how to split an array in sub arrays of the same length

P

Paolo Bacchilega

Hi,

Is there a way to split an array in sub arrays of the same length, that
is :

[ 1, 2, 3, 4, 5, 6, 7, 8 ].unknown_function(3)

that returns:

[ [1, 2, 3], [4, 5, 6], [7, 8] ]


thanks in advance.
 
W

WATANABE Hirofumi

Hi,

Paolo Bacchilega said:
Is there a way to split an array in sub arrays of the same length, that
is :

[ 1, 2, 3, 4, 5, 6, 7, 8 ].unknown_function(3)

that returns:

[ [1, 2, 3], [4, 5, 6], [7, 8] ]

% irb --prompt simple
require 'enumerator' => true
[ 1, 2, 3, 4, 5, 6, 7, 8 ].enum_slice(3).to_a
=> [[1, 2, 3], [4, 5, 6], [7, 8]]
 
P

Peña, Botp

fr paulo:
# Is there a way to split an array in sub arrays of the same=20
# length, that
# is :
#=20
# [ 1, 2, 3, 4, 5, 6, 7, 8 ].unknown_function(3)
# that returns:
# [ [1, 2, 3], [4, 5, 6], [7, 8] ]

irb(main):021:0> x
=3D> [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
irb(main):022:0> y=3D[]
=3D> []
irb(main):023:0> x.each_slice(3){|s| y << s}
=3D> nil
irb(main):024:0> y
=3D> [[0, 1, 2], [3, 4, 5], [6, 7, 8], [9]]

kind regards -botp
 
M

Morton Goldberg

Although you could use Enumerator::each_slice as others have
suggested, it might be simpler to augment Array with a partition
function.

class Array

def partition(n, r=[])
raise ArgumentError if n <= 0
if n <= size then
r << first(n)
last(size - n).partition(n, r)
else
r << self unless empty?
return r
end
end

end

p [ 1, 2, 3, 4, 5, 6, 7, 8 ].partition(1) => [[1], [2], [3], [4],
[5], [6], [7], [8]]
p [ 1, 2, 3, 4, 5, 6, 7, 8 ].partition(3) => [[1, 2, 3], [4, 5, 6],
[7, 8]]
p [ 1, 2, 3, 4, 5, 6, 7, 8 ].partition(4) => [[1, 2, 3, 4], [5, 6, 7,
8]]
p [ 1, 2, 3, 4, 5, 6, 7, 8 ].partition(6) => [[1, 2, 3, 4, 5, 6], [7,
8]]
p [ 1, 2, 3, 4, 5, 6, 7, 8 ].partition(8) => [[1, 2, 3, 4, 5, 6, 7, 8]]
p [ 1, 2, 3, 4, 5, 6, 7, 8 ].partition(10) => [[1, 2, 3, 4, 5, 6, 7, 8]]
p [ 1, 2, 3, 4, 5, 6, 7, 8 ].partition(-2) =>
untitled text:6:in `partition': ArgumentError (ArgumentError)
from untitled text:24

Hope this is helps.
Regards, Morton
 
P

Paolo Bacchilega

Il giorno gio, 20/07/2006 alle 06.53 +0000, Paolo Bacchilega ha scritto:
Hi,

Is there a way to split an array in sub arrays of the same length, that
is :

[ 1, 2, 3, 4, 5, 6, 7, 8 ].unknown_function(3)

that returns:

[ [1, 2, 3], [4, 5, 6], [7, 8] ]


thanks in advance.

thanks to all for responding.
 
W

William James

Paolo said:
Hi,

Is there a way to split an array in sub arrays of the same length, that
is :

[ 1, 2, 3, 4, 5, 6, 7, 8 ].unknown_function(3)

that returns:

[ [1, 2, 3], [4, 5, 6], [7, 8] ]


thanks in advance.

Always remember that a program without inject is a
pointless program.

[ 1, 2, 3, 4, 5, 6, 7, 8 ].inject([[]]){|a,x|
a.last.size<3 ? a.last << x : a << [x]; a }
 
L

Logan Capaldo

Paolo said:
Is there a way to split an array in sub arrays of the same length,
that

Stolen from the facets[1] extensions for Arrays[2]:

class Array
def each_slice(n=nil, &yld)
n = yld.arity.abs unless n
i=0
while i < self.length
yld.call(*self.slice(i,n))
i+=n
end
end
end

[1] http://facets.rubyforge.org/
[2] http://facets.rubyforge.org/api/core/classes/Array.html#M000163

This is in enumerator, enumerator comes with ruby. Do we really need
an "Optimized for Array" version? (That is its purpose according to
the docs.)

Well surprising to me, there really is a noticeable speed difference.
I gues sit makes sense, each_slice in enumerator.c is written to use
each, and not just call to_a first either. (Which makes sense, Files
are Enumerables after all, maybe you want it in chunks of N lines at
a time. You wouldn't want to have to slurp the whole file into memory
just to do that).

