Collapse a Data.Frame into a Vector

Collapse a data.frame into a vector

If your data.frame is named dat, try:

unlist(dat, use.names=FALSE)

You may also want to try the following:

as.character(unlist(dat, use.names=FALSE))
c(na.omit(as.character(unlist(dat, use.names=FALSE))))

which might be useful if your data are entered as factors.

Also, in your question, you make it appear that you don't actually have a basic "vector", but rather a one-column data.frame. Thus, something like dat2 <- data.frame(COL = c(na.omit(as.character(unlist(dat, use.names=FALSE))))) might be what you're looking for.

R: Collapse values in data.frame into a vector similar to paste(collapse=',')

Make cyls a list column like this:

my_df <- mtcars %>% group_by(gear) %>% summarize(cyls=list(unique(cyl)))

my_df$cyls[[1]] # boom dbl vectors for each row stored as a list
[1] 6 8 4

Collapse every series of four rows in a data frame into a single vector, overwriting missing values

How about this:

library(dplyr)
library(tidyr)
df <- df %>% mutate(obs = rep(1:(nrow(.)/4), each=4))
df <- df %>% 
  pivot_longer(-obs, names_to="var", values_to="vals") %>% 
  na.omit() %>% 
  group_by(obs) %>% 
  mutate(col = seq_along(obs)) %>% 
  select(obs, col, vals) %>% 
  pivot_wider(names_from="col", names_prefix="V", values_from="vals")
df
# # A tibble: 3 x 7
# # Groups:   obs [3]
#     obs V1    V2        V3                  V4      V5          V6             
#   <int> <chr> <chr>     <chr>               <chr>   <chr>       <chr>          
# 1     1 Buy   Completed 2021-02-11 20:49:19 0.11057 Fee1.00 USD Total199.00 USD
# 2     2 Buy   Completed 2021-02-11 20:48:03 82.146  Fee0.50 USD Total100.00 USD
# 3     3 Buy   Completed 2021-02-11 20:47:22 30.15   Fee0.64 USD Total127.00 USD 

Collapse cells across several rows to a single column in a R data frame

Try this base R approach:

#Code
data <- aggregate(x3~x1+x2,data,function(x) paste0(x,collapse = ','))

Output:

  x1 x2  x3
1  1  A 6,7
2  3  C 8,9
3  5  E  10

R dataframe - Collapse multiple columns into a single numeric vector, row-by-row

I don't think davo1979 is looking to transform his data to a long format, therefore, I don't think tidyr::gather is the tool he wants.

The answer, which I think, most closestly address his question is

lapply(1:nrow(mtcars), function(x) as.numeric(mtcars[x,2:8]))

This will store each row as a numeric vector, contained within a list.

However, davo1979 it is important to know what is your ultimate goal with this data. I ask because your desired outcome of a bunch of numeric vectors is a clumsy way to handle and store data in R.

Aggregate and collapse a vector based while maintaing order

Using DF shown reproducibly in the Note at the end, sort by from giving DF2 and then iterate through its rows removing any row with a duplicate. We need a loop here since each removal depends on the prior ones. Finally summarize the result.

library(dplyr)

DF2 <- arrange(DF, from)

i <- 1
while(i <= nrow(DF2)) {
  ix <- seq_len(i-1)
  dup <- with(DF2, (to[i] %in% c(to[ix], from[ix])) | (from[i] %in% to[ix]))
  if (dup) DF2 <- DF2[-i, ] else i <- i + 1
}

DF2 %>%
  group_by(from) %>%
  summarize(to = toString(to), priority = sum(priority)) %>%
  ungroup

giving:

# A tibble: 4 x 3
   from to    priority
  <int> <chr>    <int>
1     1 8, 7         6
2     2 6            1
3     3 4            1
4     9 5            3

Note

Lines <- "from | to  | priority
   1 |   8 |        1
   2 |   6 |        1
   3 |   4 |        1
   4 |   5 |        3
   5 |   6 |        4
   6 |   2 |        5
   7 |   8 |        2
   4 |   3 |        5
   2 |   1 |        1
   6 |   6 |        4
   1 |   7 |        5
   8 |   4 |        6
   9 |   5 |        3"
DF <- read.table(text = Lines, header = TRUE, sep = "|", strip.white = TRUE)

collapse a dataframe in R that contains both numeric and character variables

Is this what you are looking for?

library(dplyr)

data %>%
  dplyr::group_by(ag, date) %>%
  summarise(across(everything(), ~
                     if(is.numeric(.x)) mean(.x) else first(.x)))

#> `summarise()` has grouped output by 'ag'. You can override using the `.groups` argument.
#> # A tibble: 12 x 6
#> # Groups:   ag [4]
#>    ag     date num_var1 num_var2 alpha_var1 alpha_var2
#>    <chr> <int>    <dbl>    <dbl> <chr>      <chr>     
#>  1 A         1        3       22 A          Y         
#>  2 A         2       11       14 I          Q         
#>  3 A         3       19        6 Q          I         
#>  4 B         1        4       21 B          X         
#>  5 B         2       12       13 J          P         
#>  6 B         3       20        5 R          H         
#>  7 C         1        5       20 C          W         
#>  8 C         2       13       12 K          O         
#>  9 C         3       21        4 S          G         
#> 10 D         1        6       19 D          V         
#> 11 D         2       14       11 L          N         
#> 12 D         3       22        3 T          F

^{Created on 2022-03-03 by the reprex package (v2.0.1)}

Merging data frame columns into a vector

We can have a list column. If we specify the simplify = FALSE in combn, it would return a list of vectors, which can be wrapped in tibble to create a list column

combn(5, 2, simplify = FALSE) %>% 
      tibble(vector = .)

---

Or another option is to make use of unnest_wider with unite

library(tidyr)
library(dplyr)
library(stringr)
combn(5, 2, simplify = FALSE) %>% 
    tibble(vector = ., vector2 = vector) %>% 
    unnest_wider(c(vector2), names_repair =
      ~ c('vector', str_c('V', 1:2))) %>% 
    unite(string, V1, V2, sep=";", remove = FALSE)

---

If we use the OP's method, an option is map2 which returns a list after concatenate both the columns elementwise

library(purrr)
combn(5, 2) %>% 
  t %>% 
  as.data.frame %>% 
  mutate(
  string = paste(V1, V2, sep = ';'),
  vector =  map2(V1, V2, c))
#  V1 V2 string vector
#1   1  2    1;2   1, 2
#2   1  3    1;3   1, 3
#3   1  4    1;4   1, 4
#4   1  5    1;5   1, 5
#5   2  3    2;3   2, 3
#6   2  4    2;4   2, 4
#7   2  5    2;5   2, 5
#8   3  4    3;4   3, 4
#9   3  5    3;5   3, 5
#10  4  5    4;5   4, 5

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