Make a Consecutive row calculation in R

Question

Suppose I have the following data:

a<- c(1:10)
b<- c(10:1)

Now I want to make a Consecutive calculation (of variable length in this example 2)on both rows (a and b) and save the output in two separate lists(a and b).

The calculation should look like the following:

for a:

(1+2)/2; (2+3)/2; (3+4)/2;...; (9+10)/2

for b(the same):

(10+9)/2; (9+8)/2; (8+7)/2;...;(2+1)/2

a
1,5 2,5 3,5 ... 9,5
b
9,5 8,5 7,5 ... 1,5

I found this function in StackOverflow:

v <- c(1, 2, 3, 10, 20, 30)
grp <- 3

res <- sapply(1:(length(v)-grp+1),function(x){sum(v[x:(x+grp-1)])})

Which pretty much does what i need but I would prefer a function which does that without using sapply and just base R. Any Help would be appreciated!

Colonel Beauvel · Accepted Answer · 2015-10-06T14:17:10.030

5

You can do base R:

f = function(x) (head(x,-1) + tail(x,-1))/2

list(a=f(a), b=f(b))
#$a
#[1] 1.5 2.5 3.5 4.5 5.5 6.5 7.5 8.5 9.5

#$b
#[1] 9.5 8.5 7.5 6.5 5.5 4.5 3.5 2.5 1.5

Or if you want to use the apply family:

library(zoo)

list(a=rollapply(a,2, mean), b=rollapply(b,2, mean))

sapply is really not recommended but if you want to use it (just for test!):

sapply(1:(length(a)-1), function(i) mean(a[i:(i+1)]))
#[1] 1.5 2.5 3.5 4.5 5.5 6.5 7.5 8.5 9.5
#same for b

edited Oct 06 '15 at 14:17

answered Oct 06 '15 at 13:59

Colonel Beauvel

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Pretty quick draw there – David Arenburg Oct 06 '15 at 14:01
1

Too much vitamin this mornin'. Just luck actually. – Colonel Beauvel Oct 06 '15 at 14:02
Thank you! Just one question is it possible to adjust your fist answer so that I could choose how many Consecutive rows to sum up? So for example doing the same calculation like: `(1+2+3)2 ; (3+4+5)/2 etc.` . I tried by setting -1 to -2 but that did not yield the right result. – Max Mustermann Oct 07 '15 at 07:59
use the `rollapply` function, it is far more simple in this case :) – Colonel Beauvel Oct 07 '15 at 08:02

score 3 · Answer 2 · answered Oct 06 '15 at 14:20

3

na.omit(filter(a, c(1,1))/2)
na.omit(filter(b, c(1,1))/2)

answered Oct 06 '15 at 14:20

jogo

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RHertel · Answer 3 · 2015-10-06T14:37:42.163

1

You could try this:

d1 <- ((a + a[seq(a)+1])/2)[-length(a)]
#[1] 1.5 2.5 3.5 4.5 5.5 6.5 7.5 8.5 9.5

and

d2 <- ((b + b[seq(b)+1])/2)[-length(b)]
#[1] 9.5 8.5 7.5 6.5 5.5 4.5 3.5 2.5 1.5

The last part [-length(a)] and [-length(b)] removes NA entries at the end of the sequence.

edited Oct 06 '15 at 14:37

answered Oct 06 '15 at 14:07

RHertel

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score 0 · Answer 4 · answered Oct 06 '15 at 14:07

If the length of a and b is same

for(i in 1:(length(a) - 1))
{
 list1[i] <- (a[i] + a[i+1])/2
 list2[i] <- (b[i] + b[i+1])/2
}
> list1
#[1] 1.5 2.5 3.5 4.5 5.5 6.5 7.5 8.5 9.5

> list2
#[1] 9.5 8.5 7.5 6.5 5.5 4.5 3.5 2.5 1.5

Or else write two different loops for both

for(i in 1:(length(a) - 1))
{
  list1[i] <- ((a[i] + a[i+1])/2)
}

for(i in 1:(length(b) - 1))
{
  list2[i] <- ((b[i] + b[i+1])/2)
}

Make a Consecutive row calculation in R

4 Answers4