I have an array called #results that is made up only of arrays. I want to iterate through #results and permanently delete any of the inner arrays that are smaller than a given size:
My code:
def check_results limit
#results.each_with_index do |result, index|
#results.delete_at(index) if result.size < limit
end
end
Unfortunately, this only deletes the first item where the array length is less than limit. For example if limit = 4 and #results = [[1, 1, 1, 1], [1, 1, 1, 1], [1, 1, 1], [1, 1]] then check_results returns [[1, 1, 1, 1], [1, 1, 1, 1], [1, 1]]
I can't figure out why this is happening. Am I using the wrong loop?
You should do this, as delete_at modifies the array, and you will get unexpected behavior if you are deleting elements while iterating it
#results.reject { |i| i.size < limit }
Above code will exclude all array elements whose size is smaller than limit
It's not a good idea to modify the #results array in place as that will conflict with the outer iteration.
What you should do instead is use select to build a new array.
def check_results(limit)
#result.select { |result| result.size > limit }
end
As per the documentation, #delete_at returns the element at that index.
a = ["ant", "bat", "cat", "dog"]
a.delete_at(2) #=> "cat"
a #=> ["ant", "bat", "dog"]
a.delete_at(99) #=> nil
I added some debug statements to show you what is happening at each step, assuming limit is 4:
#results = [[1, 1, 1, 1], [1, 1, 1, 1], [1, 1, 1], [1, 1]]
#results.each_with_index do |r, i|
puts "RESULT: #{r.to_s}"
puts "INDEX: #{i}"
#results.delete_at(i) if r.size < 4
puts "ARRAY: #{#results.to_s}"
end
RESULT: [1, 1, 1, 1]
INDEX: 0
ARRAY: [[1, 1, 1, 1], [1, 1, 1, 1], [1, 1, 1], [1, 1]]
RESULT: [1, 1, 1, 1]
INDEX: 1
ARRAY: [[1, 1, 1, 1], [1, 1, 1, 1], [1, 1, 1], [1, 1]]
RESULT: [1, 1, 1]
INDEX: 2
ARRAY: [[1, 1, 1, 1], [1, 1, 1, 1], [1, 1]]
# #results == [[1, 1, 1, 1], [1, 1, 1, 1], [1, 1]]
As you can see, the element originally at index 2 has been removed. Because you are modifying #results while you are iterating through it, an index of 3 no longer exists, and an index of 2 has already been analyzed. This is why you should not modify an object while iterating through it.
Ideally, you want to use #delete_if. Similar to methods ending in !, #delete_if will modify the array (not return a copy of the result), based on conditions from a block (as an argument). The following would be how you would implement the method:
def check_results(limit)
#results.delete_if { |arr| arr.length < limit }
end
#results = [ ['foo', 'bar'], ['bizz', 'bazz'], ['kaboom'] ]
check_results(2)
# => #results == [ ['foo', 'bar'], ['bizz', 'bazz'] ]
If you do not want to modify #results, then I suggest a similar method, #reject. Again, #results will not be modified, and instead a copy of the results will be returned.
def check_results(limit)
#results.reject { |arr| arr.length < limit }
end
#results = [ ['foo', 'bar'], ['bizz', 'bazz'], ['kaboom'] ]
check_results(2)
# => [ ['foo', 'bar'], ['bizz', 'bazz'] ]
# => #results == [ ['foo', 'bar'], ['bizz', 'bazz'], ['kaboom'] ]
Related
Normally in backtracking, we take a helper function which takes in an initial state and each recursive call takes care of its own computation and pass the result to the next recursion call. Theoretically, we denote this through unseen and seen variable.
For example, in permutation for a string we will use this program:
def permute(str)
return str if str.length < 2
permute_helper(str, "")
end
def permute_helper(unseen, seen)
#base case
if unseen.length <= 0
p seen
return
else
(0..unseen.length-1).each do |i|
buffer = unseen
buffer = buffer.split('')
buffer.delete_at(i)
buffer = buffer.join('')
permute_helper(buffer, seen+unseen[i])
end
end
end
permute('abc')
Thie will print out the required results.
