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I want to know how we can identify that descending order Range is given.
I have a test case where range is given in (1..-1) and to_a method gives empty array for same.
(1..5).to_a
=> [1, 2, 3, 4, 5]
(5..1).to_a
=> []
Help me write a condition which can confirm descending order Range is given.
Use Range#size:
(5..1).size.zero?
#⇒ true
(1..5).size.zero?
#⇒ false
Range has Range#first and Range#last methods.
r = (5..1)
r.first > r.last # => true
If you want just to check for a reverse range:
(5..1).to_a.empty? #=> true
(1..5).to_a.empty? #=> false
If you want to have an array as output, you could do
r.to_a.then { |ary| ary.empty? ? (r.end..r.begin).to_a.reverse : ary }
The output for the two cases are:
r = (5..1) #=> [5, 4, 3, 2, 1]
r = (1..5) #=> [1, 2, 3, 4, 5]
(1..5).to_a.present? ? "Ascending" : "Descending"
if value present in array means it Ascending or Descending cause if you give (5..1) range it create blank Array.
User must enter the array as input. Hash has to accept the input array elements as values.
Create a Ruby program for this by using loops. If array completed print this statement “All array elements are assigned to keys in the hash”
A = [1, 6, 4, 5]
H = {“k1” => 1
“k2” => 6
“k3” => 4
“k4” => 5}
Another solution can be,
a.each_with_index.inject({}) { |m,(a,i)| m["k#{i+1}"] = a; m }
Update: answering to your question
puts 'Enter number of hash elements'
n = gets.to_i
n.times |i|
hash["k#{i+1}"] = gets.to_i
end
puts 'All array elements are assigned to keys in the hash'
A = [1, 6, 4, 5]
Hash[[*'k1'.."k#{A.length}"].zip(A)]
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Is it possible to check inclusion array inside array?
i want to check if
primary = [1,2,3] includes secondary = [2,3]
i have tried primary.include?(secondary) => false
need to return bool value
Solution
Without duplicates
If there aren't any duplicate, you can calculate the Array difference, and check if it is empty :
(secondary-primary).empty?
#=> true
General case
subset_of? checks that for every unique element in secondary, there are at least as many elements in primary :
class Array
def count_by
each_with_object(Hash.new(0)) { |e, h| h[e] += 1 }
end
def subset_of?(superset)
superset_counts = superset.count_by
count_by.all? { |k, count| superset_counts[k] >= count }
end
end
Example :
secondary.subset_of?(primary)
#=> true
[2,2].subset_of?([1, 2, 3])
#=> false
This should work for any Array, and it should be faster than other answers for big arrays.
Test
([1,2,3] - [3,4,5,2,1]).empty?
#=> true
([1,2,3,'a'] - [3,4,5,2,1]).empty?
#=> false
test = primary.dup
secondary.all? { |e| test.index(e).tap { |i| test.delete_at(i) if i } }
primary, secondary = [1, 2, 3], [2, 2]
#⇒ false
primary, secondary = [1, 2, 2, 3], [2, 2, 1]
#⇒ true
What’s being done here:
we iterate secondary, claiming that all blocks should return true
on each subsequent iteration we
immediately return false breaking the loop if there is no such element in primary
otherwise we mutate the copy of primary, removing the element already checked.
The only trick here is using Object#tap to always return true when element found. The element in the primary might be falsey, Array#delete returns the element deleted and we might accidentally return falsey, mistakenly breaking a loop in such a case. We should return true to all? loop as soon as we have the element found, hence tap.
The input to verify that this is the only correct answer here so far:
primary, secondary = [1, 2, 2, 3, nil, nil], [2, 2, 1, nil, nil]
#⇒ true
primary, secondary = [1, 2, 2, 3, nil], [2, 2, 1, nil, nil]
#⇒ false
primary = [1,2,3]
secondary = [2,3]
primary.each_cons(secondary.size).include?(secondary)
Enumerable#each_cons takes chunks of the array and iterates by one for each group.
Enumerable is awesome!
Read the Enumerable docs, I learn something new every time.
a = [5, 1, 6, 14, 2, 8]
b = [2, 6, 15]
a - b
=> [5, 1, 14, 8]
b - a
=> [15]
#this is what you need
(b - a).empty?
