CODEWITHSUNDEEP
r
Chargaff
Chargaff

Reputation: 1572

R Aggregate with rle function

struggling with this one :
Sample data : 

    foo <- structure(c("1", "1", "1", "1", "1", "1", "2", "2", "2", "2",
"2", "2", "2", "2", "2", "C", "C", "C", NA, NA, NA, NA, "C",
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, "C",
"C", NA, NA, NA, NA, NA, "C", "C", "C", "C", "C", "C", "C", "C",
"C", "C", "C", NA, NA, NA, NA, "C", "C", "C", "C", "C", "C",
"C", "C", NA, NA, NA, NA, NA, NA, NA, NA, "C", "C", "C", NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, "C", "C", "C", NA, NA,
NA, NA, NA, "C", "C", NA, NA, NA, NA, NA, "C", "C", "C", NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, "C", "C", NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, "C", "C", "C", "C",
"C", "C", "C", "C", "C", "C", "C", "C", "C", "C", "C", "C", "C",
"C", NA, NA, "C", "C", "C", "C", "C", NA, "C", "C", "C", "C",
"C", "C", "C", "C", "C", "C", "C", "C", "C", "C", "C", "C", "C",
"C", "C", NA, NA, "C", "C", NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA), .Dim = c(15L, 13L), .Dimnames = list(NULL, c("weeks",
"vv", "rv", "ja", "aa", "bv", "aj", "vb", "rj", "rr", "vr", "bb",
"jr")))

How to group theses datas by weeks and apply rle function ? Expected output would be : 

out <- structure(c("weeks", "1", "2", "vv", "1", "1", "rv", "0", "1",
"ja", "1", "1", "aa", "1", "1", "bv", "1", "0", "aj", "1", "1",
"vb", "1", "0", "rj", "1", "0", "rr", "1", "1", "vr", "2", "2",
"bb", "1", "1", "jr", "1", "0"), .Dim = c(3L, 13L), .Dimnames = list(
    NULL, NULL))

Tried so far :

test <- aggregate(foo, by=data.frame(foo[,1]), function(x,na.rm=T) rle(as.numeric(x))$values)

Gives me a list (as expected with rle) which I could then compute and try to get the expected output. But I'm sure there is a cleaner way to accomplish that... Any ideas ?

Thanks for looking !

Upvotes: 2

Views: 598

Answers (2)

Aaron - mostly inactive
Aaron - mostly inactive

Reputation: 37754

Apparently a bunch of ways to process the rle output; here's another.

d <- data.frame(foo)
aggregate(!is.na(d[,-1]), d[,1,drop=FALSE], function(x) sum(rle(x)$values))
#   weeks vv rv ja aa bv aj vb rj rr vr bb jr
# 1     1  1  0  1  1  1  1  1  1  1  2  1  1
# 2     2  1  1  1  1  0  1  0  0  1  2  1  0

Upvotes: 4

Justin
Justin

Reputation: 43255

I wound up with something very similar... not sure which answer you're looking for:

aggregate(list(foo[, -1]), list(weeks = foo[, 1]), 
          function(x) length(na.omit(rle(x)$values)))

  weeks vv rv ja aa bv aj vb rj rr vr bb jr
1     1  1  0  1  1  1  1  1  1  1  2  1  1
2     2  1  1  1  1  0  1  0  0  1  2  1  0

Upvotes: 4

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