Related
I have a complex problem of string manipulation at hand.
I have a string in which I will have cycles, as well as recurrences which I need to identify and list down.
'abcabcabcabcabcdkkabclilabcoabcdieabcdowabcdppabzabx'
Following are the possible patterns ->
Actual indexes not used
abc -> 0,3,6,9,12,15,17, ..... (occurence index for recurring string),
0,3,6,9 (unique_occurence index for recurring string, 12, 15, 17
disqualified as there abc was a part of longer repeating substring)
abcd -> 12, 15, 17 (occurence index for recurring string), 12, 15, 17
(unique occurence index for recurring string)
bcda -> 13, 16, 18.. (occurence index for recurring string), (unique occurence index for recurring string) as it is an overlap for
the string abcd Hence it is something not required ab ->
0,3,6,9,12,15,17, 25, 27 ...(occurence index for recurring string),
25, 27(unique occurence index for recurring string). .....
I want to find all unique recurring occurences/recurrences, i.e. All Unique, Non-Overlapping values of recurring string. As mentioned above. And the input string may contain,
ALL cyclic patterns(abcabcabcdefdefdeflkjlkjlkj => abc, def, lkj are recurrences in cycle, but bc, ab, bcab are not expected as they are outcomes of false positives)
OR
Separately recurring patterns(abcxabcdabcm => abc is recurrence but not cycle, i.e. they are not adjecent)
Or
A mix of both(abcabcabcabcabclkabcdokabcdhuabcd => abc is a cyclic recurrence, and abcd is a non cyclic recurrence and we need to find both -> only abcd, abc are recurring, not bc, ab, bcda, etc)
Can someone propose a solution algo for this problem statement. I am trying using suffix_arrays which is not finding overlapping results as well.
A hash is constructed whose keys consist of all unique substrings of a given string that appear at least twice in the string (not overlapping) and, for each key, the value is an array of all offsets into the string where the value of the key (a substring) begins.
Code
def recurring_substrings(str)
arr = str.chars
(1..str.size/2).each_with_object({}) do |n,h|
arr.each_cons(n).map { |b| b.join }.uniq.each do |s|
str.scan(Regexp.new(s)) { (h[s] ||= []) << Regexp.last_match.begin(0) }
end
end.reject { |_,v| v.size == 1 }
end
Examples
recurring_substrings 'abjkabrjkab'
#=> {"a"=>[0, 4, 9], "b"=>[1, 5, 10], "j"=>[2, 7], "k"=>[3, 8], "ab"=>[0, 4, 9],
# "jk"=>[2, 7], "ka"=>[3, 8], "jka"=>[2, 7], "kab"=>[3, 8], "jkab"=>[2, 7]}
recurring_substrings "abcabcabcabcabcdkkabclilabcoabcdieabcdowabcdppabzabx"
#=> {"a"=>[0, 3, 6, 9, 12, 18, 24, 28, 34, 40, 46, 49],
# "b"=>[1, 4, 7, 10, 13, 19, 25, 29, 35, 41, 47, 50],
# "c"=>[2, 5, 8, 11, 14, 20, 26, 30, 36, 42], "d"=>[15, 31, 37, 43],
# "k"=>[16, 17], "l"=>[21, 23], "i"=>[22, 32], "o"=>[27, 38], "p"=>[44, 45],
# "ab"=>[0, 3, 6, 9, 12, 18, 24, 28, 34, 40, 46, 49],
# "bc"=>[1, 4, 7, 10, 13, 19, 25, 29, 35, 41], "ca"=>[2, 5, 8, 11],
# "cd"=>[14, 30, 36, 42],
# "abc"=>[0, 3, 6, 9, 12, 18, 24, 28, 34, 40], "bca"=>[1, 4, 7, 10],
# "cab"=>[2, 5, 8, 11], "bcd"=>[13, 29, 35, 41],
# "abca"=>[0, 6], "bcab"=>[1, 7], "cabc"=>[2, 8], "abcd"=>[12, 28, 34, 40],
# "abcab"=>[0, 6], "bcabc"=>[1, 7], "cabca"=>[2, 8],
# "abcabc"=>[0, 6], "bcabca"=>[1, 7], "cabcab"=>[2, 8]}
Explanation
For the first example above, the steps are as follows.
