CODEWITHSUNDEEP
r
user18224546
user18224546

Reputation:

Do subtraction between columns of a data.frame

The database below presents the ranking of three methods, which I called M1, M2 and M3. Therefore, I would like to make a new dataset with the difference between the ranks of two methods. In this case, the following relation would be, M1 and M2, M1 and M3, M2 and M3. Therefore, the dataset will have three columns (n, M1-M2, M1-M3 and M2-M3).

result<-structure(list(n = c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 
12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 
28, 29), M1 = c(1L, 29L, 28L, 27L, 25L, 26L, 24L, 20L, 21L, 
22L, 23L, 15L, 12L, 17L, 18L, 19L, 16L, 13L, 14L, 5L, 6L, 7L, 
8L, 9L, 10L, 11L, 4L, 2L, 3L), M2 = c(1, 29, 28, 27, 26, 25, 
24, 23, 22, 21, 20, 15, 12, 19, 18, 17, 16, 14, 13, 11, 10, 9, 
8, 7, 6, 5, 4, 3, 2), M3 = c(1L, 29L, 28L, 27L, 25L, 26L, 24L, 
20L, 21L, 22L, 23L, 15L, 12L, 17L, 18L, 19L, 16L, 13L, 14L, 5L, 
6L, 7L, 8L, 9L, 10L, 11L, 4L, 2L, 3L)), class = "data.frame", row.names = c(NA,-29L))
  
    > result
    n M1 M2 M3
1   1  1  1  1
2   2 29 29 29
3   3 28 28 28
4   4 27 27 27
5   5 25 26 25
6   6 26 25 26
7   7 24 24 24
8   8 20 23 20
9   9 21 22 21
10 10 22 21 22
11 11 23 20 23
12 12 15 15 15
13 13 12 12 12
14 14 17 19 17
15 15 18 18 18
16 16 19 17 19
17 17 16 16 16
18 18 13 14 13
19 19 14 13 14
20 20  5 11  5
21 21  6 10  6
22 22  7  9  7
23 23  8  8  8
24 24  9  7  9
25 25 10  6 10
26 26 11  5 11
27 27  4  4  4
28 28  2  3  2
29 29  3  2  3

Upvotes: 1

Views: 132

Answers (4)

Alison Moss
Alison Moss

Reputation: 11

Its a dataframe so you should be able to do something like

result$M1_M2 <- result$M1 - result$M2

or something similar

Upvotes: -1

rdelrossi
rdelrossi

Reputation: 1154

For a tidyverse solution you can use mutate() from the dplyr package to create new columns in your data.frame:

library(dplyr)

newDF <- result %>% 
  mutate(D1 = M1-M2, D2 = M1-M3, D3 = M2 - M3)

Output:

    n M1 M2 M3 D1 D2 D3
1   1  1  1  1  0  0  0
2   2 29 29 29  0  0  0
3   3 28 28 28  0  0  0
4   4 27 27 27  0  0  0
5   5 25 26 25 -1  0  1
6   6 26 25 26  1  0 -1
7   7 24 24 24  0  0  0
8   8 20 23 20 -3  0  3
9   9 21 22 21 -1  0  1
10 10 22 21 22  1  0 -1
11 11 23 20 23  3  0 -3
12 12 15 15 15  0  0  0
13 13 12 12 12  0  0  0
14 14 17 19 17 -2  0  2
15 15 18 18 18  0  0  0
16 16 19 17 19  2  0 -2
17 17 16 16 16  0  0  0
18 18 13 14 13 -1  0  1
19 19 14 13 14  1  0 -1
20 20  5 11  5 -6  0  6
21 21  6 10  6 -4  0  4
22 22  7  9  7 -2  0  2
23 23  8  8  8  0  0  0
24 24  9  7  9  2  0 -2
25 25 10  6 10  4  0 -4
26 26 11  5 11  6  0 -6
27 27  4  4  4  0  0  0
28 28  2  3  2 -1  0  1
29 29  3  2  3  1  0 -1

Upvotes: 1

langtang
langtang

Reputation: 24722

result = cbind(n=result$n,(result[,c(2:4)] - result[,c(3,4,2)])[,c(1,3,2)])
names(result)=c("n", "M1-M2", "M1-M3", "M2-M3")

Output:

    n M1-M2 M1-M3 M2-M3
1   1     0     0     0
2   2     0     0     0
3   3     0     0     0
4   4     0     0     0
5   5    -1     0     1
6   6     1     0    -1
7   7     0     0     0
8   8    -3     0     3
9   9    -1     0     1
10 10     1     0    -1
11 11     3     0    -3
12 12     0     0     0
13 13     0     0     0
14 14    -2     0     2
15 15     0     0     0
16 16     2     0    -2
17 17     0     0     0
18 18    -1     0     1
19 19     1     0    -1
20 20    -6     0     6
21 21    -4     0     4
22 22    -2     0     2
23 23     0     0     0
24 24     2     0    -2
25 25     4     0    -4
26 26     6     0    -6
27 27     0     0     0
28 28    -1     0     1
29 29     1     0    -1

Upvotes: 1

ThomasIsCoding
ThomasIsCoding

Reputation: 101139

We can try

cbind(
  result,
  `colnames<-`(
    combn(result[-1], 2, function(v) v[[1]] - v[[2]]),
    combn(names(result)[-1], 2, function(v) paste0(v, collapse = "-"))
  )
)

which gives

    n M1 M2 M3 M1-M2 M1-M3 M2-M3
1   1  1  1  1     0     0     0
2   2 29 29 29     0     0     0
3   3 28 28 28     0     0     0
4   4 27 27 27     0     0     0
5   5 25 26 25    -1     0     1
6   6 26 25 26     1     0    -1
7   7 24 24 24     0     0     0
8   8 20 23 20    -3     0     3
9   9 21 22 21    -1     0     1
10 10 22 21 22     1     0    -1
11 11 23 20 23     3     0    -3
12 12 15 15 15     0     0     0
13 13 12 12 12     0     0     0
14 14 17 19 17    -2     0     2
15 15 18 18 18     0     0     0
16 16 19 17 19     2     0    -2
17 17 16 16 16     0     0     0
18 18 13 14 13    -1     0     1
19 19 14 13 14     1     0    -1
20 20  5 11  5    -6     0     6
21 21  6 10  6    -4     0     4
22 22  7  9  7    -2     0     2
23 23  8  8  8     0     0     0
24 24  9  7  9     2     0    -2
25 25 10  6 10     4     0    -4
26 26 11  5 11     6     0    -6
27 27  4  4  4     0     0     0
28 28  2  3  2    -1     0     1
29 29  3  2  3     1     0    -1

Upvotes: 3

Related Questions