Sorting character vector containing semantic versions

Question

Seems like a pretty basic question, but I can't really figure out an "easy" way to do it.

I'd like to sort a character vector containing semantic version numbers with base R functionality:

vsns  <- c("1", "10", "1.1", "1.10", "1.2", "1.1.1", 
           "1.1.10", "1.1.2", "1.1.1.1", "1.1.1.10", "1.1.1.2")

It should look like this after sorting:

# [1] "1"        "1.1"      "1.1.1"    "1.1.1.1"  "1.1.1.2"  "1.1.1.10"
# [7] "1.1.2"    "1.1.10"   "1.2"      "1.10"     "10"

This doesn't get me what I want, of cours, as R simply sorts the whole thing alphabetically:

sort(vsns)
# [1] "1"        "1.1"      "1.1.1"    "1.1.1.1"  "1.1.1.10" "1.1.1.2"  "1.1.10"  
# [8] "1.1.2"    "1.10"     "1.2"      "10"    
vsns[order(vsns)]
# [1] "1"        "1.1"      "1.1.1"    "1.1.1.1"  "1.1.1.10" "1.1.1.2"  "1.1.10"  
# [8] "1.1.2"    "1.10"     "1.2"      "10"

Trying normalizing it (somewhat along this post), but I can't think of a matching/substitution scheme that would fit the structure of semantic versions:

tmp <- gsub("\\.", "", vsns)
# [1] "011"  "021"  "0101" "0201"
tmp_nchar <- sapply(tmp, nchar)
to_add <- max(tmp_nchar) - tmp_nchar
tmp <- sapply(1:length(tmp), function(ii) {
  paste0(tmp[ii], paste(rep("A", to_add[ii]), collapse = ""))
})
# [1] "10"       "1.10"     "1.1.10"   "1.1.1.10" "1.1.1.1"  "1.1.1.2"  "1.1.1"   
# [8] "1.1.2"    "1.1"      "1.2"      "1"   
vsns[order(tmp)]
#  [1] "1AAAA" "10AAA" "11AAA" "110AA" "12AAA" "111AA" "1110A" "112AA" "1111A" "11110"
# [11] "1112A"

The best I could come up with so far is this, but it seems pretty... Involved ;-)

sortVersionNumbers <- function(x, decreasing = FALSE) {
  tmp <- strsplit(x, split = "\\.")  
  tmp_l <- sapply(tmp, length)  
  idx_max <- which.max(tmp_l)[1]
  tmp_l_max <- tmp_l[idx_max]
  tmp_n <- lapply(tmp, function(ii) {
    ii_l <- length(ii)
    if (ii_l < tmp_l_max) {
      c(ii, rep(NA, (tmp_l_max - ii_l)))
    } else {
      ii
    }
  })
  tmp <- matrix(as.numeric(unlist(tmp_n)), nrow = length(tmp_n), byrow = TRUE)
  tmp_cols <- ncol(tmp)
  expr <- paste0("order(", paste(paste0("tmp[,", 1:tmp_cols, "]"), 
    collapse = ", "), ", na.last = FALSE",
    ifelse(decreasing, ", decreasing = FALSE)", ")"))
  idx <- eval(parse(text = expr))
  tmp_2 <- tmp[idx,]  
  sapply(1:nrow(tmp_2), function(ii) {
    paste(na.omit(tmp_2[ii,]), collapse = ".")
  })
}
sortVersionNumbers(vsns)
# [1] "1"        "1.1"      "1.1.1"    "1.1.1.1"  "1.1.1.2"  "1.1.1.10" "1.1.2"   
# [8] "1.1.10"   "1.2"      "1.10"     "10" 
sortVersionNumbers(sort(vsns))
# [1] "1"        "1.1"      "1.1.1"    "1.1.1.1"  "1.1.1.2"  "1.1.1.10" "1.1.2"   
# [8] "1.1.10"   "1.2"      "1.10"     "10"