I was asked to do this without using two variables in a recent interview. without storing state in the seen variable. I couldn't think through the whole at that time but I would like to ask how to do backtracking without storing states ?
The permutations of the string "cd" is ["cd", "dc"]. If we now wish to obtain the permutations of the string "bcd" we simply replace each element of this array with three strings, each having "b" at a different position. "cd" becomes "bcd", "cbd" and "cdb" and "dc" becomes "bdc", "dbc" and "dba". The permutations of "bcd" are therefore
["bcd", "cbd", "cdb", "bdc", "dbc", "dba"]
If we now wish to obtain the permutations of "abcd", we replace each element of the above six-element array with four strings, each with "a" in a different position. For example, "bcd" becomes "abcd", "bacd", "bcad" and "bcda". The structure of the recursion should now be obvious.
def permute(str)
case str.length
when 0, 1
str
when 2
[str, str.reverse]
else
first = str[0]
sz = str.size-1
permute(str[1..-1]).flat_map { |s| (0..sz).map { |i| s.dup.insert(i,first) } }
end
end
permute('')
#=> ""
permute('a')
#=> "a"
permute('ab')
#=> ["ab", "ba"]
permute('abc')
#=> ["abc", "bac", "bca", "acb", "cab", "cba"]
permute('abcd')
#=> ["abcd", "bacd", "bcad", "bcda", "acbd", "cabd", "cbad", "cbda",
# "acdb", "cadb", "cdab", "cdba", "abdc", "badc", "bdac", "bdca",
# "adbc", "dabc", "dbac", "dbca", "adcb", "dacb", "dcab", "dcba"]
str is of course the "unseen" variable.
#CarySwoveland's answer an explanation is awesome, per usual. For those looking to permute an array, consider this functional approach. While this uses an auxiliary lambda all_pos, no extra state parameter is used to accumulate the result.
def permute ((x, *xs))
all_pos = lambda do |(y,*ys)|
if y.nil?
[[ x ]]
else
[[ x, y, *ys ]] + (all_pos.call ys) .map { |rest| [ y, *rest ] }
end
end
if x.nil? or xs.empty?
[[x]]
else
(permute xs) .flat_map &all_pos
end
end
permute [1,2,3,4]
# [ [1, 2, 3, 4]
# , [2, 1, 3, 4]
# , [2, 3, 1, 4]
# , [2, 3, 4, 1]
# , [1, 3, 2, 4]
# , [3, 1, 2, 4]
# , [3, 2, 1, 4]
# , [3, 2, 4, 1]
# , [1, 3, 4, 2]
# , [3, 1, 4, 2]
# , [3, 4, 1, 2]
# , [3, 4, 2, 1]
# , [1, 2, 4, 3]
# , [2, 1, 4, 3]
# , [2, 4, 1, 3]
# , [2, 4, 3, 1]
# , [1, 4, 2, 3]
# , [4, 1, 2, 3]
# , [4, 2, 1, 3]
# , [4, 2, 3, 1]
# , [1, 4, 3, 2]
# , [4, 1, 3, 2]
# , [4, 3, 1, 2]
# , [4, 3, 2, 1]
# ]
I'm a beginner and I want to create a matrix. For example:
0 1 1
1 1 1
1 1 2
irb(main):001:0> t = [[1]*3]*3
=> [[1, 1, 1], [1, 1, 1], [1, 1, 1]]
irb(main):002:0> (0...3).each do |x| t[x][x]=x end
=> 0...3
irb(main):003:0> t
=> [[0, 1, 2], [0, 1, 2], [0, 1, 2]] # why all values changed?
What's wrong?
