I'm looking for a method that is similar to Array#combination, but the order matters.
Given this input:
array = ['a','b','c','d','e']
I'd like to get:
[['a','b','c'],['b','c','d'],['c','d','e']]
I'm trying to find the method that does this:
array = ['a','b','c','d','e']
x,a = 3,[]
until x > (ary.size) do
a += (0.upto(ary.size - x).map{|i| ary[i..(x-1)+i]} )
x += 1
end
The Enumerable documentation is your friend:
array = ['a','b','c','d','e']
array.each_cons(3).to_a
# => [["a", "b", "c"], ["b", "c", "d"], ["c", "d", "e"]]
each_cons(n) { ... }
Iterates the given block for each array of consecutive elements. If no block is given, returns an enumerator.
Related
In Ruby 2.4, how do I find the earliest index of an element of an array in another array? That is, if any element of an array occurs in the other array, I want to get the first index. I thought find_index might do it, but
a = ["a", "b", "c"]
# => ["a", "b", "c"]
a.find_index("a")
# => 0
a.find_index(["b", "c"])
# => nil
In the above example, I would expect to see the output "1" because the element "b" occurs at index 1 in the array "a".
find_index takes a single element. You could find the minimum by doing something like
a = ["a", "b", "c"]
to_find = ["b", "c"]
to_find.map {|i| a.find_index(i) } .compact.min # => 1
You can use find_index and pass the needed value from the array:
a = ["a", "b", "c"]
p a.find_index('a')
p a.find_index('b')
p a.find_index('c')
# => 0
# => 1
# => 2
You can use map to get every element inside your a array and then to get the index corresponding to each element:
p a.map{|e| a.find_index(e)}
# => [0, 1, 2]
Another possible way to handle it could be to use the Enumerable#each_with_index:
a.each_with_index{|e,i| puts "Element: #{e}, Index: #{i}"}
# => Element: a, Index: 0
# => Element: b, Index: 1
# => Element: c, Index: 2
If you want to check the indexes for each element in ["b", "c"] using the ["a", "b", "c"] array, you can map the first one, get the array values, and then use the a,b,c to check those indexes:
p ["b", "c"].map{|e| ["a", "b", "c"].find_index(e) }
# => [1, 2]
You can also see Array#index and Enumerable#find_index.
You can use Array#index with a block.
a = ['a', 'b', 'c']
a.index { |x| ['b', 'c'].include?(x) }
#=> 1
Quote from the docs:
If a block is given instead of an argument, returns the index of the first object for which the block returns true. Returns nil if no match is found.
As Cary pointed out in his comment it is not the most performant algorithm to compare all elements in a against all elements in ['b', 'c'] (this would lead to O(n*m)). Depending on the size of both arrays if might sense to build a more efficient data structure first. Using a Set instead of an Array has some cost in creating the set upfront, but makes the comparison in the block much faster (overall O(n+m)):
require 'set'
a = ['a', 'b', 'c']
set = Set.new(['b', 'c'])
a.index { |x| set.include?(x) }
#=> 1
You could find index of all elements of array b in the array a, and find the min index to find the index at which an element from array b occurred first in array a.
Something like below:
a = ["a", "b", "c"]
b = ["b", "c"]
b.map { |x| a.find_index(x) }.min
#=> 1
I am writing some code to find the longest palindrome in a string. I want to start at index 0 and then push the increasing length of the substring to an array:
ex:
string = "ababa"
[["a", "b"], ["a", "b", "a"], ["a", "b", "a", "b"], ["a", "b", "a", "b", "a"]]
It should then start on index 1 and do the same:
ex:
string = "ababa"
[["b","a"],["b","a","b"],["b","a","b","a"]
This should continue until the index is length -1 .However, the following code stops after it has gone through all iterations beginning with the first index and only returns:
[["a", "b"], ["a", "b", "a"], ["a", "b", "a", "b"], ["a", "b", "a", "b", "a"]]
What is the flaw in my logic? Code below
def longest_palindrome(s)
array = s.chars
start = 0
place = 1
output = []
while start < s.length - 1
while place < s.length
output << array[start..place]
place += 1
end
start += 1
end
return output
end
I believe this is what you are after:
def longest_palindrome(s)
arr = s.chars
output = []
(0...s.length).each do |start|
(start + 1...s.length).each do |place|
output << arr[start..place]
end
end
output
end
longest_palindrome("ababa")
=> [["a", "b"], ["a", "b", "a"], ["a", "b", "a", "b"], ["a", "b", "a", "b", "a"], ["b", "a"], ["b", "a", "b"], ["b", "a", "b", "a"], ["a", "b"], ["a", "b", "a"], ["b", "a"]]
Using iterators really simplifies things. Here is a more concise version:
def longest_palindrome(s)
output = (0...s.length).flat_map do |start|
(start + 1...s.length).map do |place|
s[start..place]
end
end
end
longest_palindrome("ababa")
=> ["ab", "aba", "abab", "ababa", "ba", "bab", "baba", "ab", "aba", "ba"]
A working, non-optimized, written on-the-fly 30 minutes, who knows how awfully slow longest palindrome finder:
def longest_palindrome(string, min_size = 2)
string = string.downcase # we will not be taking