str = 'abjkabrjkab'
arr = str.chars
#=> ["a", "b", "j", "k", "a", "b", "r", "j", "k", "a", "b"]
q = str.size/2 # max size for string to repeat at least once
#=> 5
b = (1..q).each_with_object({})
#=> #<Enumerator: 1..5:each_with_object({})>
We can see which elements will be generated by this enumerator by converting it to an array. (I will do this a few more times below.)
b.to_a
#=> [[1, {}], [2, {}], [3, {}], [4, {}], [5, {}]]
The empty hashes will be built up as calculations progress.
Next pass the first element to the block and set the block variables to it using parallel assignment (sometimes called multiple assignment).
n,h = b.next
#=> [1, {}]
n #=> 1
h #=> {}
c = arr.each_cons(n)
#=> #<Enumerator: ["a", "b", "j", "k", "a", "b", "r", "j", "k", "a", "b"]:each_cons(1)>
c is an array of all substrings of length 1. At the next iteration it will be an array of all substrings of length 2 and so on. See Emumerable#each_cons.
c.to_a # Let's see which elements will be generated.
#=> [["a"], ["b"], ["j"], ["k"], ["a"], ["b"], ["r"], ["j"], ["k"], ["a"], ["b"]]
d = c.map { |b| b.join }
#=> ["a", "b", "j", "k", "a", "b", "r", "j", "k", "a", "b"]
e = d.uniq
#=> ["a", "b", "j", "k", "r"]
At the next iteration this will be
r = arr.each_cons(2)
#=> #<Enumerator: ["a", "b", "j", "k", "a", "b", "r", "j", "k", "a", "b"]:
# each_cons(2)>
r.to_a
#=> [["a", "b"], ["b", "j"], ["j", "k"], ["k", "a"], ["a", "b"],
# ["b", "r"], ["r", "j"], ["j", "k"], ["k", "a"], ["a", "b"]]
s = r.map { |b| b.join }
#=> ["ab", "bj", "jk", "ka", "ab", "br", "rj", "jk", "ka", "ab"]
s.uniq
#=> ["ab", "bj", "jk", "ka", "br", "rj"]
Continuing,
f = e.each
#=> #<Enumerator: ["a", "b", "j", "k", "r"]:each>
f.to_a # Let's see which elements will be generated.
#=> ["a", "b", "j", "k", "r"]
s = f.next
#=> "a"
r = (Regexp.new(s))
#=> /a/
str.scan(r) { (h[s] ||= []) << Regexp.last_match.begin(0) }
If h does not yet have a key s, h[s] #=> nil. h[s] ||= [], which expands to h[s] = h[s] || [], converts h[s] to an empty array before executing h[s] << Regexp.last_match.begin(0). That is, h[s] = h[s] || [] #=> nil || [] #=> [].
Within the block the MatchData object is retrieved with the class method Regexp::last_match. (Alternatively, one could substitute the global variable $~ for Regexp.last_match. For details, search for "special global variables" at Regexp.) MatchData#begin returns the index of str at which the current match begins.
Now
h #=> {"a"=>[0, 4, 9]}
The remaining calculations are similar, adding key-value pairs to h until the has given in the example has been constructed.
For further processing after #CarySwoveland's excellent answer :
def ignore_smaller_substrings(hash)
found_indices = []
new_hash = {}
hash.sort_by{|s,_| [-s.size,s]}.each{|s,indices|
indices -= found_indices
found_indices |= indices
new_hash[s]=indices unless indices.empty?
