How to sort an array of strings containing numbers and letters in ruby

Question

I have an array of strings, each element contains two or three numbers, one space and one letter.

array = [ "293 C",
 "401 B","421 B","428 B","439 B","315 C","529 B","560 B","566 B",
 "567 B","616 B","39 C","28 C","30 C","698 B","719 B","722 B",
 "640 B","786 B","645 B","236 B","255 B","442 C","446 C","477 C",
 "368 C","381 C","399 C","406 C","504 C","505 C","515 C","116 C",
 "138 C","147 C","174 C"]

What I need as a result is something like this.

["236 B","255 B","401 B","421 B","428 B","439 B","529 B","560 B",
"566 B","567 B","616 B","640 B","645 B","698 B","719 B","722 B",
"786 B","28 C","30 C", "39 C","116 C","138 C","147 C","174 C", "293 C",
"315 C", "368 C", "381 C","399 C","406 C","442 C","446 C",
"477 C","504 C","505 C","515 C"]

In other words, I need to sort the list of strings using the letter suffix as the major sort key, and the integer value of the digit prefix as the minor key.

I've tried with sort_by method, but it only allows me to sort the array by the letter (if I split each element first)

Amadan · Accepted Answer · 2017-07-21 17:16:02Z

5

array.sort_by { |item|
  number, letter = *item.split
  [letter, number.to_i]
}

Arrays compare as their first element; in case an element is equal, next element is compared.

answered Jul 21, 2017 at 17:16

Amadan

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2 Comments

Aleksei Matiushkin Over a year ago

splat is redundant.

Amadan Over a year ago

@mudasobwa: It is. And I am inconsistent in applying it. But the rule that autosplats arrays in multiple assignments is probably the one part of Ruby that I truly dislike.

dawg · Accepted Answer · 2017-07-21 19:08:11Z

2

Another alternative:

array.sort_by {|s| [s[/[a-zA-Z]+/], s.to_i]}

                        ^                      grab the letter or letters and the end
                                      ^        convert the digits at the front

Or, as point out in comments, it is better not to use a latin based character class.

You can do:

array.sort_by { |s| [s[s.index(/[\t ]\S/)+1..-1], s.to_i] }

                                 ^       tab, space followed by not a space

Or,

array.sort_by { |s| [s[/[^\d\t ]+/], s.to_i] }

                            ^             skip digits and spaces -- the rest

Or,

array.sort_by { |s| [s[/\p{L}+/], s.to_i] }

                           ^ code point in the category "letter".

edited Jul 21, 2017 at 19:08

answered Jul 21, 2017 at 17:54

dawg

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9 Comments

Cary Swoveland Over a year ago

...or { |s| [s[/\D+/], s.to_i] }.

dawg Over a year ago

s[/\D+/] picks up leading spaces before the letters. If there ever where 'noise' in the input data, such as an inconsistent number of spaces between the two columns, it would change the sort. s[s.index(/\s\S/)+1..-1] probably better.

Aleksei Matiushkin Over a year ago

@dawg well, it’s a naïve approach to think that letters means latin. I could bring more examples on the table if we were tête-à-tête, но вокруг, к сожалению, много наблюдателей. :)

Aleksei Matiushkin Over a year ago

\p{L} should do.

dawg Over a year ago

Google translate say Следуйте за деньгами => Follow the money

|

Gerry · Accepted Answer · 2017-07-21 17:20:40Z

1

Here is one way to do it:

array.sort_by { |e| [e[-1], e.to_i] }

answered Jul 21, 2017 at 17:20

Gerry

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2 Comments

Aleksei Matiushkin Over a year ago

How is this answer different compared to the one by @Amadan?

Gerry Over a year ago

@mudasobwa Not much, only differs in how the array is formed; and it can be used in a single line without the need of ; (BTW, I wasn't aware of Amadan's answer while posting).

Cary Swoveland · Accepted Answer · 2017-07-21 18:32:21Z

1

array.sort_by { |s| s[-1] << s[/\d+/].rjust(3, '0') }
  #=> ["236 B", "255 B", "401 B",..., "722 B", "786 B",
  #     "28 C",  "30 C",  "39 C",..., "505 C", "515 C"]

Note:

array.map { |s| s[-1] << s[/\d+/].rjust(3, '0') }.sort
  #=> ["B236", "B255", "B401",..., "B722", "B786",
  #    "C028", "C030", "C039",..., "C505", "C515"]

answered Jul 21, 2017 at 18:32

Cary Swoveland

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Comments

Aleksei Matiushkin · Accepted Answer · 2017-07-21 17:48:54Z

0

arr.sort_by { |e| e.split.reverse.map { |s| s.to_i(36) } }

answered Jul 21, 2017 at 17:48

Aleksei Matiushkin

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