return string if string == string.reverse # skip all calculaions of the passed string itself is a palindrome in original order
letters = string.chars # Convert string to array of bytes
combinations = [] # Initialize all letter combinations
(min_size..letters.size).each do |n| # min_size is the shortest length a palindrome is allowed to be, default 2
combinations.concat(letters.combination(n).to_a) # concat all combinations for n amount of characters in the string
end
palindromes = [] # Initialize array for all palindromes
combinations.each do |combo| # interate every combo
combo.size.times do # for every letter ordering is done via size of this letter combo
palindromes << combo.dup if combo == combo.reverse # add to list of palindromes if the combinations is the same backwards
combo.rotate! # rotate the letters for next order checking
end
end
palindromes.sort {|a, b| a.size <=> b.size }.last.join # sort the palidromes by length, take the biggest one, and return it as a full string
end
p longest_palindrome("racecar") #=> racecar
p longest_palindrome("applesauce") #=> pecep
p longest_palindrome("madam im adam") #=> mada m adam
p longest_palindrome("madamimadam") #=> madamimadam
but it works, sort of.
As you can see if sort of gets odd when spaces/puncuation are added to the mix.
You can always clean the string of it in the first line if need be.
Viva la algorithmless coding!
I need to render the contents of the first 5 elements of an array and display "And X more" on a web page. Is there a built-in method on Array (or Enumerable) that easily separates one array into two sub-arrays: the first consisting of up to a fixed size and the second consisting of the array remainder?
I'm looking for one simple method call that will do this for me. Most of the methods that I looked at (like Enumerable#partition) use a logical condition to divide the array and don't supply the index to the block.
I just wrote the following code to do what I want. Please save me from myself and direct me to a method that already does it.
class Array
def bifurcate(size=length)
if size < 0
raise ArgumentError, "attempt to bifurcate using negative size"
end
remainder_size = length - size
if remainder_size < 0
remainder_size = 0
end
[
first(size),
last(remainder_size)
]
end
end
('a'..'g').to_a.bifurcate(2)
# => [["a", "b"], ["c", "d", "e", "f", "g"]]
('a'..'g').to_a.bifurcate(20)
# => [["a", "b", "c", "d", "e", "f", "g"], []]
('a'..'g').to_a.bifurcate()
# => [["a", "b", "c", "d", "e", "f", "g"], []]
('a'..'g').to_a.bifurcate(0)
# [[], ["a", "b", "c", "d", "e", "f", "g"]]
('a'..'g').to_a.bifurcate(-1)
# ArgumentError: attempt to bifurcate using negative size
Also, let me qualify that I want one simple method call to do what I want. Also consider that the starting array may contain duplicate values and this method needs to respect the original array and return duplicates.
You can use Enumerable#partition along with Enumerator#with_index method, as shown below:
size = 2
(1..6).partition.with_index { |_,i| i < size }
#=> [[1, 2], [3, 4, 5, 6]]
Alternatively, if your input array can be mutated, then, following will also do the trick
[array.shift(size), array]
[array.take(3), array.drop(3)]
# [["a", "b", "c"], ["d", "e", "f", "g"]]
Hope it helps :)
Use Array#[]:
[arr[0,size_of_first], arr[size_of_first..-1] || []]
I am trying to delete elements from an array if its index is greater than a certain value. I am looking to do something like this:
a = ["a", "b", "c"]
b = a.delete_if {|x| x.index > 1 }
I took a look at drop, delete_if, etc. I tried completing this using each_with_index like this:
new_arr = []
a.each_with_index do |obj, index|
if index > 1
obj.delete
end
new_arry << obj
end
How can I delete an array element if it's array position is greater than a certain value?
Here are some other ways to return a sans elements at indices >= index, which is probably better expressed as "returning the first index elements". All below return ["a", "b"]).
a = ["a", "b", "c", "d", "e"]
index = 2
Non-destructive (i.e., a is not altered)
a[0,index]
index.times.map { |i| a[i] }
Destructive (a is modified or "mutated")
a.object_id #=> 70109376954280
a = a[0,index]
a.object_id #=> 70109377839640
a.object_id #=> 70109377699700
a.replace(a.first(index))
a.object_id #=> 70109377699700
You can use slice! and give it a range. It is a destructive method as indicated by the !, so it will mutate your array.
a = [1, 2, 3, 4]
a.slice!(2..-1)
a = [1, 2]
Array#first gives you the first n elements.
b = a.first(1)
# => ["a"]
If you want to do it in a destructive way, then this will do:
a.pop(a.length - 1)
a # => ["a"]
You can append with_index:
a = ["a", "b", "c"]
a.delete_if.with_index { |x, i| i > 1 }
a #=> ["a", "b"]
Another example:
a = ("a".."z").to_a
a.delete_if.with_index { |x, i| i.odd? }
#=> ["a", "c", "e", "g", "i", "k", "m", "o", "q", "s", "u", "w", "y"]
Going by your question, "How can I delete an array element if it's array position is greater than a certain value?".
