tidyverse solution for multiplying columns by a vector

Question

I looked for solutions here: Multiply columns in a data frame by a vector and here: What is the right way to multiply data frame by vector?, but it doesn't really work.

What I want to do is a more or less clean tidyverse way where I multiply columns by a vector and then add these as new columns to the existing data frame. Taking teh data example from the first link:

c1 <- c(1,2,3)
c2 <- c(4,5,6)
c3 <- c(7,8,9)
d1 <- data.frame(c1,c2,c3)

  c1 c2 c3
1  1  4  7
2  2  5  8
3  3  6  9

v1 <- c(1,2,3)

my desired result would be:

  c1 c2 c3 pro_c1 pro_c2 pro_c3
1  1  4  7      1      8     21
2  2  5  8      2     10     24
3  3  6  9      3     12     27

I tried:

library(tidyverse)
d1 |>
  mutate(pro = sweep(across(everything()), 2, v1, "*"))

But here the problem is the new columns are actually a data frame within my data frame. And I'm struggling with turning this data frame-in-data frame into regular columns. I assume, I could probably first setNames on this inner data frame and then unnest, but wondering if there's a more direct way by looping over each column with across and feed it with the first/second/third element of v1?

(I know I could probably also first create a standalone data frame with the three new multiplied columns, give them a unique name and then bind_cols on both, d1 and the df with the products.)

@akrun if "by row" means that I want to multiply all c1 rows with the first value of v1, then all rows of c2 with the second value of v2 and so on, then yes, by row. — deschen, Mar 31 '22 at 21:53
c1 is multiplied with 1. c2 is multiploed with 2. c3 is multiplied with 3. Does this make sense? — deschen, Mar 31 '22 at 21:55

score 4 · Answer 1 · answered Mar 31 '22 at 22:03

4

This is perhaps ridiculous, but you could use

library(dplyr)

d1 %>% 
  mutate(across(everything(), 
                ~.x * v1[which(names(d1) == cur_column())],
                .names = "pro_{.col}"))

which returns

  c1 c2 c3 pro_c1 pro_c2 pro_c3
1  1  4  7      1      8     21
2  2  5  8      2     10     24
3  3  6  9      3     12     27

answered Mar 31 '22 at 22:03

Martin Gal

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Thank you my dear friend, I hope you are also doing great. – Anoushiravan R Mar 31 '22 at 22:09
Interesting one. – deschen Mar 31 '22 at 22:14

deschen · Answer 2 · 2022-04-01T09:28:45.077

Just for the fun part, I trialed & errored a bit more after seeing some of your solutions. Since I started treating myself to the pain of using the base R native pipe which doesn't yet allow for passing a "." argument silently as the first argument, I had to fiddle around with it a bit more:

library(tidyverse)
d1 |> 
  (\(x)(bind_cols(x, x |>
                       map2_dfc(v1, `*`) |> 
                       rename_with(.cols = everything(),
                                   .fn   = ~paste0("pro_", .)))))()

  c1 c2 c3 pro_c1 pro_c2 pro_c3
1  1  4  7      1      8     21
2  2  5  8      2     10     24
3  3  6  9      3     12     27

Found an even easier solution:

d1 |> 
  add_column(d1 |> 
               map2_dfc(v1, `*`) |> 
               rename_with(.cols = everything(),
                           .fn   = ~paste0("pct_", .)))

akrun · Accepted Answer · 2022-03-31T22:15:59.210

If it is by row, then one option is c_across

library(dplyr)
library(stringr)
library(tibble)
new <- as_tibble(setNames(as.list(v1), names(d1)))
d1 %>% 
  rowwise %>% 
  mutate(c_across(everything()) * new) %>%
  rename_with(~ str_c("pro_", .x), everything()) %>%
  bind_cols(d1, .)

-output

   1 c2 c3 pro_c1 pro_c2 pro_c3
1  1  4  7      1      8     21
2  2  5  8      2     10     24
3  3  6  9      3     12     27

Or another option is map2

library(purrr)
map2_dfc(d1, v1,  `*`) %>%
   rename_with(~ str_c("pro_", .x), everything()) %>%
   bind_cols(d1, .)

-output

 c1 c2 c3 pro_c1 pro_c2 pro_c3
1  1  4  7      1      8     21
2  2  5  8      2     10     24
3  3  6  9      3     12     27

Also, with the OP' approach, it is a data.frame column. It can be unpacked

library(tidyr)
d1 |> 
    mutate(pro = sweep(cur_data(), 2, v1, `*`)) |> 
    unpack(pro, names_sep = "_")

-output

# A tibble: 3 × 6
     c1    c2    c3 pro_c1 pro_c2 pro_c3
  <dbl> <dbl> <dbl>  <dbl>  <dbl>  <dbl>
1     1     4     7      1      8     21
2     2     5     8      2     10     24
3     3     6     9      3     12     27

EDIT: Based on @deschen comments with names_sep

Awesome, the unpack solution is great. BTW, you could simplify this a bit more by replacing the whole setNames part by unpacks `names_sep = "_"`. — deschen, Mar 31 '22 at 22:12

IRTFM · Answer 4 · 2022-04-01T01:29:14.067

Here is a dplyr-ized version of the usual apply(. , 1, fun) paradigm:

d1 %>% apply(1, "*", v1) %>% t %>% cbind(d1, .)

  c1 c2 c3 c1 c2 c3
1  1  4  7  1  8 21
2  2  5  8  2 10 24
3  3  6  9  3 12 27

It gets a bit hackish if you want to assign column names to the matrix before binding back to the starting dataframe:

d1 %>% apply(1, "*", v1) %>% t %>% `colnames<-`(.,  paste0("pro_", colnames(.))) %>% cbind(d1, .)
  c1 c2 c3 pro_c1 pro_c2 pro_c3
1  1  4  7      1      8     21
2  2  5  8      2     10     24
3  3  6  9      3     12     27

score 0 · Answer 5 · answered Apr 01 '22 at 08:21

Similar to @IRTFM's solution, but does not need apply(...)

 cbind(d1, t(t(d1)*v1))

##   c1 c2 c3 c1 c2 c3
## 1  1  4  7  1  8 21
## 2  2  5  8  2 10 24
## 3  3  6  9  3 12 27

Or,

result <- cbind(d1, t(t(d1)*v1))
colnames(result) <- c(colnames(d1), paste0('pro_', colnames(d1)))
result

which gives the column names you want.

tidyverse solution for multiplying columns by a vector

5 Answers5