}
new_hash
end
pp ignore_smaller_substrings(recurring_substrings('abcabcabcabcabcdkkabclilabcoabcdieabcdowabcdppabzabx'))
Hash is sorted by decreasing string length (and then alphabetically), and indices are only allowed to appear once.
It outputs
{"abcabc"=>[0, 6],
"bcabca"=>[1, 7],
"cabcab"=>[2, 8],
"abcd"=>[12, 28, 34, 40],
"abc"=>[3, 9, 18, 24],
"bca"=>[4, 10],
"bcd"=>[13, 29, 35, 41],
"cab"=>[5, 11],
"ab"=>[46, 49],
"bc"=>[19, 25],
"cd"=>[14, 30, 36, 42],
"b"=>[47, 50],
"c"=>[20, 26],
"d"=>[15, 31, 37, 43],
"i"=>[22, 32],
"k"=>[16, 17],
"l"=>[21, 23],
"o"=>[27, 38],
"p"=>[44, 45]}
It doesn't answer the question exactly, but it comes a bit closer.
I'm writing a poker program, and I can't figure out how to handle straights.
Straight: All cards in a hand of 5 cards are consecutive values.
ex. 2..6, 3..7, 4..8, 5..9, 6..T, 7..J, 8..Q, 9..K, T..A
cards = [2, 3, 4, 5, 6, 7, 8, 9, "T", "J", "Q", "K", "A"]
How can I check a hand, which is an array, for these combinations? Preferably I can check it to see if it's 5 in a row in the cards array.
Edit 2: This is my absolutely final solution:
require 'set'
STRAIGHTS = ['A',*2..9,'T','J','Q','K','A'].each_cons(5).map(&:to_set)
#=> [#<Set: {"A", 2, 3, 4, 5}>, #<Set: {2, 3, 4, 5, 6}>,
# ...#<Set: {9, "T", "J", "Q", "K"}>, #<Set: {"T", "J", "Q", "K", "A"}>]
def straight?(hand)
STRAIGHTS.include?(hand.to_set)
end
STRAIGHTS.include?([6,3,4,5,2].to_set)
# STRAIGHTS.include?(#<Set: {6, 3, 4, 5, 2}>)
#=> true
straight?([6,5,4,3,2]) #=> true
straight?(["T","J","Q","K","A"]) #=> true
straight?(["A","K","Q","J","T"]) #=> true
straight?([2,3,4,5,"A"]) #=> true
straight?([6,7,8,9,"J"]) #=> false
straight?(["J",7,8,9,"T"]) #=> false
Edit 1: #mudasobwa upset the apple cart by pointing out that 'A',2,3,4,5 is a valid straight. I believe I've fixed my answer. (I trust he's not going to tell me that 'K','A',2,3,4 is also valid.)
I would suggest the following:
CARDS = [2, 3, 4, 5, 6, 7, 8, 9, "T", "J", "Q", "K", "A"]
STRAIGHTS = CARDS.each_cons(5).to_a
#=>[[2, 3, 4, 5, 6], [3, 4, 5, 6, 7], [4, 5, 6, 7, 8],
# [5, 6, 7, 8, 9], [6, 7, 8, 9, "T"], [7, 8, 9, "T", "J"],
# [8, 9, "T", "J", "Q"], [9, "T", "J", "Q", "K"],
# ["T", "J", "Q", "K", "A"]]
def straight?(hand)
(hand.map {|c| CARDS.index(c)}.sort == [0,1,2,3,12]) ||
STRAIGHTS.include?(hand.sort {|a,b| CARDS.index(a) <=> CARDS.index(b)})
end
If we map each card to a value (9 is 9, "T" is 10, "J" is 11, etc.), then there are two facts that are true of all straights that we can use to solve our problem:
All straights have exactly five unique card values
The difference between the last and first cards' values is always 4
And so:
CARD_VALUES = {
2 => 2, 3 => 3, 4 => 4,
5 => 5, 6 => 6, 7 => 7,
8 => 8, 9 => 9, "T" => 10,
"J" => 11, "Q" => 12, "K" => 13,
"A" => 14
}
def is_straight?(hand)
hand_sorted = hand.map {|card| CARD_VALUES[card] }
.sort.uniq
hand_sorted.size == 5 &&
(hand_sorted.last - hand_sorted.first) == 4
end
This method (1) converts each card to its numeric value with map, then (2) sorts them, and then (3) throws out duplicates with uniq. To illustrate with various hands:
hand | 4 A T A 2 | 2 2 3 3 4 | 5 6 4 8 7 | 3 6 2 8 7
---------+--------------------+--------------------+--------------------+----------------
1. map | 4 14 10 14 2 | 2 2 3 3 4 | 5 6 4 8 7 | 3 6 2 8 7
2. sort | 2 4 10 14 14 | 2 2 3 3 4 | 4 5 6 7 8 | 2 3 6 7 8
3. uniq | 2 4 10 14 | 2 3 4 | 4 5 6 7 8 | 2 3 6 7 8
Alternatively...