The way you construct the array does not cretae new arrays for each row but references the same array for all rows:
t.each do |row|
p row.object_id
end
# 70325094342320
# 70325094342320
# 70325094342320
It is the same as:
a = [1, 1, 1]
t = [a, a, a]
Try this to see the difference:
t = [[1] * 3, [1] * 3, [1] * 3]
Say I have an array [1,2,3] and I want every combination of these numbers that don't exceed 4. So I would have [1,2,3].someMethod(4) and it would give me:
[1,1,1,1]
[1,1,2]
[1,3]
[2,2]
So far I have:
(1..4).flat_map{|size| [1,2,3].repeated_combination(size).to_a }
but this gives me every possible combinations, including the ones that exceed my given limit. Is there an good way to either only get combinations that add up to my limit?
arr = [1,2,3]
(arr+[0]).repeated_combination(4).select { |a| a.reduce(:+) == 4 }.map { |a| a - [0] }
#=> [[1, 3], [2, 2], [1, 1, 2], [1, 1, 1, 1]]
Change == to <= if desired.
This answer, like the others, assumes arr contains natural numbers, including 1.
results = (1..4).each.with_object([]) do |size, results|
[1,2,3].repeated_combination(size) do |combo|
results << combo if combo.reduce(:+) == 4
end
end
p results
--output:--
[[1, 3], [2, 2], [1, 1, 2], [1, 1, 1, 1]]
Parameterizing the algorithm:
def do_stuff(values, target_total)
(1..target_total).each.with_object([]) do |size, results|
values.repeated_combination(size) do |combo|
results << combo if combo.reduce(:+) == 4
end
end
end
p do_stuff([1, 2, 3], 4)
You can filter out the arrays you don't want by using the select method. Just select all the arrays that have a sum == 4 (the sum is calculated by the inject method).
all_arrs = (1..4).flat_map do |size|
[1,2,3].repeated_combination(size).to_a
end
valid_arrs = all_arrs.select do |arr|
arr.inject { |a, b| a + b } == 4
end
print valid_arrs
# Output:
# [[1, 3], [2, 2], [1, 1, 2], [1, 1, 1, 1]]
A recursive approach.
def some_method(a, n)
return [[]] if n == 0
a.select { |e| e <= n }.\
flat_map { |e| some_method(a,n-e).map { |es| ([e] + es).sort } }.\
sort.\
uniq
end
p some_method([1,2,3], 4)
# => [[1, 1, 1, 1], [1, 1, 2], [1, 3], [2, 2]]
EDIT: Here is another recursive version without filtering duplicates but with opposite order. I added comments to make it clearer.
def some_method(a, n)
return [[]] if n == 0 # bottom (solution) found
return [] if a.empty? || n < 0 # no solution
max = a.max
# search all solutions with biggest value
l = some_method(a, n-max).map { |e| [max] + e }
# search all solutions without biggest value
r = some_method(a-[max],n)
l + r
end
p some_method([1,2,3], 4)
# => [[3, 1], [2, 2], [2, 1, 1], [1, 1, 1, 1]]
Given:
a = [[1,2,3,4],
[1,2,3,7],
[1,2,3,4]]
What do I need to do to output the location of the 7 as (1,3)?
I've tried using .index to no avail.
require 'matrix'
a = [[1, 2, 3, 4],
[1, 2, 3, 7],
[1, 2, 3, 4]]
Matrix[*a].index(7)
=> [1, 3]
If your sub-arrays are all the same width, you can flatten it into a single array and think of the position as row_num * row_width + col_num:
idx = a.flatten.index(7)
row_width = a[0].length
row = idx / row_width
col = idx - (row * row_width)
puts [row, col] # => [1, 3]
Or you could just iterate it to find all matches:
def find_indices_for(array, value)
array.with_object([]).with_index do |(row, matches), row_index|
matches << [row_index, row.index(value)] if row.index(value)
end
end
find_indices_for(a, 7) # => [[1, 3]]
find_indices_for(a, 2) # => [[0, 1], [1, 1], [2, 1]]
each_with_index works pretty well here:
def locator(array, number)
locations = Array.new
array.each_with_index do |mini_array, index|
mini_array.each_with_index do |element, sub_index|
locations << [index, sub_index] if element == number
end
end
locations
end
Now, locator(array, number) will return an array of containing all the locations of number in array.
