I have 2 vectors as below
Vec1 = c(1,2,3,4)
names(Vec1) = c('Val1', 'Val2', 'Val3', 'Val4')
Vec2 = c(10, 11)
names(Vec2) = c('Val2', 'Val4')
Is there any easy way to subtract Vec2 from Vec1 based on same names in the vector element? The names which are not available in Vec2 can be assumed to have zero value.
`[<-`(Vec1, names(Vec2), Vec1[names(Vec2)] - Vec2)
## OR using replace() without creating a helper variable
Vec1 |> replace(names(Vec2), Vec1[names(Vec2)] - Vec2)
Result:
#> Val1 Val2 Val3 Val4
#> 1 -8 3 -7
If Vec1 always has the full length (and Vec2 may or may not have missing entries) an approach using replace
Vecs <- Vec2[names(Vec1)]
Vec1 - replace(Vecs, is.na(Vecs), 0)
Val1 Val2 Val3 Val4
1 -8 3 -7
R makes things easy when your objects are in a data frame. If we create a column containing -Vec2 we can take the rowSums() to get the desired output:
rowSums(data.frame(Vec1, -Vec2[names(Vec1)]), na.rm = TRUE)
Output:
Val1 Val2 Val3 Val4
1 -8 3 -7
Alternatively, and probably slightly more efficiently, you can create a matrix:
rowSums(cbind(Vec1, -Vec2[names(Vec1)]), na.rm = TRUE)
# ^^ same output
You might be tempted to think that this is less efficient, as a matrix is a memory contiguous array, whereas a data frame is a list of memory-contiguous vectors with copy-on-modify semantics. This means data.frame(Vec1, Vec2) does not copy Vec1 or Vec2, whereas cbind(Vec1, Vec2) copies both.
This is true, but the source for rowSums() contains the line if(is.data.frame(x)) x <- as.matrix(x), so you end up creating a matrix anyway. If your data is very large and you want to avoid copies as much as possible you should use the answer by Andre Wildberg, which at least avoids copying Vec1. However, it's usually better to optimise for readability and (particularly if you might expand this to more vectors) my view is that putting the data in tabular form makes the most sense.
Using match() to Align Indices:
idx <- match(names(Vec2), names(Vec1))
Vec1[idx] <- Vec1[idx] - Vec2
Val1 Val2 Val3 Val4
1 -8 3 -7
%in% and match is "%in%" <- function(x, table) match(x, table, nomatch = 0) > 0.We can use merge with some cosmeticts (not needed)
merge(data.frame(Vec1, n=names(Vec1)),
data.frame(-Vec2, n=names(Vec2)), by='n', all.x=TRUE) |>
subset(select=-n) |>
rowSums(na.rm=TRUE) |> # or replace
setNames(names(Vec1))
Val1 Val2 Val3 Val4 1 -8 3 -7
Didn't know this is about speed. As mentioned in a now deleted comment posted shortly after the question: "use match with names and simplest indexing.". {collapse} should be hard to beat:
Vec1[i] = Vec1[i<-collapse::fmatch(names(Vec2), names(Vec1))]-Vec2
Val1 Val2 Val3 Val4 1 -8 3 -7
ifelse(hasName(Vec2, names(Vec1)), Vec1 - Vec2[names(Vec1)], Vec1)
# [1] 1 -8 3 -7
EDIT: hasName() replaced %in%.
You can try
> library(data.table)
> Vec1 - nafill(Vec2[names(Vec1)], fill = 0)
Val1 Val2 Val3 Val4
1 -8 3 -7
set.seed(0)
n1 <- 1e5
Vec1 <- setNames(sample.int(n1), paste0("v", 1:n1))
n2 <- 1e2
Vec2 <- setNames(sample.int(n2), paste0("v", sample(n1, n2)))
microbenchmark(
`M--1` = `[<-`(Vec1, names(Vec2), Vec1[names(Vec2)] - Vec2),
`M--2` = Vec1 |> replace(names(Vec2), Vec1[names(Vec2)] - Vec2),
`SamR` = rowSums(data.frame(Vec1, -Vec2[names(Vec1)]), na.rm = TRUE),
`Andre` = {
Vecs <- Vec2[names(Vec1)]
Vec1 - replace(Vecs, is.na(Vecs), 0)
},
`SAL` = {
idx <- match(names(Vec2), names(Vec1))
Vec1[idx] <- Vec1[idx] - Vec2
},
`Friede` = {
merge(data.frame(Vec1, n = names(Vec1)),
data.frame(-Vec2, n = names(Vec2)),
by = "n", all.x = TRUE
) |>
subset(select = -n) |>
rowSums(na.rm = TRUE) |> # or replace
setNames(names(Vec1))
},
`s_baldur` = ifelse(names(Vec1) %in% names(Vec2), Vec1 - Vec2[names(Vec1)], Vec1),
`Thomas` = Vec1 - data.table::nafill(Vec2[names(Vec1)], fill = 0),
unit = "relative",
times = 50L
)
and you will see the base R solution provide by @SAL is the most efficient
Unit: relative
expr min lq mean median uq max
M--1 2.513836 2.490961 2.909901 2.534369 2.738404 3.010798
M--2 2.616862 2.362815 2.629789 2.403732 2.533827 3.291382
SamR 10.188135 8.827689 10.617243 9.646134 10.883201 16.071042
Andre 4.191419 4.021243 5.773643 4.326937 5.428873 18.314273
SAL 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000
Friede 232.612582 192.581732 180.411511 187.376767 177.789439 123.654579
s_baldur 7.437139 6.561556 7.051161 7.389931 7.969211 5.450089
Thomas 4.000498 3.631732 4.174856 3.677764 5.209227 3.736972
neval
50
50
50
50
50
50
50
50
{collapse} option to my answer.
match(names(Vec2), names(Vec1)) |> (\(idx) {Vec1[idx] <- Vec1[idx] - Vec2; Vec1})()