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How can I ensure uniqueness in this array while maintaining its length at 5?
def fixed
5.times.collect { SecureRandom.random_number(10) }
end
This behaviour seems odd:
5.times.collect.uniq { SecureRandom.random_number(10) }
# => [0, 2, 3, 4]
5.times.collect.uniq { SecureRandom.random_number(10) }
# => [0, 1, 3]
5.times.collect.uniq { SecureRandom.random_number(10) }
# => [0, 1, 2, 3, 4]
5.times.collect.uniq { SecureRandom.random_number(10) }
# => [0, 1, 2, 4]
5.times.collect.uniq { SecureRandom.random_number(10) }
# => [0, 1, 2, 3]
When the number of possible values is small – like 10 in your example – then I would generate an array with all options and just pick a random sample of entries:
(0..9).to_a.sample(5)
If the number of possible values is huge then generation all values first is certainly not an option. Then I would generate a random value as long as the array doesn't contain enough entries:
require 'set'
values = Set.new
until values.length == 5 do
values.add(SecureRandom.random_number(1_000_000))
end
values.to_a
Note the I am using a Set to ensure the uniqueness of the values in the second version.
Using SecureRandom
def fixed
unique_numbers = []
5.times.collect do
loop do
number = SecureRandom.random_number(10)
break number unless unique_numbers.include?(number)
end
end
end
And if you want to generate unique numbers between 1 to 10, then you can create array of 1 to 10 and use shuffle or sample to get random numbers.
Using shuffle
> (0...10).to_a.shuffle.take(5)
=> [4, 0, 1, 3, 7]
> (0...10).to_a.shuffle.take(5)
=> [6, 2, 3, 9, 1]
> (0...10).to_a.shuffle.take(5)
=> [9, 2, 5, 8, 4]
> (0...10).to_a.shuffle.take(5)
=> [5, 0, 6, 8, 7]
> (0...10).to_a.shuffle.take(5)
=> [2, 7, 1, 5, 0]
Using sample
> (1..10).to_a.sample(5)
=> [4, 6, 3, 2, 7]
> (1..10).to_a.sample(5)
=> [5, 8, 2, 3, 7]
> (1..10).to_a.sample(5)
=> [2, 5, 6, 1, 3]
> (1..10).to_a.sample(5)
=> [8, 5, 10, 9, 3]
> (1..10).to_a.sample(5)
=> [8, 1, 5, 3, 4]
You can also pass SecureRandom custom random generator as an argument with sample
> (1..10).to_a.sample(5, random: SecureRandom)
=> [6, 3, 4, 7, 10]
> (1..10).to_a.sample(5, random: SecureRandom)
=> [7, 4, 8, 1, 5]
> (1..10).to_a.sample(5, random: SecureRandom)
=> [8, 3, 9, 5, 10]
> (1..10).to_a.sample(5, random: SecureRandom)
=> [6, 8, 9, 2, 1]
> (1..10).to_a.sample(5, random: SecureRandom)
=> [9, 10, 1, 8, 2]
Just out of curiosity, using Enumerable#cycle infinite generator.
MAX = 10
SIZE = 5
[MAX].cycle.inject(Set.new) do |acc, max|
break acc if acc.size >= SIZE
acc << SecureRandom.random_number(max)
end
#⇒ #<Set: {2, 1, 7, 0, 9}>
or even with generic loop:
loop.each_with_object(Set.new) do |_, acc|
break acc if acc.size >= SIZE
acc << SecureRandom.random_number(10)
end
#⇒ #<Set: {2, 6, 7, 1, 3}>
One way would be to generate a range of numbers from 0 to 10 and
then shuffle them to get the unique random numbers.
You can convert that range to Array using to_a and shuffle them using shuffle
You can do something like this:
(0..10).to_a.shuffle[0..4] # => [8, 6, 1, 9, 10]
[0..4] will give you the first 5 shuffled elements.
This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
diff a ruby string or array
I have an old array: [1, 2, 3, 4, 5], and new: [1, 2, 4, 6]
How to get difference with Ruby: that 5, 3 was removed and 6 was added?
irb(main):001:0> a = [1, 2, 3, 4, 5]
=> [1, 2, 3, 4, 5]
irb(main):002:0> b = [1, 2, 4, 6]
=> [1, 2, 4, 6]
irb(main):003:0> a - b
=> [3, 5]
irb(main):005:0> b - a
=> [6]
irb(main):006:0>
I have an array like this: [7, 8, 9, 10, 11, 12, 1, 2, 3, 4, 5, 6]
What's the simplest way to return each item in the array from position 6 until 0 where the resulting array looks like: [1,2,3,4,5,6,7]
This positions in the array can be dynamic, for example passing in 4 and 9 should return [11,12,1,2,3,4]
I'm wondering if there's a method that accomplishes this in Rails api.
