Grouping elements of an array

I have an array of records that contain timestamps at random intervals.
The records are ordered by timestamp.

I would like to convert the array into an array of arrays; each subarray
would contain “grouped records.” Grouping would occur if the timestamp
of the next element in the original array is within thirty seconds of
the current element.

Example (second column is timestamp in seconds starting from zero).

A 0
B 15
C 35
D 100
E 205
F 215
G 300

would result in

[[A, B, C], [D], [E, F], [G]]

Any help on how to do this in the “Ruby Way” would be appreciated.

  • Steve W.

On Thu, Mar 18, 2010 at 4:50 PM, Steve W. [email protected]
wrote:

A 0

Any help on how to do this in the “Ruby Way” would be appreciated.

What about
A 0
B 20
C 40

Does that become
[[A,B],[B,C]] or [[A,B,C]] or something else? The congruence class here
is
unclear.

Josh C. wrote:

On Thu, Mar 18, 2010 at 4:50 PM, Steve W. [email protected]
wrote:

A 0

Any help on how to do this in the “Ruby Way” would be appreciated.

What about
A 0
B 20
C 40

Does that become
[[A,B],[B,C]] or [[A,B,C]] or something else? The congruence class here
is
unclear.

It would be [[A,B,C]].

  • Steve W.

Hi

On Thu, Mar 18, 2010 at 11:50 PM, Steve W. [email protected]
wrote:

A 0

Any help on how to do this in the “Ruby Way” would be appreciated.

How about this:

1 arr = [0,15,35,100,205,300]
2 arr2 = [0, 20, 40]
3
4 def group(array)
5 array.map!{|e| [e]}
6 array.inject([]) do |r, e|
7 if r == [] or e[0] - r.last.last > 30 then
8 r.push(e)
9 else
10 r[-1].push(e[0])
11 end
12 r
13 end
14 end
15
16 puts arr.inspect
17 puts group(arr).inspect
18 puts arr2.inspect
19 puts group(arr2).inspect

On Thu, Mar 18, 2010 at 4:50 PM, Steve W. [email protected]
wrote:

A 0

Any help on how to do this in the “Ruby Way” would be appreciated.

Here is my solution, it’s conceptually similar to Roger’s, though
differs in
implementation http://gist.github.com/337195

On Thu, Mar 18, 2010 at 9:29 PM, Urabe S.
[email protected]wrote:

Roger B. wrote:

How about this:

You should really know about Enumerable#group_by.

irb(main):001:0> [0,15,35,100,205,300].group_by {|i| i/100 }
=> {0=>[0, 15, 35], 1=>[100], 2=>[205], 3=>[300]}

Your results are correct only because of a happenstance of the data. Add
99
in there, it should group with 100, but it groups with the 0…100
congruence class

ruby-1.9.1-p378 > [0,15,35,99,100,205,300].group_by {|i| i/100 }
=> {0=>[0, 15, 35, 99], 1=>[100], 2=>[205], 3=>[300]}

Because these groups are relative to each other, I think you must do
something like Roger or I did, where you iterate through the list and
compare it to the groups.

Roger B. wrote:

How about this:

You should really know about Enumerable#group_by.

irb(main):001:0> [0,15,35,100,205,300].group_by {|i| i/100 }
=> {0=>[0, 15, 35], 1=>[100], 2=>[205], 3=>[300]}

On Fri, Mar 19, 2010 at 4:29 AM, Urabe S. [email protected]
wrote:

Roger B. wrote:

How about this:

You should really know about Enumerable#group_by.

irb(main):001:0> [0,15,35,100,205,300].group_by {|i| i/100 }
=> {0=>[0, 15, 35], 1=>[100], 2=>[205], 3=>[300]}

This does not solve the problem.

irb(main):011:0> [0,15,35,99,100,205,300].group_by{|i| i/100}
=> {0=>[0, 15, 35, 99], 1=>[100], 2=>[205], 3=>[300]}

but should be

[[0, 15, 35], [99, 100], [205], [300]]

at least if I understood the problem correctly.

I have an array of records that…

Thank you all for your solutions.

The time they saved me, I’ll spend learning about the techniques you
employed.

  • Steve W.

Roger B. wrote:

This does not solve the problem.

OK OK, but edge detection is also possible.

irb(main):001:0> a = [0,15,35,99,100,205,300]
=> [0, 15, 35, 99, 100, 205, 300]
irb(main):002:0> edges = a.each_cons(2).group_by {|(x, y)|
irb(main):003:1* y - x > 30
irb(main):004:1> }[true].map {|i|
irb(main):005:1* i[0]
irb(main):006:1> }
=> [35, 100, 205]
irb(main):007:0> a.group_by {|i| edges.find_index {|j| i <= j } }
=> {0=>[0, 15, 35], 1=>[99, 100], 2=>[205], nil=>[300]}

Hello, I am new to Ruby. Below is my solution:

a=[0, 15, 35, 100, 205, 215, 300]
b=[]
c=[]
d=a[0]
a.each do |i|
if i - d < 30
c << i
else
b << c
c=[i]
end
d =i
end
if c.size > 0
b << c
end

p b

On Thu, Mar 18, 2010 at 6:50 PM, Steve W. [email protected]
wrote:

A 0, B 15, C 35, D 100, E 205, F 215, G 300

would result in

[[A, B, C], [D], [E, F], [G]]

Any help on how to do this in the “Ruby Way” would be appreciated.