I originally posted the following solution, which isn't bad, but is definitely more convoluted:
If the hand is sorted, this is easy. You can use Enumerable#each_cons to check each possible straight.
CARDS = [ 2, 3, 4, 5, 6, 7, 8, 9, "T", "J", "Q", "K", "A" ]
hand = [ 4, 5, 6, 7, 8 ]
def is_straight?(hand)
CARDS.each_cons(5).any? do |straight|
hand == straight
end
end
if is_straight?(hand)
puts "Straight!"
else
puts "Not straight!"
end
# => Straight!
each_cons(5) returns each consecutive set of 5 items, so in the above example hand is first compared to [ 2, 3, 4, 5, 6 ], then [ 3, 4, 5, 6, 7 ], and then [ 4, 5, 6, 7, 8 ], which is a match, so any? returns true.
Note that this is not the most efficient solution, but unless you need to check many thousands of hands per second, this is more than adequately performant.
If your hands aren't sorted yet, you'll need to do that first. The simplest way to do that is create a Hash that maps cards to a numeric value (as above) and then use sort_by:
def sort_hand(hand)
hand.sort_by {|card| CARD_VALUES[card] }
end
hand = [ 4, "A", 2, "A", "T" ]
sort_hand(hand)
# => [ 2, 4, "T", "A", "A" ]
I did not want to participate, but I can’t keep silence looking at all these oversophisticated solutions around.
hand = [2, 5, 7, 'A', 'J'].map(&:to_s)
'23456789TJQKA' =~ hand.sort_by{|hc| '23456789TJQKA'.index(hc)}.join ||
'A23456789TJQK' =~ hand.sort_by{|hc| 'A23456789TJQK'.index(hc)}.join
In a not lame hardcoded manner:
suit = '23456789TJQKA'
suit =~ hand.sort_by{|hc| suit.index(hc)}.join ||
suit.rotate(-1) =~ hand.sort_by{|hc| suit.rotate(-1).index(hc)}.join
Generate list of valid hands:
valid_hands = cards[0..8].each_with_index.map{|b,i| cards[i..i+4]}
#=> [[2, 3, 4, 5, 6], [3, 4, 5, 6, 7], [4, 5, 6, 7, 8], [5, 6, 7, 8, 9], [6, 7, 8, 9, "T"], [7, 8, 9, "T", "J"], [8, 9, "T", "J", "Q"], [9, "T", "J", "Q", "K"], ["T", "J", "Q", "K", "A"]]
Once you have the list of all valid hands, you can now check if provided hand is among any? of them (valid ones) or not:
if valid_hands.any? { |h| (h - hand).empty? }
puts "Valid hand"
else
puts "Not Valid"
end
UPDATE
In-case 2, 3, 4, 5, "A", 2, 3, 4, "K", "A", 2, 3, "Q", "K", "A", 2, "J", "Q", "K", "A" are also considered as valid hands, calculate them as follows:
valid_hands = cards.each_with_index.map { |b,i| i < 9 ? cards[i..i+4] : cards[0..i-9] + cards[i..-1] }
# => [[2, 3, 4, 5, 6], [3, 4, 5, 6, 7], [4, 5, 6, 7, 8], [5, 6, 7, 8, 9], [6, 7, 8, 9, "T"], [7, 8, 9, "T", "J"], [8, 9, "T", "J", "Q"], [9, "T", "J", "Q", "K"], ["T", "J", "Q", "K", "A"], [2, "J", "Q", "K", "A"], [2, 3, "Q", "K", "A"], [2, 3, 4, "K", "A"], [2, 3, 4, 5, "A"]]
I recommend writing classes to represent a Card (and maybe Deck and Hand too). Aim for an interface like this:
deck = Deck.new.shuffle!