Thanks in advance
EDIT
Let's assume that no negative numbers, so doing array[2..-2] wont work.
Array#splice almost works for this, but if the second position is less than the first, it returns nil.
def foo a, min, max
a.rotate(min).first((max - min) % a.length + 1)
end
a = [7, 8, 9, 10, 11, 12, 1, 2, 3, 4, 5, 6]
foo(a, 6, 0) # => [1, 2, 3, 4, 5, 6, 7]
foo(a, 4, 9) # => [11, 12, 1, 2, 3, 4]
class Array
def get_sub_array(start,last)
(start > last) ? (self[start..-1] + self[0..last]) : self[start..last]
end
end
Then
a = [7, 8, 9, 10, 11, 12, 1, 2, 3, 4, 5, 6]
a.get_sub_array(6,0)
#[1, 2, 3, 4, 5, 6, 7]
Or if you don't want to monkey patch
You could have a method like
def get_sub_array(array, start,last)
(start > last) ? (array[start..-1] + array[0..last]) : array[start..last]
end
a = [7, 8, 9, 10, 11, 12, 1, 2, 3, 4, 5, 6]
get_sub_array(a,6,0)
#[1, 2, 3, 4, 5, 6, 7]
def some_function(some_array,start_val=6, end_val=0)
if end_val > start_val
some_array[start_val,(end_val - start_val)]
else
(some_array[start_val, some_array.size] << some_array[0, (end_val)]).flatten
end
end
You can use ternary operator to make it one liner too:
def some_function(some_array,start_val=6, end_val=0)
end_val > start_val ? some_array[start_val,(end_val - start_val)] : (some_array[start_val, some_array.size] << some_array[0, (end_val)]).flatten
end
a = [7, 8, 9, 10, 11, 12, 1, 2, 3, 4, 5, 6]
some_function(a) # => [1, 2, 3, 4, 5, 6, 7]
some_function(a, 4, 9) # => [11, 12, 1, 2, 3, 4]
min=6
max=0
arr = [7, 8, 9, 10, 11, 12, 1, 2, 3, 4, 5, 6]
result = []
if max<min
result << arr[min..arr.length]
result << arr[0..max]
else
result << arr[min..max]
end
A couple more ways (my preference being for #1).
a = [7, 8, 9, 10, 11, 12, 1, 2, 3, 4, 5, 6]
#1
def foo a, min, max
as = a.size
max += as if max < min
(min..max).map { |i| a[i%as] }
end
foo(a, 6, 0) # => [ 1, 2, 3, 4, 5, 6, 7]
foo(a, 4, 9) # => [11, 12, 1, 2, 3, 4]
#2
def foo a, min, max
max += a.size if max < min
e = a.cycle
min.times { e.next }
(max-min+1).times.map { e.next }
end
foo(a, 6, 0) # => [ 1, 2, 3, 4, 5, 6, 7]
foo(a, 4, 9) # => [11, 12, 1, 2, 3, 4]
def foo a, s, e
a = e < s ? (a[s,a.size] << a[0..e]).flatten : a[s..e]
end
a = [7, 8, 9, 10, 11, 12, 1, 2, 3, 4, 5, 6]
a = foo(a, 6, 0) # => [1, 2, 3, 4, 5, 6, 7]
a = foo(a, 4, 9) # => [11, 12, 1, 2, 3, 4]
myArray = [7, 8, 9, 10, 11, 12, 1, 2, 3, 4, 5, 6]
myArray[6..-1] returns [1, 2, 3, 4, 5, 6]
myArray[4..9] returns [11,12,1,2,3,4]
This question already has answers here:
How to chunk an array in Ruby
(2 answers)
Closed 4 years ago.
I need a way to split an array in to an exact number of smaller arrays of roughly-equal size. Anyone have any method of doing this?
For instance
a = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]
groups = a.method_i_need(3)
groups.inspect
=> [[1,2,3,4,5], [6,7,8,9], [10,11,12,13]]
Note that this is an entirely separate problem from dividing an array into chunks, because a.each_slice(3).to_a would produce 5 groups (not 3, like we desire) and the final group may be a completely different size than the others:
[[1,2,3], [4,5,6], [7,8,9], [10,11,12], [13]] # this is NOT desired here.
