_along(1, ncol(x), 4), seq_along(4, ncol(x), 4))
return(united_cols)
}
```
and
```{r}
unite_columns <-function(x){
united_cols <-unite(x, seq(1, ncol(x), 4), seq(4, ncol(x), 4))
return(united_cols)
}
```
I was thinking I could use a similar tactic that is used to merge strings but it did not work.
Any help would be greatly appreciated. TIA

Answer 1

You can use split.default to split the columns in every 4 columns and paste the values rowwise using do.call.
result <- data.frame(sapply(split.default(new, ceiling(seq_along(new)/4)),
function(x) do.call(paste0, x)))
# X1 X2
#1 0F2A 8A4R
#2 3F4C 9B5K
data
new <- data.frame(V1 = c(0,3), V2 = c("F","F"), V3 = c(2,4), V4 = c("A","C"),
V5 = c(8, 9), V6 = c("A", "B"), V7 = c(4, 5), V8 = c("R", "K"))
new
# V1 V2 V3 V4 V5 V6 V7 V8
#1 0 F 2 A 8 A 4 R
#2 3 F 4 C 9 B 5 K

Answer 2

Using Reduce.
sapply(1:(ncol(new)/4), function(f) Reduce(paste0, new[1:4*f]))
# [,1] [,2] [,3]
# [1,] "0F2A" "FAAR" "2A1D"
# [2,] "3F4C" "FCBK" "4B2S"
If you want a data frame:
as.data.frame(sapply(1:(ncol(new)/4), function(f) Reduce(paste0, new[1:4*f])))
# V1 V2 V3
# 1 0F2A FAAR 2A1D
# 2 3F4C FCBK 4B2S
Data
new <- structure(list(V1 = c(0, 3), V2 = c("F", "F"), V3 = c(2, 4),
V4 = c("A", "C"), V5 = c(8, 9), V6 = c("A", "B"), V7 = c(4,
5), V8 = c("R", "K"), V9 = c(1, 2), V10 = c("C", "D"), V11 = c(9,
8), V12 = c("D", "S")), class = "data.frame", row.names = c(NA,
-2L))

Answer 3

Another purrr and tidyr option could be:
imap_dfc(.x = split.default(df, ceiling(1:ncol(df)/4)),
~ .x %>%
unite(col = !!paste0("X", .y), everything(), sep = ""))
X1 X2
1 0F2A 8A4R
2 3F4C 9B5K

Answer 4

You can also do like this, with tidyverse only. I have used purrr::map2dfc to do this.
First argument of map is a sequence of length n/4 (needless to say you may use ncol(new) instead of storing n separately.
and second argument is names of columns to be generated.
At each iteration, the map function will take out four columns as per integer division function used,
name it as per second argument
and then select that column only.
all the lists generated at each iteration of map function will be col-bind and therefore map2_dfc has been used.
I think that is pretty clear.
library(tidyverse)
#say n is 148
n <- 148
map2_dfc(seq_len(n/4), paste0("X", seq_len(n/4)), ~new %>%
unite(!!.y,
seq_along(new)[(3 + seq_along(new)) %/% 4 == .x],
sep = "") %>% select(all_of(.y))
)
Check it for data generated by #Ronak
n <- 8
map2_dfc(seq_len(n/4), paste0("X", seq_len(n/4)), ~new %>%
unite(!!.y,
seq_along(new)[(3 + seq_along(new)) %/% 4 == .x],
sep = "") %>% select(all_of(.y))
)
X1 X2
1 0F2A 8A4R
2 3F4C 9B5K
Or on data generated by #jay.sf
n <- 12
map2_dfc(seq_len(n/4), paste0("X", seq_len(n/4)), ~new %>%
unite(!!.y,
seq_along(new)[(3 + seq_along(new)) %/% 4 == .x],
sep = "") %>% select(all_of(.y))
)
X1 X2 X3
1 0F2A 8A4R 1C9D
2 3F4C 9B5K 2D8S