CODEWITHSUNDEEP
rvectorcountrank
Paul Manning
Paul Manning

Reputation: 1

ordering a vector in R while ignoring yet keeping NAs

If I have a vector a = c(1300,NA,NA,NA,NA,1500,NA,NA,6000,NA,NA,900)

How can I this to result in:

b=[2,NA,NA,NA,NA,3,NA,NA,4,NA,NA,5]?

Sidenote :I tried to make them repeat so it was

a=[1300,1300,1300,1300,1300,1500,1500,1500,6000,6000,6000,900]

But when I use rank its gets some crazy half numbers, any ideas? I'm at wits end for figuring this out.

Keeping the amount of NAs after a number is very important here so I cant just only ignore them

Upvotes: 0

Views: 190

Answers (3)

Rich Scriven
Rich Scriven

Reputation: 99331

Take a ^ is.na(a) and multiply it by rank(a). We use ties="first" to ensure we get increasing values at each index, not averages.

rank(a, ties="first") * a ^ is.na(a)
# [1]  2 NA NA NA NA  3 NA NA  4 NA NA  1

Upvotes: 1

d.b
d.b

Reputation: 32548

replace(a, !is.na(a), rank(a[!is.na(a)], ties.method = "first"))
# [1]  2 NA NA NA NA  3 NA NA  4 NA NA  1

Upvotes: 1

Gregor Thomas
Gregor Thomas

Reputation: 145775

The dplyr::dense_rank function behaves as you want:

library(dplyr)
dense_rank(a)
# [1]  2 NA NA NA NA  3 NA NA  4 NA NA  1

It also works on the dense vector:

b = c(1300,1300,1300,1300,1300,1500,1500,1500,6000,6000,6000,900)
dense_rank(b)
# [1] 2 2 2 2 2 3 3 3 4 4 4 1

Upvotes: 1

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