I wonder if we can get this each_slice stuck in the standard lib for
Array.



% cat enumerator_vs_facets.rb
#!/usr/bin/env ruby
require 'enumerator'
require 'benchmark'

class Array
def facets_each_slice(n=nil, &yld)
n = yld.arity.abs unless n
i=0
while i < self.length
yld.call(*self.slice(i,n))
i+=n
end
end
end

arrays = [ (1..97).to_a, (0..99).to_a, ["hello", "world"] ]
N = 1000
Benchmark.bmbm do |bm|
bm.report("Facets: ") do
N.times do
arrays.each do |array|
array.facets_each_slice(3) { |*x| "#{x}" }
end
end
end

bm.report("Enumerator: ") do
N.times do
arrays.each do |array|
array.each_slice(3) { |*x| "#{x}" }
end
end
end
end



% ruby enumerator_vs_facets.rb
Rehearsal ------------------------------------------------
Facets: 1.250000 0.010000 1.260000 ( 1.364671)
Enumerator: 1.380000 0.010000 1.390000 ( 1.442132)
--------------------------------------- total: 2.650000sec

user system total real
Facets: 1.250000 0.010000 1.260000 ( 1.315879)
Enumerator: 1.390000 0.010000 1.400000 ( 1.447592)
 
B

bbiker

Logan said:
% cat enumerator_vs_facets.rb
#!/usr/bin/env ruby
require 'enumerator'
require 'benchmark'

class Array
def facets_each_slice(n=nil, &yld)
n = yld.arity.abs unless n
i=0
while i < self.length
yld.call(*self.slice(i,n))
i+=n
end
end
end

arrays = [ (1..97).to_a, (0..99).to_a, ["hello", "world"] ]
N = 1000
Benchmark.bmbm do |bm|
bm.report("Facets: ") do
N.times do
arrays.each do |array|
array.facets_each_slice(3) { |*x| "#{x}" }
end
end
end

bm.report("Enumerator: ") do
N.times do
arrays.each do |array|
array.each_slice(3) { |*x| "#{x}" }
end
end
end
end



% ruby enumerator_vs_facets.rb
Rehearsal ------------------------------------------------
Facets: 1.250000 0.010000 1.260000 ( 1.364671)
Enumerator: 1.380000 0.010000 1.390000 ( 1.442132)
--------------------------------------- total: 2.650000sec

user system total real
Facets: 1.250000 0.010000 1.260000 ( 1.315879)
Enumerator: 1.390000 0.010000 1.400000 ( 1.447592)

I believe the OP asked how to split an array in "x" sub arrays of equal
length.
To me, it appears that the response show how to partition an array into
sub arrays of "y" elements.

The following method does what the OP asked for, as far as I read it.

class Array
def split(n, b=[], d=[])
#########################################################################
# evenly split array 'a' into 'n' sub arrays
#########################################################################
if self.size == 0 or n <= 0
b = nil
elsif n == 1
b = *self
else
# determine how many elements of the array should go in each sub
arrays
buckets = n
length = self.size
while buckets > 0
elements = length / buckets + (length % buckets > 0 ? 1 : 0)
d << elements
length -= elements
buckets -= 1
end
# p d

# evenly distribute array elements into an array with 'n' sub arrays
start = 0 # start
0.upto(n-1) do |idx|
len = d[idx]
b[idx] = *self.slice(start,len)
start += len
end
end
b
end
end
 
B

bbiker

bbiker said:
I believe the OP asked how to split an array in "x" sub arrays of equal
length.
To me, it appears that the response show how to partition an array into
sub arrays of "y" elements.

The following method does what the OP asked for, as far as I read it.

class Array
def split(n, b=[], d=[])
#########################################################################
# evenly split array 'a' into 'n' sub arrays
#########################################################################
if self.size == 0 or n <= 0
b = nil
elsif n == 1
b = *self
else
# determine how many elements of the array should go in each sub
arrays
buckets = n
length = self.size
while buckets > 0
elements = length / buckets + (length % buckets > 0 ? 1 : 0)
d << elements
length -= elements
buckets -= 1
end
# p d

# evenly distribute array elements into an array with 'n' sub arrays
start = 0 # start
0.upto(n-1) do |idx|
len = d[idx]
b[idx] = *self.slice(start,len)
start += len
end
end
b
end
end

Sorry, but I pressed the Post message button too soon.

c =
[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23].split(5)
=>

[[1, 2, 3, 4, 5], [6, 7, 8, 9, 10], [11, 12, 13, 14, 15], [16, 17, 18,
19], [20, 21, 22, 23]]

Note that the first 3 sub arrays contain 5 elements and the last 2 sub
arrays contain 4 elements.

I am a Ruby newbie and would appreciate a critique on the method I
posted.

Thank You
 
W

William James

bbiker said:
bbiker said:
I believe the OP asked how to split an array in "x" sub arrays of equal
length.
To me, it appears that the response show how to partition an array into
sub arrays of "y" elements.