hand = Hand.new(deck.draw 5)
hand.straight?
#=>false
puts hand
8♣ 8♦ T♠ 2♦ 7♦
The encapsulation of functionality gives you readability and makes it easy to extend (i.e. with suits)
Here's a more simplistic version, implemented as a single Card class. I did add suits though.
class Card
include Enumerable #enables sorting
attr_accessor :value, :suit
#values = [2, 3, 4, 5, 6, 7, 8, 9, "T", "J", "Q", "K", "A"]
#suits = ["♣","♦","♥","♠"]
def self.all
#values.product(#suits).map{|c| Card.new c}
end
def self.straight?(cards)
["A", *#values].each_cons(5).include?(cards.map(&:value))
end
def self.flush?(cards)
cards.map(&:suit).uniq.size == 1
end
def initialize(v)
#value, #suit = *v
end
def <=>(other) #for sorting
#values.index(value) <=> #values.index(other.value)
end
def to_s
"#{value}#{suit}"
end
end
This works as follows
deck = Card.all
puts deck
#=> 2♣ 2♦ 2♥ 2♠ 3♣ 3♦ 3♥ 3♠ 4♣ 4♦ 4♥ 4♠ 5♣ 5♦ 5♥ 5♠ 6♣ 6♦ 6♥ 6♠ 7♣ 7♦ 7♥ 7♠ 8♣ 8♦ 8♥ 8♠ 9♣ 9♦ 9♥ 9♠ T♣ T♦ T♥ T♠ J♣ J♦ J♥ J♠ Q♣ Q♦ Q♥ Q♠ K♣ K♦ K♥ K♠ A♣ A♦ A♥ A♠
hand = deck.sample 5
puts hand
#=> Q♥ 6♦ 2♣ T♠ Q♦
Card.straight?(hand)
#=>false
Step 0: Let's start with an empty class
class CardUtils
end
Step 1: Store values of card in Hash
Hash allows fast referencing of values of a card.
##card_values = {
'A' => 1, 2 => 2, 3 => 3, 4 => 4, 5 => 5,
6 => 6, 7 => 7, 8 => 8, 9 => 9, 'T' => 10,
'J' => 11, 'Q' => 12, 'K' => 13
}
Thus, you can reference the card value simply as below.
##card_values['A']
# => 1
##card_values[8]
# => 8
Step 2: Sort the hand
Apply sort! method to the hand with reference to the card values.
def self.sort(hand)
hand.sort {|x,y| ##card_values[x] <=> ##card_values[y]}
end
# => ["A", 2, 3, 4, 5, 6, 7, 8, 9, "T", "J", "Q", "K"]
Step 3: Function that tells whether two cards are consecutive
def self.is_consecutive(x, y)
val_x = ##card_values[x]
val_y = ##card_values[y]
val_x == val_y - 1 || val_x + 13 == val_y
end
# is_consecutive('A', 2)
# => true
# is_consecutive('K', 'A')
# => true
# is_consecutive('A', 3)
# => false
Step 4: Check for 'straight'
It could be done with simple iteration.
