# Partitioning with Set.divide

**URL:** <https://www.ruby-forum.com/t/partitioning-with-set-divide/93214>\
**Category:** Ruby\
**Created:** [April 29, 2007, 2:07pm UTC](https://www.ruby-forum.com/t/partitioning-with-set-divide/93214 "2007-04-29T14:07:49Z")\
**Posts on this page:** 7\
**Page:** 1

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**Author:** ![splattael](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/s/858c86/32.png) [@splattael](https://www.ruby-forum.com/u/splattael)\
**Post date:** [April 29, 2007, 2:07pm UTC](https://www.ruby-forum.com/t/partitioning-with-set-divide/93214/1 "2007-04-29T14:07:49Z")

</div>

Hello all,

I have been playing with partitioning a set recently and I am stuck with  
an issue. The whole story is here:

[http://www.rubyrailways.com/partitioning-sets-in-ruby/](http://www.rubyrailways.com/partitioning-sets-in-ruby/)

A quick version for those who would not like to read the article:

Consider this input:

a 53 2 3  
b 8 62 1 23  
a 9 0 31  
b 4 45 4 16 7  
b 1 23  
c 3 42 2 31 4 6  
a 1 3 22  
a 7 83 1 23 3  
b 1 14 4 15 16 2  
c 5 16 2 34

the goal is to create a partition based on the character in the first  
column, i.e.:

\<Set: \<Set: {“a 9 0 31”, “a 7 83 1 23 3”, “a 53 2 3”, "a 1 3 22 "}\>,  
\<Set: {"b 1 23 ", “b 1 14 4 15 16 2”, “b 8 62 1 23”, “b 4 45 4 16 7”}\>,  
\<Set: {“c 5 16 2 34”, “c 3 42 2 31 4 6”}\>}\>

Which is exactly what Set.divide does. However, there is one problem: I  
would like to know if there are duplicate lines. I.e. divide returns the  
same result, no matter that the input is this:

c 5 16 2 34  
c 5 16 2 34  
c 5 16 2 34

or this:

c 5 16 2 34

What I would need is a modified divide which returns also the count of  
the elements in the input set (at least for those elements which are  
more than once in the set). Is this doable or do I have to roll some  
code to do this for me additionally?

Cheers,  
Peter

\_\_  
[http://www.rubyrailways.com](http://www.rubyrailways.com) :: Ruby and Web2.0 blog  
[http://scrubyt.org](http://scrubyt.org) :: Ruby web scraping framework  
[http://rubykitchensink.ca/](http://rubykitchensink.ca/) :: The indexed archive of all things Ruby

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<div class="post-metadata">

**Author:** ![splattael](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/s/858c86/32.png) [@splattael](https://www.ruby-forum.com/u/splattael)\
**Post date:** [April 29, 2007, 2:57pm UTC](https://www.ruby-forum.com/t/partitioning-with-set-divide/93214/2 "2007-04-29T14:57:56Z")

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Well, you could first count how many repetitions there are of each  
line and then partition the set of pairs [line, count].

e.g.

h = Hash.new {0} # is this how I set a default value?

STDIN.each\_line {|l| h[l] += 1}

partitioning = Set.new(h.to\_a).divide{|a| a[0][0]}

Aur

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<div class="post-metadata">

**Author:** ![splattael](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/s/858c86/32.png) [@splattael](https://www.ruby-forum.com/u/splattael)\
**Post date:** [April 29, 2007, 3:02pm UTC](https://www.ruby-forum.com/t/partitioning-with-set-divide/93214/3 "2007-04-29T15:02:56Z")

</div>

On 4/29/07, SonOfLilit [removed\_email\_address@domain.invalid](mailto:removed_email_address@domain.invalid) wrote:

> > A quick version for those who would not like to read the article:  
> > a 7 83 1 23 3  
> > Which is exactly what Set.divide does. However, there is one problem: I
> > 
> > [http://scrubyt.org](http://scrubyt.org) :: Ruby web scraping framework  
> > [http://rubykitchensink.ca/](http://rubykitchensink.ca/) :: The indexed archive of all things Ruby

Even better:  
h = Hash.new {0} # is this how I set a default value?  
STDIN.each\_line {|l| h[l] += 1}  
partitioning = h.to\_set.divide{|a| a[0][0]} # changed to .to\_set,  
which is great. might need to be .to\_set{|k,v| k,v}

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<div class="post-metadata">

**Author:** ![splattael](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/s/858c86/32.png) [@splattael](https://www.ruby-forum.com/u/splattael)\
**Post date:** [April 29, 2007, 3:42pm UTC](https://www.ruby-forum.com/t/partitioning-with-set-divide/93214/4 "2007-04-29T15:42:01Z")

