CODEWITHSUNDEEP
r
rnso
rnso

Reputation: 24623

Creating combination of sequences

I am trying to solve following problem:

Consider 5 simple sequences: 0:100, 100:0, rep(0,101), rep(50,101), rep(100,101)

I need sets of 3 numeric variables, which have above sequences in all combinations. Since there are 5 sequences and 3 variables, there can be 5*5*5 combinations, hence total of 12625 (5*5*5*101) numbers in each variable (101 for each sequence).

These can be grouped in a data.frame of 12625 rows and 4 columns. First column (V) will simply have seq(1:12625) (rownumbers can be used in its place). Other 3 columns (A,B,C) will have above 5 sequences in different combinations. For example, the first 101 rows will have 0:100 in all 3 A,B and C. Next 101 rows will have 0:100 in A and B, and 100:0 in C. And so on...

I can create sequences as:

s = list()
s[[1]] = 0:100
s[[2]] = 100:0
s[[3]] = rep(0,101)
s[[4]] = rep(50,101)
s[[5]] = rep(100,101)

But how to proceed further? I do not really need the data frame but I need a function that returns a list containing the values of c(A,B,C) for the number (first or V column) sent to it. The number can obviously vary from 1 to 12625.

How can I create such a function. I will prefer a vector solution or one using apply family functions to optimize the speed.

Upvotes: 2

Views: 1251

Answers (2)

David Arenburg
David Arenburg

Reputation: 92300

You asked for a vectorized solution, so here's one using only data.table (similar to @SimonGs methodology) 

library(data.table)
grd <- CJ(A = seq_len(5), B = seq_len(5), C = seq_len(5))
res <- grd[, lapply(.SD, function(x) unlist(s[x]))]
res
#          A   B   C
#     1:   0   0   0
#     2:   1   1   1
#     3:   2   2   2
#     4:   3   3   3
#     5:   4   4   4
#  ---            
# 12621: 100 100 100
# 12622: 100 100 100
# 12623: 100 100 100
# 12624: 100 100 100
# 12625: 100 100 100

Upvotes: 4

SimonG
SimonG

Reputation: 4881

I came up with two solutions. I find this hard to do with apply and the likes since they tend to give an output that is not so nice to handle (maybe someone can "tame" them better than I can :D)

First solution uses seperate calls to lapply, second one uses a for loop and some programming No-No's. Personally I prefer the second one, first one is faster though...

grd <- expand.grid(a=1:5,b=1:5,c=1:5)

# apply-ish
A <- lapply(grd[,1], function(z){ s[[z]] })
B <- lapply(grd[,2], function(z){ s[[z]] })
C <- lapply(grd[,3], function(z){ s[[z]] })
dfr <- data.frame(A=do.call(c,A), B=do.call(c,B), C=do.call(c,C))

# for-ish
mat <- NULL
for(i in 1:nrow(grd)){
 cur <- grd[i,]
 tmp <- cbind(s[[cur[,1]]],s[[cur[,2]]],s[[cur[,3]]])
 mat <- rbind(mat,tmp)
}

The output of both dfr and mat seem to be what you describe.

Cheers!

Upvotes: 2

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