I assume what you want is that the final array you have should contain only elements before the specified index.
You can just do this:
your_array.select { |element| your_array.index(element) < max_index }
E.g
figures = [1,2,3,4,5,6]
figures.select{ |fig| figures.index(fig) < 3 }
# => [1, 2, 3]
I am trying to count the number of similar prefix beginnings to a string in Ruby. e.g; input "ababaa" should output 11;
ababaa = 6
babaa = 0
abaa = 3
baa = 0
aa = 1
a = 1
I have got as far as the code below, using a nested loop to go through each of the above as an array, however it looks as though Ruby is currently outputting the count of just the first Array object, "ababaa".
Solved, thanks :)
def string_suffix(string)
num = 0
ary = []
string.length.times do
ary << string[num..string.length]
num = num + 1
end
result = 0
ary.each do |x| # ["ababaa", "babaa", "abaa", "baa", "aa", "a"]
x.chars.each_with_index do |c,index|
break unless c == string[index]
result = result + 1
end
end
return result
end
I have looked far and wide and still cannot solve the issue, It looks like the (final, nested) array is breaking after the first iteration of the 'ary' Array and just returning that output.
You are returning the result while you are still in the loop. You need to move result = 0 out of the loop, and move the return result statement outside of the loop too. At the moment the function is going through the first iteration of the loop ("ababaa", for which all characters match), but you want result to equal the sum of all results.
Additionally, instead of doing:
count = 0
x.chars.each do |x|
if x == string[count]
count = count + 1
result = result + 1
else
count = count + 1
end
end
You could use the function each_with_index, to get
x.chars.each_with_index do |c,index|
if c == string[index]
result = result + 1
end
end
However, since you are trying to count how many characters in the substring are a prefix of string, you want to break when you first find a character c that is not equal to string[index], so that you don't end up counting extra characters. The loop then becomes:
x.chars.each_with_index do |c,index|
if c == string[index]
result = result + 1
else
break
end
end
I noticed you are returning the result inside your second loop, at the end. This means that after you've gone through the first item in your array the function returns just the result for the first item. Move your return statement to outside the loop.
As I understand, the problem is this: given a string s, for each i = 0..s.size-1, compute the number of leading characters of s[0..-i-1] that match the corresponding characters (i.e., at same offsets) of s[i..-1], and sum these s.size subtotals.
Here's a Ruby-like way to do that, using Enumerable#reduce (aka inject) and Enumerable#take_while:
str = "ababaa"
arr = str.chars
(0...arr.size).reduce(0) do |tot,i|
tot + arr[0..-i-1].zip(arr[i..-1]).take_while { |x,y| x == y }.size
end
#=> 11
The steps:
arr = str.chars
#=> ["a", "b", "a", "b", "a", "a"]
r = 0...arr.size
#=> 0...6
When the first element of r is passed to the block, the block variables are set to:
tot = 0
i = 0
The block calculation is therefore as follows:
a = arr[0..-i-1].zip(arr[i..-1])
#=> arr[0..-1].zip(arr[0..-1])
#=> arr.zip(arr)
#=> ["a", "b", "a", "b", "a", "a"].zip(["a", "b", "a", "b", "a", "a"])
#=> [["a", "a"], ["b", "b"], ["a", "a"], ["b", "b"], ["a", "a"], ["a", "a"]]
b = a.take_while { |x,y| x == y }
#=> [["a", "a"], ["b", "b"], ["a", "a"], ["b", "b"], ["a", "a"], ["a", "a"]]
tot + b.size
#=> 0 + 6
#=> 6
Note that this calculation will always equal arr.size for the first element of arr passed to the block.
When the next element of arr is passed to the block, the block variable i is set to 1. tot, which we just computed, equals 6. The block calculation is therefore:
a = arr[0..-i-1].zip(arr[i..-1])
#=> arr[0..-2].zip(arr[1..-1])
#=> ["a", "b", "a", "b", "a"].zip(["b", "a", "b", "a", "a"])
#=> [["a", "b"], ["b", "a"], ["a", "b"], ["b", "a"], ["a", "a"]]
b = a.take_while { |x,y| x == y }
#=> []
tot + b.size
#=> 6 + 0
#=> 6
The remaining calculations are similar. After all elements of arr have been sent to the block, reduce returns the value of tot.