def self.has_straight(hand)
hand = sort(hand)
max_consecutive_count = 0
consecutive_count = 0
hand.each_with_index do |curr, i|
prev = hand[i - 1]
if is_consecutive(prev, curr) then
consecutive_count += 1
else
consecutive_count = 0
end
if consecutive_count > max_consecutive_count then
max_consecutive_count = consecutive_count
end
end
max_consecutive_count >= 5
end
# hand = [2, 3, 4, 5, 6, 7, 8, 9, "T", "J", "Q", "K", "A"]
# CardUtils.has_straight(hand)
# => true
Final Result
class CardUtils
##card_values = {
'A' => 1, 2 => 2, 3 => 3, 4 => 4, 5 => 5,
6 => 6, 7 => 7, 8 => 8, 9 => 9, 'T' => 10,
'J' => 11, 'Q' => 12, 'K' => 13
}
def self.is_consecutive(x, y)
val_x = ##card_values[x]
val_y = ##card_values[y]
val_x == val_y - 1 || val_x + 13 == val_y
end
def self.sort(hand)
hand.sort {|x,y| ##card_values[x] <=> ##card_values[y]}
end
def self.has_straight(hand)
hand = sort(hand)
max_consecutive_count = 0
consecutive_count = 0
hand.each_with_index do |curr, i|
prev = hand[i - 1]
if is_consecutive(prev, curr) then
consecutive_count += 1
else
consecutive_count = 0
end
if consecutive_count > max_consecutive_count then
max_consecutive_count = consecutive_count
end
end
max_consecutive_count >= 5
end
end
This is how I would write it:
hand = [3,4,5,2,'A']
def is_straight(hand)
# No need to check further if we do not have 5 unique cards.
return false unless hand.uniq.size == 5
# Note the A at beginning AND end to count A as 1 or 14.
list_of_straights = 'A23456789TJQKA'.chars.each_cons(5)
sorted_hand = hand.map(&:to_s).sort
list_of_straights.any? do |straight|
straight.sort==sorted_hand
end
end
puts is_straight(hand) #=> true
Alternatively if you do not like all the sorting you could exchange the last part to:
hand_as_stings = hand.map(&:to_s)
list_of_straights.any? do |straight|
(straight-hand_as_stings).empty?
end
I'm having an issue that I can't seem to find documented or explained anywhere so I'm hoping someone here can help me out. I've verified the unexpected behavior on three versions of Ruby, all 2.1+, and verified that it doesn't happen on an earlier version (though it's through tryruby.org and I don't know which version they're using). Anyway, for the question I'll just post some code with results and hopefully someone can help me debug it.
arr = %w( r a c e c a r ) #=> ["r","a","c","e","c","a","r"]
arr.select { |c| arr.count(c).odd? } #=> ["e"]
arr.select! { |c| arr.count(c).odd? } #=> ["e","r"] <<<<<<<<<<<<<<< ??????
I think the confusing part for me is clearly marked and if anyone can explain if this is a bug or if there's some logic to it, I'd greatly appreciate it. Thanks!
You're modifying the array while you're read from it while you iterate over it. I'm not sure the result is defined behavior. The algorithm isn't required to keep the object in any kind of sane state while it's running.
Some debug printing during the iteration shows why your particular result happens:
irb(main):005:0> x
=> ["r", "a", "c", "e", "c", "a", "r"]
irb(main):006:0> x.select! { |c| p x; x.count(c).odd? }
["r", "a", "c", "e", "c", "a", "r"]
["r", "a", "c", "e", "c", "a", "r"]
["r", "a", "c", "e", "c", "a", "r"]
["r", "a", "c", "e", "c", "a", "r"] # "e" is kept...
