R中的缩短向量（在data.frame内）多重化代码

Question

I have this code to do a series of multiplication and addition in R. Could someone give me a suggestion to use *apply to make it neater and shorter?

I looked around with keywords such as "series", "multiplication" but did not get anywhere. If this question has been posted before, please let me know the link. Thank you.

df1n is a data.frame with >78 variables and 215 observation.

dff[,3]<-(df1n[5]*((df1n[25]*df1n[26]*df1n[27]) + (df1n[28]*df1n[29]*df1n[30]) +
         (df1n[31]*df1n[32]*df1n[33]) + (df1n[34]*df1n[35]*df1n[36]) + 
         (df1n[37]*df1n[38]*df1n[39]) + (df1n[40]*df1n[41]*df1n[42]) +
         (df1n[61]*df1n[62]*df1n[63]) + (df1n[64]*df1n[65]*df1n[66]) + 
         (df1n[67]*df1n[68]*df1n[69]) + (df1n[70]*df1n[71]*df1n[72]) + 
         (df1n[73]*df1n[74]*df1n[75]) + (df1n[76]*df1n[77]*df1n[78]))
         )

Regards, ikel

Answer 1

I'd use lapply() to construct a list of indices for each of the (A*B*C) pieces. Then, pass those indices to sapply , extracting each set of elements from df1n and multiplying them with prod() . The rest is self-explanatory:

df1n <- 1:100
ll <- lapply(c(seq(25, 40, by=3), seq(61, 76, by=3)), 
             function(X) seq(X, by=1, length.out=3))
df1n[5] * sum(sapply(ll, function(i) prod(df1n[i])))
# [1] 11439180

---

EDIT : Now that I know that each element of df1n is a length-215 vector, here's the code I'd suggest instead:

# Example data (a list in which each element is a vector of length 215).
x <- replicate(100, 1:215, simplify=FALSE) 
ll <- lapply(seq(25, 76, by=3), function(X) seq(X, by=1, length.out=3))
res <- x[[5]] * rowSums(sapply(ll, function(i) Reduce("*", x[i])))

str(res)
# num [1:215] 18 288 1458 4608 11250 ...

Answer 2

Here's a modification of Josh's solution that produces the same ouput as your original code on the sample data below.

# Some sample data
df1n <- as.data.frame( lapply(1:80, function(i) seq(i, len=5)) )

df1n[5] * rowSums(sapply(c(seq(25,40,3), seq(61,76,3)), 
    function(i) df1n[[i]]*df1n[[i+1]]*df1n[[i+2]]))
#  c.5..6..7..8..9.
#1         11439180
#2         14386680
#3         17580528
#4         21032640
#5         24755220