The following method does what the OP asked for, as far as I read it.

class Array
def split(n, b=[], d=[])
#########################################################################
# evenly split array 'a' into 'n' sub arrays
#########################################################################
if self.size == 0 or n <= 0
b = nil
elsif n == 1
b = *self
else
# determine how many elements of the array should go in each sub
arrays
buckets = n
length = self.size
while buckets > 0
elements = length / buckets + (length % buckets > 0 ? 1 : 0)
d << elements
length -= elements
buckets -= 1
end
# p d

# evenly distribute array elements into an array with 'n' sub arrays
start = 0 # start
0.upto(n-1) do |idx|
len = d[idx]
b[idx] = *self.slice(start,len)
start += len
end
end
b
end
end

Sorry, but I pressed the Post message button too soon.

c =
[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23].split(5)
=>

[[1, 2, 3, 4, 5], [6, 7, 8, 9, 10], [11, 12, 13, 14, 15], [16, 17, 18,
19], [20, 21, 22, 23]]

Note that the first 3 sub arrays contain 5 elements and the last 2 sub
arrays contain 4 elements.

I am a Ruby newbie and would appreciate a critique on the method I
posted.

Thank You

class Array
def split n
count , fat_ones = self.size / n , self.size % n
self.inject( [[]] ){ |a,e|
a.last.size < count + ( a.size <= fat_ones ? 1 : 0 ) ?
a.last << e : a << [e] ; a }
end
end

a = (1..23).to_a
p a.split(5)
p a
 
W

William James

William said:
bbiker said:
bbiker said:
I believe the OP asked how to split an array in "x" sub arrays of equal
length.
To me, it appears that the response show how to partition an array into
sub arrays of "y" elements.

The following method does what the OP asked for, as far as I read it.

class Array
def split(n, b=[], d=[])
#########################################################################
# evenly split array 'a' into 'n' sub arrays
#########################################################################
if self.size == 0 or n <= 0
b = nil
elsif n == 1
b = *self
else
# determine how many elements of the array should go in each sub
arrays
buckets = n
length = self.size
while buckets > 0
elements = length / buckets + (length % buckets > 0 ? 1 : 0)
d << elements
length -= elements
buckets -= 1
end
# p d

# evenly distribute array elements into an array with 'n' sub arrays
start = 0 # start
0.upto(n-1) do |idx|
len = d[idx]
b[idx] = *self.slice(start,len)
start += len
end
end
b
end
end

Sorry, but I pressed the Post message button too soon.

c =
[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23].split(5)
=>

[[1, 2, 3, 4, 5], [6, 7, 8, 9, 10], [11, 12, 13, 14, 15], [16, 17, 18,
19], [20, 21, 22, 23]]

Note that the first 3 sub arrays contain 5 elements and the last 2 sub
arrays contain 4 elements.

I am a Ruby newbie and would appreciate a critique on the method I
posted.

Thank You

class Array
def split n
count , fat_ones = self.size / n , self.size % n
self.inject( [[]] ){ |a,e|
a.last.size < count + ( a.size <= fat_ones ? 1 : 0 ) ?
a.last << e : a << [e] ; a }
end
end

a = (1..23).to_a
p a.split(5)
p a

Perhaps faster:

class Array
def split n
count , fat_ones = self.size / n , self.size % n
cp, out = self.dup, []
out << cp.
slice!(0, count + (out.size < fat_ones ? 1 : 0 )) while cp!=[]
out
end
end
 
B

bbiker

William said:
I am a Ruby newbie and would appreciate a critique on the method I
posted.

Thank You

class Array
def split n
count , fat_ones = self.size / n , self.size % n
self.inject( [[]] ){ |a,e|
a.last.size < count + ( a.size <= fat_ones ? 1 : 0 ) ?
a.last << e : a << [e] ; a }
end
end

a = (1..23).to_a
p a.split(5)
p a

Yes! it is a much better way !!!
Since obviously I am not familiar with the inject method, I took to the
pick axe and irb to work it out until I understood exactly what you
were doing.

The only question left is how to best handle bad inputs?

n <= 0, n not specified, and an empty array, n > [1,2,3,4,5].size

for [1,2,3,4,5].split(0) => Exception: divided by 0

for [1,2,3,4,5].split(-n) => [[], [1], [2], [3], [4], [5]]

for [].split(+/-n) => [[]] where n != 0
for [].split(0) => Exception: divided by 0

for [1,2,3,4,5].split() =>
Exception: wrong number of arguments (0 for 1)

for [1,2,3,4,5].split(8) => [[1], [2], [3], [4], [5]]

I would guess this might be the correct way. This could cause a problem
if the user were to check the size of the returned array For the above
example it would be5 instead of the expected 8.

You could say that for n == 0 or n not specifiend are already handled
by existing Exceptions. Do need to handle negative values of n, n >
array.size and array.size == 0
 

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