Can we assume that all version numbers would be nicely organized and have the same number of `.`s? — A5C1D2H2I1M1N2O1R2T1, Sep 25 '14 at 08:56
No, as I'd like to leave me the option to have a {major}.{minor}.{patch}.{dev} scheme. So it can be anything from 1 to 4 "semantic blocks". — Rappster, Sep 25 '14 at 08:58
Can you give a couple of different possible inputs and expected outputs? — A5C1D2H2I1M1N2O1R2T1, Sep 25 '14 at 09:00
Yeah, sorry. Just realized that the example doesn't stress that point yet. I'll update it — Rappster, Sep 25 '14 at 09:01

Martin Morgan · Accepted Answer · 2014-09-25T11:29:24.140

From ?numeric_version

> sort(numeric_version(vsns))
 [1] '1'        '1.1'      '1.1.1'    '1.1.1.1'  '1.1.1.2'  '1.1.1.10'
 [7] '1.1.2'    '1.1.10'   '1.2'      '1.10'     '10'

It's relatively interesting to see how this is implemented. numeric_version splits a single version string into integer parts, and stores the vector of versions as a list of integer vectors. A method on xtfrm (which is used by sort()) transforms the vector of integers making up each version string into a numeric value, with the guts being

base <- max(unlist(x), 0, na.rm = TRUE) + 1                                 
x <- vapply(x, function(t) sum(t/base^seq.int(0, length.out = length(t))), 
    1)

the result is a numeric vector that can be used to order the original vector in a standard way. Thus an ad hoc solution is

xtfrm.my_version <- function(x) {
    x <- lapply(strsplit(x, ".", fixed=TRUE), as.integer)
    base <- max(unlist(x), 0, na.rm = TRUE) + 1
    vapply(x, function(t) sum(t/base^seq.int(0, length.out = length(t))), 1)
}

vsns  <- c("1", "10", "1.1", "1.10", "1.2", "1.1.1",
           "1.1.10", "1.1.2", "1.1.1.1", "1.1.1.10", "1.1.1.2")
class(vsns) = "my_version"
sort(vsns)

Yes!! Those days when you find out about a little helper that was always there but that you never knew about ;-) Thanks a lot, exactly what I was looking for! — Rappster, Sep 25 '14 at 09:35

score 2 · Answer 2 · answered Sep 25 '14 at 09:26

2

Does this work

vsns  <- c("1", "10", "1.1", "1.10", "1.2", "1.1.1", 
       "1.1.10", "1.1.2", "1.1.1.1", "1.1.1.10", "1.1.1.2")
x <- strsplit(vsns, "\\.")
max.length <- max(sapply(x, function(i) max(nchar(i))))
y <- lapply(x, function(i) sprintf(as.numeric(i), fmt = paste0("%0", max.length, "d")))
y <- sapply(y, paste, collapse = ".")
vsns[order(y)]
# [1] "1"        "1.1"      "1.1.1"    "1.1.1.1"  "1.1.1.2"  "1.1.1.10"
# [7] "1.1.2"    "1.1.10"   "1.2"      "1.10"     "10"

answered Sep 25 '14 at 09:26

konvas

14,126
2
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Looks good! `[1] "1" "1.1" "1.1.1" "1.1.1.1" "1.1.1.2" "1.1.1.10" "1.1.2" [8] "1.1.10" "1.2" "1.10" "10"` – Rappster Sep 25 '14 at 09:28
Yeah I wasn't sure how you wanted to order it exactly. But if you add zeros before numbers with fewer digits then sorting the string works :) – konvas Sep 25 '14 at 09:29

score 1 · Answer 3 · answered Sep 25 '14 at 09:18

1

Try:

