R - multiply dataframe by vectors with different sizes, tidyverse/pipe style-CodePudding

I have the following code:

dfl <- group_split(mtcars, am)

v <- list()

v[[1]] <- seq(1:6)

v[[2]] <- seq(1:8)

x <- list()

for(i in 1:2){
  x[[i]] <- list()
for (j in 1:length(v[[i]])){
  x[[i]][[j]] <- dfl[[i]]$qsec*v[[i]][[j]]
}
}

x[[1]] <- reduce(x[[1]], bind_cols)
x[[2]] <- reduce(x[[2]], bind_cols)

I wish to achieve this same output, but in tidyverse style and all within a single pipeline without breaking the pipe. This has proved to be more difficult because vectors v[[1]] and v[[2]] are of different length. I tried using the map2 function, but it fails for this precise reason. What would be the best way to approach this?

CodePudding user response：

I'm not exactly sure what you are trying to return - it looks like you just want multiples of the qsec column.. You can do that like this:

mtcars %>% 
  group_split(am) %>%
  map2(.x=.,.y=list(1:6,1:8),~as.data.frame(sapply(.y, \(y) .x$qsec*y)))

Output (same as your x[[1]] and x[[2]])

[[1]]
      V1    V2    V3    V4     V5     V6
1  19.44 38.88 58.32 77.76  97.20 116.64
2  17.02 34.04 51.06 68.08  85.10 102.12
3  20.22 40.44 60.66 80.88 101.10 121.32
4  15.84 31.68 47.52 63.36  79.20  95.04
5  20.00 40.00 60.00 80.00 100.00 120.00
6  22.90 45.80 68.70 91.60 114.50 137.40
7  18.30 36.60 54.90 73.20  91.50 109.80
8  18.90 37.80 56.70 75.60  94.50 113.40
9  17.40 34.80 52.20 69.60  87.00 104.40
10 17.60 35.20 52.80 70.40  88.00 105.60
11 18.00 36.00 54.00 72.00  90.00 108.00
12 17.98 35.96 53.94 71.92  89.90 107.88
13 17.82 35.64 53.46 71.28  89.10 106.92
14 17.42 34.84 52.26 69.68  87.10 104.52
15 20.01 40.02 60.03 80.04 100.05 120.06
16 16.87 33.74 50.61 67.48  84.35 101.22
17 17.30 34.60 51.90 69.20  86.50 103.80
18 15.41 30.82 46.23 61.64  77.05  92.46
19 17.05 34.10 51.15 68.20  85.25 102.30

[[2]]
      V1    V2    V3    V4    V5     V6     V7     V8
1  16.46 32.92 49.38 65.84 82.30  98.76 115.22 131.68
2  17.02 34.04 51.06 68.08 85.10 102.12 119.14 136.16
3  18.61 37.22 55.83 74.44 93.05 111.66 130.27 148.88
4  19.47 38.94 58.41 77.88 97.35 116.82 136.29 155.76
5  18.52 37.04 55.56 74.08 92.60 111.12 129.64 148.16
6  19.90 39.80 59.70 79.60 99.50 119.40 139.30 159.20
7  18.90 37.80 56.70 75.60 94.50 113.40 132.30 151.20
8  16.70 33.40 50.10 66.80 83.50 100.20 116.90 133.60
9  16.90 33.80 50.70 67.60 84.50 101.40 118.30 135.20
10 14.50 29.00 43.50 58.00 72.50  87.00 101.50 116.00
11 15.50 31.00 46.50 62.00 77.50  93.00 108.50 124.00
12 14.60 29.20 43.80 58.40 73.00  87.60 102.20 116.80
13 18.60 37.20 55.80 74.40 93.00 111.60 130.20 148.80

CodePudding user response：

You can also use outer.

library(dplyr)
library(purrr)

mtcars %>% 
  group_split(am) %>% 
  map2(list(6, 8), ~ as.data.frame(outer(.x$qsec, seq.int(.y))))

CodePudding user response：

A direct interpretation would be

v <- list()
v[[1]] <- seq(1:6)
v[[2]] <- seq(1:8)
names(v) <- levels(factor(mtcars$am))
v["0"]

mtcars %>% 
  group_by(am) %>% 
  group_map(function(value, key) {
    t(tcrossprod(
      v[[as.character(key)]],
      value$qsec
    ))
  }) %>% 
  identity()

Now, a few thing: am is a numeric, so to index it, we used names property. Hence why the key has to be converted with as.character. Also, note that identity does nothing, but it is a way to end pipes, so as to be able to comment-out parts without having to remove the leading %>%.