In this problem, the desired number of chunks is specified in advance, and the sizes of each chunk will differ by 1 at most.
You're looking for Enumerable#each_slice
a = [0, 1, 2, 3, 4, 5, 6, 7]
a.each_slice(3) # => #<Enumerator: [0, 1, 2, 3, 4, 5, 6, 7]:each_slice(3)>
a.each_slice(3).to_a # => [[0, 1, 2], [3, 4, 5], [6, 7]]
Perhaps I'm misreading the question since the other answer is already accepted, but it sounded like you wanted to split the array in to 3 equal groups, regardless of the size of each group, rather than split it into N groups of 3 as the previous answers do. If that's what you're looking for, Rails (ActiveSupport) also has a method called in_groups:
a = [0,1,2,3,4,5,6]
a.in_groups(2) # => [[0,1,2,3],[4,5,6,nil]]
a.in_groups(3, false) # => [[0,1,2],[3,4], [5,6]]
I don't think there is a ruby equivalent, however, you can get roughly the same results by adding this simple method:
class Array; def in_groups(num_groups)
return [] if num_groups == 0
slice_size = (self.size/Float(num_groups)).ceil
groups = self.each_slice(slice_size).to_a
end; end
a.in_groups(3) # => [[0,1,2], [3,4,5], [6]]
The only difference (as you can see) is that this won't spread the "empty space" across all the groups; every group but the last is equal in size, and the last group always holds the remainder plus all the "empty space".
Update:
As #rimsky astutely pointed out, the above method will not always result in the correct number of groups (sometimes it will create multiple "empty groups" at the end, and leave them out). Here's an updated version, pared down from ActiveSupport's definition which spreads the extras out to fill the requested number of groups.
def in_groups(number)
group_size = size / number
leftovers = size % number
groups = []
start = 0
number.times do |index|
length = group_size + (leftovers > 0 && leftovers > index ? 1 : 0)
groups << slice(start, length)
start += length
end
groups
end
Try
a.in_groups_of(3,false)
It will do your job
As mltsy wrote, in_groups(n, false) should do the job.
I just wanted to add a small trick to get the right balance
my_array.in_group(my_array.size.quo(max_size).ceil, false).
Here is an example to illustrate that trick:
a = (0..8).to_a
a.in_groups(4, false) => [[0, 1, 2], [3, 4], [5, 6], [7, 8]]
a.in_groups(a.size.quo(4).ceil, false) => [[0, 1, 2], [3, 4, 5], [6, 7, 8]]
This needs some better cleverness to smear out the extra pieces, but it's a reasonable start.
def i_need(bits, r)
c = r.count
(1..bits - 1).map { |i| r.shift((c + i) * 1.0 / bits ) } + [r]
end
> i_need(2, [1, 3, 5, 7, 2, 4, 6, 8])
=> [[1, 3, 5, 7], [2, 4, 6, 8]]
> i_need(3, [1, 3, 5, 7, 2, 4, 6, 8])
=> [[1, 3, 5], [7, 2, 4], [6, 8]]
> i_need(5, [1, 3, 5, 7, 2, 4, 6, 8])
=> [[1, 3], [5, 7], [2, 4], [6], [8]]
I have an array of 10 items and I want to split it up into 3 sections that look like this:
[1, 2, 3, 4]
[5, 6, 7]
[8, 9, 10]
Using each_slice I can get close:
a = *(1..10)
a.each_slice(4) # use 4 so I can fit everything into 3 sections
[1, 2, 3, 4]
[5, 6, 7, 8]
[9, 10]
But I want the first format which is more evenly distributed. I can do it writing my own method. But is there a built in way to do this in ruby 1.9+?
Update:
Since there's no built in way I'd like to change my question to - how would you implement it?
Here's my implementation
def chunk(a, pieces)
size = a.size / pieces
extra = a.size % pieces
chunks = []
start = 0
1.upto(pieces) do |i|
last = (i <= extra) ? size.next : size
chunks << a.slice(start, last)
start = chunks.flatten.size
end
chunks
end
call it like so
a = *(1..10)
puts chunk(a, 3)
will output
[1, 2, 3, 4]
[5, 6, 7]
[8, 9, 10]
If piece size is too big it pads with empty arrays
a = *(1..10)
puts chunk(a, 14)
will output
[[1], [2], [3], [4], [5], [6], [7], [8], [9], [10], [], [], [], []]