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Sorry for the triple post, but I’ve read the article and propose a  
completely different approach to the problem you present (also in a  
comment there with typos):

h = {}  
open(‘input.txt’).each\_line{|l| h[l[0…0]] += l[2…-1].split(’  
').inject(0) {|c,x| c+=x.to\_i; c}}}  
p h.map{|k,v| {k =\> v}} # to turn {“a” =\> 80, “b” =\> 60} into [{“a” =\>  
80}, {b =\> 60}]

# which is a pretty weird data structure IMHO, but I’ll play by your

rules  
Let’s look at your code:  
#############  
input = Set.new open(‘input.txt’).readlines.map{|e| e.chomp}  
groups = input.divide {|x,y| x.map[0][0] == y.map[0][0] } # what’s map  
for?  
#build the array of hashes  
p groups.map.inject([]) {|a,g| # what’s map for?  
#build the hashes for the number sequences with same letters  
a \<\< g.map.inject(Hash.new(0)) {|h,v|  
#for every sequence, sum the numbers it contains  
h[v[0…0]] += v[2…-1].split(’ ').inject(0) {|c,x|  
c+=x.to\_i; c}; h  
}; a  
}  
#############  
divide() seems redundant in your code. You both 1) divide(); and 2)  
implement divide on your own; in the same code. You do four passes on  
the lines (and maybe more in the map() calls, I can’t figure those  
out), in one of them passing twice on each line’s content, forcing the  
data to stay all in memory the whole time.

My code, which isn’t debugged but shows my idea, passes once on the  
lines, and in that pass twice on each line’s contents. It doesn’t keep  
the data in memory.

BTW I couldn’t understand the use of map() without a block. Mind  
explaining?

Aur S.

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<div class="post-metadata">

**Author:** ![splattael](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/s/858c86/32.png) [@splattael](https://www.ruby-forum.com/u/splattael)\
**Post date:** [April 29, 2007, 3:44pm UTC](https://www.ruby-forum.com/t/partitioning-with-set-divide/93214/5 "2007-04-29T15:44:02Z")

</div>

Ah, sorry. [tryruby.hobix.com](http://tryruby.hobix.com) helped me find that hash.map() is  
identical to hash.to\_a

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<div class="post-metadata">

**Author:** ![splattael](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/s/858c86/32.png) [@splattael](https://www.ruby-forum.com/u/splattael)\
**Post date:** [April 29, 2007, 3:55pm UTC](https://www.ruby-forum.com/t/partitioning-with-set-divide/93214/6 "2007-04-29T15:55:02Z")

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SonOfLilit wrote:

> Ah, sorry. [tryruby.hobix.com](http://tryruby.hobix.com) helped me find that hash.map() is  
> identical to hash.to\_a

Well the problem was that I just came across Set.divide and wanted to  
demonstrate it on an example - so I dug out an older problem of mine  
where I thought I can show it off - you have seen the result.

I think this was a typical manifestation of the ‘If you have a hammer,  
everything looks like a nail’ problem.

I guess if my goal would have been solving the task at hand, rather than  
putting ‘divide’ into action, I would come up with a solution similar to  
yours…

Thanks for the suggestion!

Cheers,  
Peter  
\_\_  
[http://www.rubyrailways.com](http://www.rubyrailways.com) :: Ruby and Web2.0 blog  
[http://scrubyt.org](http://scrubyt.org) :: Ruby web scraping framework  
[http://rubykitchensink.ca/](http://rubykitchensink.ca/) :: The indexed archive of all things Ruby

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<div class="post-metadata">

**Author:** ![splattael](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/s/858c86/32.png) [@splattael](https://www.ruby-forum.com/u/splattael)\
**Post date:** [April 30, 2007, 11:55am UTC](https://www.ruby-forum.com/t/partitioning-with-set-divide/93214/7 "2007-04-30T11:55:25Z")

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On 29.04.2007 14:04, Peter S. wrote:

> more than once in the set). Is this doable or do I have to roll some  
> code to do this for me additionally?

Basically you need bags. Since a quick check does not reveal any, you  
can roll your own pretty easily with a Hash with default value 0. This  
is what I’d do: (see script at end). Of course you could save another  
line by inlining “key”.

Kind regards

robert

8\<-----------------

#!/usr/bin/ruby

require ‘pp’

parts = Hash.new {|h,k| h[k] = Hash.new(0)}

DATA.each do |line|  
line.chomp!  
key = line[/^\w+/]  
parts[key][line] += 1  
end

pp parts

**END**  
a 53 2 3  
b 8 62 1 23  
a 9 0 31  
b 4 45 4 16 7  
b 1 23  
c 3 42 2 31 4 6  
a 1 3 22  
a 7 83 1 23 3  
b 1 14 4 15 16 2  
c 5 16 2 34  
c 5 16 2 34