=> false
Here solution
primary=[1,2,3]
secondary=[2,3]
secondary.all? { |e| primary.include?(e) }
To check if an array is present inside another array, just use the include? method,
primary = [1,2,3,[1,2]]
secondary = [2,3]
result = primary.include? secondary
# returns false as [2,3] is not present
primary = [1,2,3,[1,2]]
secondary = [1,2]
result = primary.include? secondary
# returns true as [1,2] is present
To check if the elements of secondary array are present inside the primary array, takes care of the duplicate elements also:
result = (secondary.select {|i| secondary.count(i)<=primary.count(i)}.length == secondary.length)
primary, secondary = [1, 2, 3], [2, 2]
# returns false
primary, secondary = [1, 2, 2, 3], [2, 2, 1]
# returns true
a1 = [5, 3, 9, 7, 8, 7, 1, 7]
a2 = [1, 9, 5, 7, 5]
It will check elements of a2 are present in a1 or not,
a2.all?{ |i| a1.include? i } #=> true
Using
secondary.all? { |e| primary.include?(e) }
Use Intersection
(primary & secondary).size == secondary.uniq.size
If any element of secondary to be present in primary
(primary & secondary).present?
Hope it helps
OK, so I have this array of arrays. Each array within the larger array is very much the same, ten specific values. If my value at location 3 is a specific value, then I want to iterate through the rest of the remaining arrays within the larger array and see if the first 3 values at locations 0, 1, and 2 match. if they then match, I'd like to delete the original array. I'm having a hard time with it, maybe there is an easy way? I'm sure there is, I'm fairly new to this whole coding stuff =) So much appreciation in advance for your help....
here's where I'm at:
#projectsandtrials.each do |removed|
if removed[3] == ["Not Harvested"]
#arraysforloop = #projectsandtrials.clone
#arraysforloop1 = #arraysforloop.clone.delete(removed)
#arraysforloop1.each do |m|
if (m & [removed[0], removed[1], removed[2]]).any?
#projectsandtrials.delete(removed)
end
end
end
end
Lets look at your situation:
#projectsandtrials.each do |removed|
// some logic, yada yada
#projectsandtrials.delete(removed)
end
You can't just delete stuff out of an array you're iterating through. At least not until you finish iterating through it. What you should be using instead is a filtering method like reject instead of just an each.
So instead of deleting right there, you should just return true when using reject.
I think about it like this when iterating through arrays.
Do I want the array to stay the same size and have the same content?
Use each.
Do I want the array to be the same size, but have different content?
Use map.
Do I want the array to be less than or equal to the current size?
Use select or reject.
Do I want it to end up being a single value?
Use reduce.
Code
def prune(arr, val)
arr.values_at(*(0..arr.size-4).reject { |i| arr[i][3] == val &&
arr[i+1..i+3].transpose[0,3].map(&:uniq).all? { |a| a.size==1 } }.
concat((arr.size-3..arr.size-1).to_a))
end
Example
arr = [ [1,2,3,4,0],
[3,4,5,6,1],
[3,4,5,4,2],
[3,4,5,6,3],
[3,4,5,6,4],
[3,4,0,6,5],
[2,3,5,4,6],
[2,3,5,5,7],
[2,3,5,7,8],
[2,3,5,8,9],
[2,3,5,7,0]
]
Notice that the last values of the elements (arrays) of arr are consecutive. This is to help you identify the elements of prune(arr, 4) (below) that have been dropped.
prune(arr, 4)
# => [[3, 4, 5, 6, 1],
# [3, 4, 5, 4, 2],
# [3, 4, 5, 6, 3],
# [3, 4, 5, 6, 4],
# [3, 4, 0, 6, 5],
# [2, 3, 5, 5, 7],
# [2, 3, 5, 7, 8],
# [2, 3, 5, 8, 9],
# [2, 3, 5, 7, 0]]
Explanation
The arrays at indices 0 and 6 have not been included in array returned.
arr[0] ([1,2,3,4,0]) has not been included because arr[0][3] = val = 4 and arr[1], arr[2] and arr[3] all begin [3,4,5].
arr[6] ([2,3,5,4,6]) has not been included because arr[6][3] = 4 and arr[7], arr[8] and arr[9] all begin [2,3,5].
arr[2] ([3,4,5,5,2]) has been included because, while arr[2][3] = 4, arr[3][0,3], arr[4][0,3] and arr[5][0,3] all not all equal (i.e., arr[5][2] = 0).
Notice that the last three elements of arr will always be included in the array returned.
Now let's examine the calculations. First consider the following.
arr.size
#=> 11
a = (0..arr.size-4).reject { |i| arr[i][3] == val &&
arr[i+1..i+3].transpose[0,3].map(&:uniq).all? { |a| a.size==1 } }
#=> (0..7).reject { |i| arr[i][3] == val &&
arr[i+1..i+3].transpose[0,3].map(&:uniq).all? { |a| a.size==1 } }
#=> [1, 2, 3, 4, 5, 7]
Consider reject's block calculation for i=0 (recall val=4).
arr[i][3] == val && arr[i+1..i+3].transpose[0,3].map(&:uniq).all? {|a| a.size==1 }}
#=> 4 == 4 && arr[1..3].transpose[0,3].map(&:uniq).all? { |a| a.size==1 }
#=> [[3,4,5,6,1],
# [3,4,5,4,2],
# [3,4,5,6,3]].transpose[0,3].map(&:uniq).all? { |a| a.size==1 }
#=> [[3, 3, 3],
# [4, 4, 4],
# [5, 5, 5],
# [6, 4, 6],
# [1, 2, 3]][0,3].map(&:uniq).all? { |a| a.size==1 }
#=> [[3, 3, 3],
# [4, 4, 4],
# [5, 5, 5]].map(&:uniq).all? { |a| a.size==1 }
#=> [[3], [4], [5]].all? { |a| a.size==1 }
#=> true
meaning arr[0] is to be rejected; i.e., not included in the returned array.