def locs2D(a,e)
a.size.times.with_object([]) do |row,arr|
row.size.times { |col| arr << [row,col] if a[row][col] == e }
end
end
locs2D(a,7) #=> [[1, 3]]
locs2D(a,3) #=> [[0, 2], [1, 2], [2, 2]]
I'd like to get [[2, 1, 3], [1, 3, 2]] from [1, 2, 3] in Ruby.
For [1, 2, 3, 4], I'd like to get [[2, 1, 3, 4], [1, 3, 2, 4], [1, 2, 4, 3]]
Rule: Within two numbers, if left one is smaller then it swap the position.
I have the following codes so far but it returns [[2, 3, 1], [2, 3, 1]]
What am I doing wrong here? I appreciate any inputs.
In amidakuji.rb
class Amidakuji
def initialize(column, rung)
#column = column
#rung = rung
#myarr = []
#per_arr = []
#build_arr = []
end
def build_initial
#arr = (1..#column).to_a
end
def swap_element
i = 0
arr = build_initial
while i < #column - 1 do
#build_arr << swap(arr, i)
i += 1
end
#build_arr
end
def swap(arr, a)
if arr[a] < arr[a + 1]
arr[a], arr[a + 1] = arr[a + 1], arr[a]
end
arr
end
end
In amidakuji_spec.rb
it 'should create an array with swapped elements' do
expect(#kuji1.swap_element).to eq ([[2, 1, 3], [1, 3, 2]])
end
Results
Failures:
expected: [[2, 1, 3], [1, 3, 2]]
got: [[2, 3, 1], [2, 3, 1]]
You can do this quite compactly by using the methods Enumerable#each_cons and Enumerable#map.
Code
def doit(arr)
(0...arr.size).each_cons(2).map do |i,j|
a = arr.dup
a[i], a[j] = a[j], a[i]
a
end
end
Examples
doit([1,2,3]) #=> [[2, 1, 3], [1, 3, 2]]
doit([1,2,3,4]) #=> [[2, 1, 3, 4], [1, 3, 2, 4], [1, 2, 4, 3]]
doit([1,2,3,4,5]) #=> [[2, 1, 3, 4, 5], [1, 3, 2, 4, 5],
#=> [1, 2, 4, 3, 5], [1, 2, 3, 5, 4]]
Explanation
arr = [1,2,3,4]
b = (0...arr.size).each_cons(2)
#=> #<Enumerator: 0...4:each_cons(2)>
To view the contents of this enumerator:
b.to_a
#=> [[0, 1], [1, 2], [2, 3]]
Lastly
b.map do |i,j|
a = arr.dup
a[i], a[j] = a[j], a[i]
a
end
#=> [[2, 1, 3, 4], [1, 3, 2, 4], [1, 2, 4, 3]]
In the last step, consider the first element of b that is passed to map, which assigns the following values to the block variables:
i => 0
j => 1
We then make a copy of arr, swap the elements offsets 0 and 1, making
a => [2, 1, 3, 4]
and then enter a at the end of the block, causing map to replace [0, 1] with that array.
Given what you're trying to accomplish and the output you're getting, it looks like you're reusing the same array when you want distinct arrays instead. Specifically this line:
#build_arr << swap(arr, i)
is always passing the same 'arr' to swap.
So first time, it swaps the 1 and the 2 to give you [2, 1, 3]
Second time, it swaps the 1 and the 3 give you [2, 3, 1]
You push the same array onto #build_arr twice, which is why it repeats.