Here’s another way:

require ‘pp’
require ‘set’

Struct.new(“Record”, :value, :timestamp)

data = [
Struct::Record.new(‘A’, 0),
Struct::Record.new(‘B’, 15),
Struct::Record.new(‘C’, 35),
Struct::Record.new(‘D’, 100),
Struct::Record.new(‘E’, 205),
Struct::Record.new(‘F’, 215),
Struct::Record.new(‘G’, 300),
]

pp data

pp data.
to_set.
divide{|i, j| (i.timestamp - j.timestamp.abs) < 30}.
map{|s| s.to_a}

$ ruby -v z.rb
ruby 1.8.7 (2008-08-11 patchlevel 72) [universal-darwin10.0]
[#<struct Struct::Record value=“A”, timestamp=0>,
#<struct Struct::Record value=“B”, timestamp=15>,
#<struct Struct::Record value=“C”, timestamp=35>,
#<struct Struct::Record value=“D”, timestamp=100>,
#<struct Struct::Record value=“E”, timestamp=205>,
#<struct Struct::Record value=“F”, timestamp=215>,
#<struct Struct::Record value=“G”, timestamp=300>]
[[#<struct Struct::Record value=“D”, timestamp=100>],
[#<struct Struct::Record value=“G”, timestamp=300>],
[#<struct Struct::Record value=“F”, timestamp=215>,
#<struct Struct::Record value=“E”, timestamp=205>],
[#<struct Struct::Record value=“A”, timestamp=0>,
#<struct Struct::Record value=“C”, timestamp=35>,
#<struct Struct::Record value=“B”, timestamp=15>]]

(Depending on the amount of data converting from array to sets to
arrays may be expensive :slight_smile:

After reviewing the suggestions, here is my code. records is the result
of a ActiveRecord::find with a member called timestamp containing a UNIX
timestamp.

I introduced an addition of a default value for the find_index call and
included a descending sort of the groups.

edges = records.each_cons(2).group_by {|(record_x, record_y)|
record_y.timestamp - record_x.timestamp < 30 }[false].map{ |edge|
edge[0] }

groups = records.group_by { |record| edges.find_index {|edge|
record.timestamp <= edge.timestamp } || edges.count }.values.sort_by {
|e| -e.count }

Thanks again for everyone’s help.

  • Steve W.

At 2010-03-18 06:50PM, “Steve W.” wrote:

A 0

Any help on how to do this in the “Ruby Way” would be appreciated.

Lots of verbose answers. It can be quite short:

a = 

[[‘a’,0],[‘b’,15],[‘c’,35],[‘d’,100],[‘e’,205],[‘f’,215],[‘g’,300]]

a.group_by {|name,val| val/100} .
  collect do |key,list|
    list.collect {|name, time| name}
  end

# => [["a", "b", "c"], ["d"], ["e", "f"], ["g"]]

On 03/19/2010 03:33 PM, Glenn J. wrote:

[[A, B, C], [D], [E, F], [G]]
end

# => [["a", "b", "c"], ["d"], ["e", "f"], ["g"]]

Yeah, but as far as I can see that’s not a solution to the problem the
OP wanted to solve. He wants to build chains based on the delta and not
based on the fact that they fall in the same interval. Am I missing
something?

Kind regards

robert

On 03/18/2010 11:50 PM, Steve W. wrote:

A 0

Any help on how to do this in the “Ruby Way” would be appreciated.

Assuming records are ordered already - otherwise you need a sort in
between.

require ‘pp’

dat = <<DDD.each_line.map {|l|r = l.split;r[1]=r[1].to_i;r}
A 0
B 15
C 35
D 100
E 205
F 215
G 300
DDD

pp dat

gr = dat.inject [] do |agg, rec|
if agg.last && rec[1] - agg.last.last[1] <= 15
agg.last << rec
agg
else
agg << [rec]
end
end

pp gr

Note: I don’t claim that this is the Ruby way.

Kind regards

robert

At 2010-03-19 01:52PM, “Robert K.” wrote:

On 03/19/2010 03:33 PM, Glenn J. wrote:

At 2010-03-18 06:50PM, “Steve W.” wrote:

I would like to convert the array into an array of arrays; each subarray
would contain “grouped records.” Grouping would occur if the timestamp
of the next element in the original array is within thirty seconds of
the current element.
[…]
Yeah, but as far as I can see that’s not a solution to the problem the
OP wanted to solve. He wants to build chains based on the delta and not
based on the fact that they fall in the same interval. Am I missing
something?

Ah, I didn’t read the words that closely – I was looking at the input
and output data instead.