["e", "a", "c", "e", "c", "a", "r"] # ... and moved to the start of the array
["e", "a", "c", "e", "c", "a", "r"]
["e", "a", "c", "e", "c", "a", "r"] # now "r" is kept
=> ["e", "r"]
You can see by the final iteration, there is only one r, and that the e has been moved to the front of the array. Presumably the algorithm modifies the array in-place, moving matched elements to the front, overwriting elements that have already failed your test. It keeps track of how many elements are matched and moved, and then truncates the array down to that many elements.
So, instead, use select.
A longer example that matches more elements makes the problem a little clearer:
irb(main):001:0> nums = (1..10).to_a
=> [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
irb(main):002:0> nums.select! { |i| p nums; i.even? }
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
[2, 2, 3, 4, 5, 6, 7, 8, 9, 10]
[2, 2, 3, 4, 5, 6, 7, 8, 9, 10]
[2, 4, 3, 4, 5, 6, 7, 8, 9, 10]
[2, 4, 3, 4, 5, 6, 7, 8, 9, 10]
[2, 4, 6, 4, 5, 6, 7, 8, 9, 10]
[2, 4, 6, 4, 5, 6, 7, 8, 9, 10]
[2, 4, 6, 8, 5, 6, 7, 8, 9, 10]
[2, 4, 6, 8, 5, 6, 7, 8, 9, 10]
=> [2, 4, 6, 8, 10]
You can see that it does indeed move matched elements to the front of the array, overwriting non-matched elements, and then truncate the array.
Just to give you some other ways of accomplishing what you're doing:
arr = %w( r a c e c a r )
arr.group_by{ |c| arr.count(c).odd? }
# => {false=>["r", "a", "c", "c", "a", "r"], true=>["e"]}
arr.group_by{ |c| arr.count(c).odd? }.values
# => [["r", "a", "c", "c", "a", "r"], ["e"]]
arr.partition{ |c| arr.count(c).odd? }
# => [["e"], ["r", "a", "c", "c", "a", "r"]]
And if you want more readable keys:
arr.group_by{ |c| arr.count(c).odd? ? :odd : :even }
# => {:even=>["r", "a", "c", "c", "a", "r"], :odd=>["e"]}
partition and group_by are basic building blocks for separating elements in an array into some sort of grouping, so it is good to be familiar with them.
I have the following hash:
hash = {"1"=>[ 5, 13, "B", 4, 10],
"2"=>[27, 19, "B", 18, 20],
"3"=>[45, 41, "B", 44, 31],
"4"=>[48, 51, "B", 58, 52],
"5"=>[70, 69, "B", 74, 73]}
Here is my code:
if hash.values.all? { |array| array[0] == "B" } ||
hash.values.all? { |array| array[1] == "B" } ||
hash.values.all? { |array| array[2] == "B" } ||
hash.values.all? { |array| array[3] == "B" } ||
hash.values.all? { |array| array[4] == "B" }
puts "Hello World"
What my code does is iterates through an array such that if the same element appears in the same index position of each array, it will output the string "Hello World" (Since "B" is in the [2] position of each array, it will puts the string. Is there a way to condense my current code without having a bunch of or's connecting each index of the array?
Assuming all arrays are always of the same length, the following gives you the column indexes where all values are equal:
hash.values.transpose.each_with_index.map do |column, index|
index if column.all? {|x| x == column[0] }
end.compact
The result is [2] for your hash. So you know that for all arrays the index 2 has the same values.
You can print "Hello World" if the resulting array has at least one element.
How does it work?
hash.values.transpose gives you all the arrays, but with transposed (all rows are now columns) values:
hash.values.transpose
=> [[5, 27, 45, 48, 70],
[13, 19, 41, 51, 69],
["B", "B", "B", "B", "B"],
[4, 18, 44, 58, 74],
[10, 20, 31, 52, 73]]
.each_with_index.map goes over every row of the transposed array while providing an inner array and its index.
