Dplyr: group_by and convert multiple columns to a vector

Question

I have a question about how to convert multiple columns to a vector. I have the following dataset that I would like to group them by their condition and take all the position count into one vector. I know I can use as.vector() to convert them individually but I wonder if there is a dplyr way. Thank you!

test -> structure(list(gene_id = c("gene0", "gene0", "gene0", "gene0", 
"gene0", "gene0", "gene0", "gene0", "gene0", "gene0", "gene0", 
"gene0", "gene0", "gene0", "gene0", "gene0", "gene0", "gene0", 
"gene0", "gene0", "gene0", "gene0", "gene0", "gene0"), codon_index = c(1L, 
2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 
3L, 1L, 2L, 3L, 1L, 2L, 3L), position_1_count = c(2L, 7L, 8L, 
0L, 2L, 22L, 19L, 15L, 134L, 1L, 127L, 30L, 0L, 0L, 1L, 4L, 65L, 
234L, 1L, 3L, 57L, 0L, 4L, 16L), position_2_count = c(0L, 5L, 
5L, 0L, 3L, 2L, 3L, 13L, 134L, 0L, 36L, 5L, 0L, 0L, 0L, 1L, 150L, 
7L, 0L, 7L, 7L, 0L, 6L, 1L), position_3_count = c(0L, 2L, 1L, 
0L, 4L, 0L, 3L, 32L, 43L, 3L, 9L, 1L, 0L, 0L, 0L, 4L, 105L, 1L, 
0L, 14L, 5L, 0L, 6L, 1L), condition = structure(c(1L, 1L, 1L, 
7L, 7L, 7L, 3L, 3L, 3L, 5L, 5L, 5L, 8L, 8L, 8L, 2L, 2L, 2L, 4L, 
4L, 4L, 6L, 6L, 6L), .Label = c("c", "cup", "n", "nup", "p", 
"pup", "min", "rich"), class = "factor")), class = c("tbl_df", 
"tbl", "data.frame"), row.names = c(NA, -24L), .Names = c("gene_id", 
"codon_index", "position_1_count", "position_2_count", "position_3_count", 
"condition"))

> head(a)
# A tibble: 6 × 6
  gene_id codon_index position_1_count position_2_count position_3_count condition
    <chr>       <int>            <int>            <int>            <int>    <fctr>
1   gene0           1                2                0                0         c
2   gene0           2                7                5                2         c
3   gene0           3                8                5                1         c
4   gene0           1                0                0                0       min
5   gene0           2                2                3                4       min
6   gene0           3               22                2                0       min

How can we convert this dataset to (I didn't add the column names here)

2 0 0 7 5 2 8 5 1 c
0 0 0 2 3 4 22 2 0 min

Answer 1

Another alternative:

library(purrr)

test %>%
  slice_rows("condition") %>%
  by_slice(function(x) unlist(x[-(1:2)]), .to = "vec")

Which gives:

#  condition                                vec
#1         c          2, 7, 8, 0, 5, 5, 0, 2, 1
#2       cup   4, 65, 234, 1, 150, 7, 4, 105, 1
#3         n 19, 15, 134, 3, 13, 134, 3, 32, 43
#4       nup        1, 3, 57, 0, 7, 7, 0, 14, 5
#5         p      1, 127, 30, 0, 36, 5, 3, 9, 1
#6       pup         0, 4, 16, 0, 6, 1, 0, 6, 1
#7       min         0, 2, 22, 0, 3, 2, 0, 4, 0
#8      rich          0, 0, 1, 0, 0, 0, 0, 0, 0

---

As per mentioned in the comments by @advance, if you want the result rowwise:

test %>% 
  slice_rows("condition") %>% 
  by_slice(function(x) as.vector(t(x[-(1:2)])), .to = "vec")

#  condition                                vec
#1         c          2, 0, 0, 7, 5, 2, 8, 5, 1
#2       cup   4, 1, 4, 65, 150, 105, 234, 7, 1
#3         n 19, 3, 3, 15, 13, 32, 134, 134, 43
#4       nup        1, 0, 0, 3, 7, 14, 57, 7, 5
#5         p      1, 0, 3, 127, 36, 9, 30, 5, 1
#6       pup         0, 0, 0, 4, 6, 6, 16, 1, 1
#7       min         0, 0, 0, 2, 3, 4, 22, 2, 0
#8      rich          0, 0, 0, 0, 0, 0, 1, 0, 0

---

Or adapting @DavidArenburg's comment using do() instead of summarise():

test %>% 
  group_by(condition) %>% 
  select(position_1_count:condition) %>%
  do(res = c(t(.[,-4]))) 

Which gives:

#  condition                                res
#1         c          2, 0, 0, 7, 5, 2, 8, 5, 1
#2       cup   4, 1, 4, 65, 150, 105, 234, 7, 1
#3         n 19, 3, 3, 15, 13, 32, 134, 134, 43
#4       nup        1, 0, 0, 3, 7, 14, 57, 7, 5
#5         p      1, 0, 3, 127, 36, 9, 30, 5, 1
#6       pup         0, 0, 0, 4, 6, 6, 16, 1, 1
#7       min         0, 0, 0, 2, 3, 4, 22, 2, 0
#8      rich          0, 0, 0, 0, 0, 0, 1, 0, 0

Answer 2

After taking Peterson's idea, I think this code works the best:

test %>% gather(Location, Count, starts_with("position"))  %>% arrange(codon_index)  %>% group_by(condition) %>% do(count = as.vector(t(.$Count)))

The result will look like this

> ans = test %>% gather(Location, Count, starts_with("position"))  %>% arrange(codon_index)  %>% group_by(condition) %>% do(count = as.vector(t(.$Count)))

    # A tibble: 8 × 2
      condition     count
    *    <fctr>    <list>
    1         c <int [9]>
    2       cup <int [9]>
    3         n <int [9]>
    4       nup <int [9]>
    5         p <int [9]>
    6       pup <int [9]>
    7       min <int [9]>
    8      rich <int [9]>
> ans$count[[1]]
[1] 2 0 0 7 5 2 8 5 1

Thanks a lot for everyone's help!

Answer 3

Am I correct that what you want is a separate vector for all of the counts for each condition? If so, a mix of dplyr and tidyr should do it. First, I gather to put all of the counts in a single column. Then, split to separate by the condition, then use lapply to generate a list, containing a separate vector for each condition:

a %>%
  gather(Location, Count, starts_with("position")) %>%
  split(.$condition) %>%
  lapply(function(x){x$Count})

gives:

$c
[1] 2 7 8 0 5 5 0 2 1

$cup
[1]   4  65 234   1 150   7   4 105   1

$n
[1]  19  15 134   3  13 134   3  32  43

$nup
[1]  1  3 57  0  7  7  0 14  5

$p
[1]   1 127  30   0  36   5   3   9   1

$pup
[1]  0  4 16  0  6  1  0  6  1

$min
[1]  0  2 22  0  3  2  0  4  0

$rich
[1] 0 0 1 0 0 0 0 0 0

If the order matters (and is wrong above) you should be able to sort before splitting, e.g. by adding arrange(codon_index) after gather