2010/3/19 Steve W. [email protected]:

would result in

[[A, B, C], [D], [E, F], [G]]

I think this kind of problems which slices consecutive elements in an
array
is not well supported in Ruby.

Enumerable#each_slice is not usable because it slices for each fixed
number
of elements.

Ruby 1.9.2 has Enumerable#slice_before and it is usable but not so
elegant
because it needs to maintain previous element.

% ruby -e ’
a = [
[“A”, 0],
[“B”, 15],
[“C”, 35],
[“D”, 100],
[“E”, 205],
[“F”, 215],
[“G”, 300]
]
prev = nil
p a.slice_before {|s,t|
tmp, prev = prev, t
tmp && (t-tmp) > 30
}.map {|es|
es.map {|s,t| s }
}

[[“A”, “B”, “C”], [“D”], [“E”, “F”], [“G”]]

We may need Enumerable#slice_between.

% ruby -e ’
module Enumerable
def slice_between(&b)
Enumerator.new {|y|
first = true
buf = []
prev = nil
self.each {|elt|
if first
first = false
buf << elt
prev = elt
else
if b.call(prev, elt)
y << buf
buf = [elt]
else
buf << elt
end
prev = elt
end
}
if !buf.empty?
y << buf
end
}
end
end
a = [
[“A”, 0],
[“B”, 15],
[“C”, 35],
[“D”, 100],
[“E”, 205],
[“F”, 215],
[“G”, 300]
]
p a.slice_between {|(s1,t1),(s2,t2)|
(t2-t1) < 30
}.map {|es|
es.map {|s,t| s }
}

[[“A”], [“B”], [“C”, “D”, “E”], [“F”, “G”]]

On Sun, Mar 21, 2010 at 12:57 AM, Tanaka A. wrote:

I think this kind of problems which slices consecutive elements in an array
is not well supported in Ruby.

On Sun, Mar 21, 2010 at 6:42 AM, Urabe S. wrote:

+1. There seems to be some real applications where that kind of tools are nifty.

On Sun, Mar 21, 2010 at 12:57 AM, Tanaka A. also wrote:

Enumerable#each_slice is not usable because it slices for each fixed number
of elements.
Ruby 1.9.2 has Enumerable#slice_before and it is usable but not so elegant
because it needs to maintain previous element.

We may need Enumerable#slice_between.

Is a really elegant method possible for this sort of thing?
Some time ago I was trying to find duplicate files,
and set up an array of files with elements (paths, sizes, checksums),
and then wrote an Enumerable#each_group method.
The only way I could think of differentiating a yield for “comparison”
from a yield of the wanted array of “similar” objects
was to have the first item of each yield to be a boolean
with true meaning compare, and false meaning it’s the wanted array.

I’d be interested in seeing a better Enumerable#each_group method,
if one is possible.

In the meantime, below is what I wrote for Enumerable#each_group,
with its application to Steve W.'s problem.

module Enumerable

Deeming successive enumerable objects to be “equivalent”

using a block compare, yields an array of the equivalent items.

def each_group_use_block_compare()
arr = nil ; obj_g = nil
self.each do | obj |
if arr then
# first item in yield is “true” indicating a yield for
comparison
if ( yield true, obj_g, obj ) == 0 then
arr << obj
obj_g = obj # group by adjacent objects, not by first in
group
next
end
# first item in yield is “false” indicating a yield of the group
yield false, arr
end
obj_g = obj ; arr = [ obj ]
end
if arr then
# first item in yield is “false” indicating a yield of the group
yield false, arr
end
return self
end
end

for the problem of Steve W.

def group_for_sw( arr )
arrg = []
arr.each_group_use_block_compare do | q_compare, aa, bb |
if q_compare then
if bb[1] - aa[1] < 30 then 0 else -1 end
else
aa.map! { |cc| cc[0] } # to preserve A, B, C objects, omit this
arrg << aa
end
end
arrg
end

arr = [ [ :A, 0 ], [ :B, 15 ], [ :C, 35 ],
[ :D, 100 ],
[ :E, 205 ], [ :F, 215 ],
[ :G, 300 ]
]
arrg = group_for_sw( arr )
p arrg #=> [[:A, :B, :C], [:D], [:E, :F], [:G]]

2010/3/21 Colin B. [email protected]:

Is a really elegant method possible for this sort of thing?
Some time ago I was trying to find duplicate files,
and set up an array of files with elements (paths, sizes, checksums),
and then wrote an Enumerable#each_group method.
The only way I could think of differentiating a yield for “comparison”
from a yield of the wanted array of “similar” objects
was to have the first item of each yield to be a boolean
with true meaning compare, and false meaning it’s the wanted array.

We need two blocks.
One for slice condition and one for sliced array.
But Ruby’s method call can take only one block at most.

Enumerable#slice_before solves this problem by returning enumerator.

enum.slice_before {|elt| condition }.each {|ary| … }

slice_between presented in [ruby-talk:359384] is similar.

enum.slice_between {|elt1, elt2| condition }.each {|ary| … }

Another possible idea is using Proc argument.
enum.foo(lambda {|elt| condition }) {|ary| … }

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