ll  = strsplit(vsns,'\\.')
dd = data.frame(t(sapply(ll, c)))
dd = data.frame(apply(dd, 2, function(x) as.numeric(as.character(x))))
dd = with(dd, dd[order(X1,X2,X3),])
ans = apply(dd, 1, paste, collapse=".")
ans
       1        2        3        4 
 "0.1.1"  "0.2.1" "0.10.1" "0.20.1"

answered Sep 25 '14 at 09:18

rnso

23,686
25
112
234

That threw some errors, but you also had the outdated `vsns` string to start with. Sorry for that, I updated my question accordingly – Rappster Sep 25 '14 at 09:30

rnso · Answer 4 · 2014-09-25T12:05:51.923

Try, with new vsns data:

vsns  <- c("1", "10", "1.1", "1.10", "1.2", "1.1.1", "1.1.10", "1.1.2", "1.1.1.1", "1.1.1.10", "1.1.1.2")
dd = data.frame(vsns)
library(splitstackshape)
dd2 = concat.split.expanded(dd, 'vsns', '.', fill = 0, drop = TRUE)
dd3 = cbind(dd, dd2)
dd4= with(dd3, dd3[order(vsns_1, vsns_2, vsns_3, vsns_4),])
dd4[is.na(dd4)]=0
dd4
       vsns vsns_1 vsns_2 vsns_3 vsns_4
9   1.1.1.1      1      1      1      1
11  1.1.1.2      1      1      1      2
10 1.1.1.10      1      1      1     10
6     1.1.1      1      1      1      0
8     1.1.2      1      1      2      0
7    1.1.10      1      1     10      0
3       1.1      1      1      0      0
5       1.2      1      2      0      0
4      1.10      1     10      0      0
1         1      1      0      0      0
2        10     10      0      0      0
> 
apply(dd4[,2:5], 1, paste, collapse='.')
          9          11          10           6           8           7           3           5           4           1 
 " 1.1.1.1"  " 1.1.1.2" " 1.1.1.10"  " 1.1.1.0"  " 1.1.2.0" " 1.1.10.0"  " 1.1.0.0"  " 1.2.0.0" " 1.10.0.0"  " 1.0.0.0" 
          2 
 "10.0.0.0"

Nice one as well. That `splitstackshape` package might in come handy some day, didn't know that yet. Thanks! — Rappster, Sep 25 '14 at 12:08

score 0 · Answer 5 · answered Sep 25 '14 at 09:30

0

Here's a solution that generalizes to version numbers with different numbers of blocks (the indented sapply + ifelse lines) and can handle mixed numbers and letters (the mixedsort line).

library(gtools)
vsns  <- c("0.1.1", "0.10", "0.2.1", "0.2.1a", "0.20", "0.20.1.3")
v <- strsplit(vsns, "\\.")
tmp <- data.frame(sapply(1:max(sapply(v, length)), function(i){
    vv <- sapply(v, "[", i)
    ifelse(is.na(vv), "0", vv)
}), stringsAsFactors=FALSE)
vsns[do.call(mixedorder, tmp)]

[1] "0.1.1"    "0.2.1"    "0.2.1a"   "0.10"     "0.20"     "0.20.1.3"

answered Sep 25 '14 at 09:30

Backlin

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Thanks for the pointer. But I'd rather use base R functionality for that. Plus, with my `vnsn`, it threw an error. – Rappster Sep 25 '14 at 09:33
Sure thing, but weird that you get an exception with your `vsns`. It works fine for me. – Backlin Sep 25 '14 at 09:36
I get this: `Error in (function (x) : unused arguments (X1 = c("0", "0", "0", "0", "0", "0"), X2 = c("1", "10", "2", "2", "20", "20"), X3 = c("1", "0", "1", "1a", "0", "1"), X4 = c("0", "0", "0", "0", "0", "3"))` – Rappster Sep 25 '14 at 09:36
Ok, won't debug it further since you already accepted the best answer though. – Backlin Sep 25 '14 at 09:39

Sorting character vector containing semantic versions

5 Answers5

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