The remaining block calculations (for i=1,...,10) are similar.
We have computed
a #=> [1, 2, 3, 4, 5, 7]
which are the indices of all elements of arr except the last 3 that are to be retained. To a we add the indices of the last three elements of arr.
b = a.concat((arr.size-3..arr.size-1).to_a)
#=> a.concat((8..10).to_a)
#=> a.concat([8,9,10])
#=> [1, 2, 3, 4, 5, 7, 8, 9, 10]
Lastly,
arr.values_at(*b)
returns the array given in the example.
Your code snippet seems fine, although there are couple of things to note:
#arraysforloop.clone.delete(removed) removes all the occurences of removed array (not only the first one). E.g. [1,2,3,1].delete(1) would leave you with [2,3]. You could fix it with using an iterator for #projectsandtrials and delete_at method.
delete method returns the same argument you pass to it (or nil if no matches found). So #arraysforloop1 = #arraysforloop.clone.delete(removed) makes your #arraysforloop1 to contain the removed array's elements only! Removing an assignment could save you.
I see no reason to have two cloned arrays, #arraysforloop and #arraysforloop1, as the former one is not used anyhow later. May be we could omit one of them?
#projectsandtrials.delete(removed) leaves you in a strange state, as long as you're iterating the same array right now. This could end up with you missing the right next element after the removed one. Here is a simple snippet to illustrate the behaviour:
> a = [1,2,3]
> a.each{|e, index| puts("element is: #{e}"); a.delete(1);}
element is: 1
element is: 3
As you see, after deleting element 1 the loop moved to element 3 directly, omitting the 2 (as it became the first element in array and algorithm thinks it's been handled already).
One of the possibilities to make it less messy is to split it to a bundle of methods. Here is an option:
def has_searched_element? row
# I leave this method implementation to you
end
def next_rows_contain_three_duplicates?(elements, index)
# I leave this method implementation to you
end
def find_row_ids_to_remove elements
[].tap do |result|
elements.each_with_index do |row, index|
condition = has_searched_element?(row) && next_rows_contain_three_duplicates?(elements, index)
result << index if condition
end
end
end
row_ids_to_remove = find_row_ids_to_remove(#projectsandtrials)
# now remove all the elements at those ids out of #projectsandtrials
I'm coding a plugin in Google Sketchup with ruby and I faced a real problem while trying to permute two arrays that are present in an array all this depending on a user combination.
I have an array of arrays like [["1"],["lol"], ["so"]]
When we have a combination like this <[1,
2, 3] it's fine, it should stay the same : [["1"],["lol"], ["so"]]
But when we have a combination like this [2, 3, 1], the output should be : [["lol"], ["so"], ["1"]]
For [3,1,2] => [["so"], ["1"], ["lol"]]
...etc
EDIT
Sorry guys I forgot for the array I have a bit like : [["1, 2, 3"], ["lol1, lol2, lol3"], ["so1, so2, so3"]] so for the combination [2, 3, 1] the output should be : [["2, 3, 1"], ["lol2, lol3, lol1"], ["so2, so3, so1"]]
Thanks for helping me out.
You could use collect:
array = [["1"],["lol"], ["so"]]
indexes = [2, 1, 3]
indexes.collect {|i| array[i-1]} #=> [["lol"], ["1"], ["so"]]
If you set the indexes to be 0-based you could drop the -1
split and map can be used to turn your strings into values:
"1, 2, 3".split(",").map { |i| i.to_i} # [1, 2, 3]
You can then also split your strings
"lol2, lol3, lol1".split(/, /) #=> ["lol2", "lol3", "lol1"]
You should be able to put that together with the above to get what you want.
indexes = [2, 1, 3]
array = [["1"],["lol"], ["so"]]
result = indexes.map{|index| array[index-1] }
You should also take a look at active_enum
https://github.com/adzap/active_enum
You could do something like:
class YourClassName < ActiveEnum::Base
value [1] => ['1']
value [2] => ['lol']
value [3] => ['so']
end
a = [["1"], ["lol"], ["so"]]
index = [2, 1, 3]
index.collect {|i| a[i - 1]}
This outputs
[["lol"], ["1"], ["so"]]