CODEWITHSUNDEEP
rdata.tablearray-difference
Kirsten
Kirsten

Reputation: 143

Multiple pairwise differences based on column name patterns

I have a data.table, dt:

dt

Id  v1 v2 v3 x1 x2 x3
1   7  1  3  5  6  8
2   1  3  5  6  8  5
3   3  5  6  8  5  1

v1, v2, v3 and x1, x2, x3 are numeric variables

I want to subtract the 'x' columns from the 'v' columns, i.e. calculate the differences v1 - x1, v2 - x2, etc. In my real data I may have 100s of such pair of variables.

Desired output:

dt

Id  v1 v2 v3 x1 x2 x3 diff1 diff2 diff3
1   7  1  3  5  6  8   -2     -4    -3
2   1  3  5  6  8  5   -5     -5     0
3   3  5  6  8  5  1   -3      0     5

I've tried out the following:

newnames <- paste0("diff", 1:3)
v <- paste0("v", 1:3)
x <- paste0("x", 1:3)
dt[ , c(newnames) := get(v) - get(x)]

However, this results in 3 identical columns all containing the difference v1 - x1.

I am aware that a possible solution is something like

dt[ , .(v1 - x1, v2 - x2, v3 - x3)]

However this is quite a long code with possible many typing errors if I have to put in 100 names not as simple as v1 and x1.

I hope you can help me.

Upvotes: 2

Views: 276

Answers (3)

MichaelChirico
MichaelChirico

Reputation: 34733

Your data looks like it belongs in a long format, for which the calculation you're after would become trivial:

# reshape
DT_long = melt(DT, id.vars='Id', measure.vars = patterns(v = '^v', x = '^x'))
DT_long
#       Id variable     v     x
# 1:     1        1     7     5
# 2:     2        1     1     6
# 3:     3        1     3     8
# 4:     1        2     1     6
# 5:     2        2     3     8
# 6:     3        2     5     5
# 7:     1        3     3     8
# 8:     2        3     5     5
# 9:     3        3     6     1

Now it's easy:

DT_long[ , diff := v - x][]
#       Id variable     v     x  diff
# 1:     1        1     7     5     2
# 2:     2        1     1     6    -5
# 3:     3        1     3     8    -5
# 4:     1        2     1     6    -5
# 5:     2        2     3     8    -5
# 6:     3        2     5     5     0
# 7:     1        3     3     8    -5
# 8:     2        3     5     5     0
# 9:     3        3     6     1     5

You can then use dcast to reshape back to wide, but it's usually worth considering whether keeping the dataset in this long form is better for the whole analysis.

Upvotes: 5

Frank Zhang
Frank Zhang

Reputation: 1688

You can use set in the loop.

library(data.table)

DT <- fread('Id  v1 v2 v3 x1 x2 x3
1   7  1  3  5  6  8
2   1  3  5  6  8  5
3   3  5  6  8  5  1')

for (i in 1:3) {
  set(DT,j=paste0("Diff_",i),value = DT[[paste0("v",i)]]-DT[[paste0("x",i)]])
}

DT
#>    Id v1 v2 v3 x1 x2 x3 Diff_1 Diff_2 Diff_3
#> 1:  1  7  1  3  5  6  8      2     -5     -5
#> 2:  2  1  3  5  6  8  5     -5     -5      0
#> 3:  3  3  5  6  8  5  1     -5      0      5

Created on 2020-05-27 by the reprex package (v0.3.0)

Upvotes: 2

IceCreamToucan
IceCreamToucan

Reputation: 28695

You could split by whether the column contains x and then take the difference of the resulting data tables.

new_cols <- 
  do.call('-', split.default(dt[,-1], grepl('x', names(dt)[-1])))

dt[, paste0('diff', seq_along(new_cols)) := new_cols]

dt
#    Id v1 v2 v3 x1 x2 x3 diff1 diff2 diff3
# 1:  1  7  1  3  5  6  8     2    -5    -5
# 2:  2  1  3  5  6  8  5    -5    -5     0
# 3:  3  3  5  6  8  5  1    -5     0     5

Or using similar logic to the code snippet in the question you could do

newnames <- paste0("diff",1:3)
v <- paste0("v",1:3)
x <- paste0("x",1:3)

dt[, (newnames) := Map('-', mget(v), mget(x))]

dt
#    Id v1 v2 v3 x1 x2 x3 diff1 diff2 diff3
# 1:  1  7  1  3  5  6  8     2    -5    -5
# 2:  2  1  3  5  6  8  5    -5    -5     0
# 3:  3  3  5  6  8  5  1    -5     0     5

Upvotes: 4

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