We look at every inner array, yielding the column index only if all elements are equal using all?.
hash.values.transpose.each_with_index.map {|column, index| index if column.all? {|x| x == column[0] }
=> [nil, nil, 2, nil, nil]
Finally, we compact the result to get rid of the nil values.
Edit: First, I used reduce to find the column with identical elements. #Nimir pointed out, that I re-implemented all?. So I edited my anwer to use all?.
From #tessi brilliant answer i though of this way:
hash.values.transpose.each_with_index do |column, index|
puts "Index:#{index} Repeated value:#{column.first}" if column.all? {|x| x == column[0]}
end
#> Index:2 Repeated value:B
How?
Well, the transpose already solves the problem:
hash.values.transpose
=> [[5, 27, 45, 48, 70],
[13, 19, 41, 51, 69],
["B", "B", "B", "B", "B"],
[4, 18, 44, 58, 74],
[10, 20, 31, 52, 73]
]
We can do:
column.all? {|x| x == column[0]}
To find column with identical items
Assuming that all the values of the hash will be arrays of the same size, how about something like:
hash
=> {"1"=>[5, 13, "B", 4, 10], "2"=>[27, 19, "B", 18, 20], "3"=>[45, 41, "B", 44, 31], "4"=>[48, 51, "B", 58, 52], "5"=>[70, 69, "B", 74, 73]}
arr_of_arrs = hash.values
=> [[5, 13, "B", 4, 10], [27, 19, "B", 18, 20], [45, 41, "B", 44, 31], [48, 51, "B", 58, 52], [70, 69, "B", 74, 73]]
first_array = arr_of_arrs.shift
=> [5, 13, "B", 4, 10]
first_array.each_with_index do |element, index|
arr_of_arrs.map {|arr| arr[index] == element }.all?
end.any?
=> true
This is not really different from what you have now, as far as performance - in fact, it may be a bit slower. However, it allows for a dynamic number of incoming key/value pairs.
I ended up using the following:
fivebs = ["B","B","B","B","B"]
if hash.values.transpose.any? {|array| array == fivebs}
puts "Hello World"
If efficiency, rather than readability, is most important, I expect this decidedly un-Ruby-like and uninteresting solution probably would do well:
arr = hash.values
arr.first.size.times.any? { |i| arr.all? { |e| e[i] == ?B } }
#=> true
Only one intermediate array (arr) is constructed (e.g, no transposed array), and it quits if and when a match is found.
More Ruby-like is the solution I mentioned in a comment on your question:
hash.values.transpose.any? { |arr| arr.all? { |e| e == ?B } }
As you asked for an explanation of #Phrogz's solution to the earlier question, which is similar to this one, let me explain the above line of code, by stepping through it:
a = hash.values
#=> [[ 5, 13, "B", 4, 10],
# [27, 19, "B", 18, 20],
# [45, 41, "B", 44, 31],
# [48, 51, "B", 58, 52],
# [70, 69, "B", 74, 73]]
b = a.transpose
#=> [[ 5, 27, 45, 48, 70],
# [ 13, 19, 41, 51, 69],
# ["B", "B", "B", "B", "B"],
# [ 4, 18, 44, 58, 74],
# [ 10, 20, 31, 52, 73]]
In the last step:
b.any? { |arr| arr.all? { |e| e == ?B } }
#=> true
(where ?B is shorthand for the one-character string "B") an enumerator is created:
c = b.to_enum(:any?)
#=> #<Enumerator: [[ 5, 27, 45, 48, 70],
# [ 13, 19, 41, 51, 69],
# ["B", "B", "B", "B", "B"],
# [ 4, 18, 44, 58, 74],
# [ 10, 20, 31, 52, 73]]:any?>
When the enumerator (any enumerator) is acting on an array, the elements of the enumerator are passed into the block (and assigned to the block variable, here arr) by Array#each. The first element passed into the block is:
arr = [5, 27, 45, 48, 70]
and the following is executed:
arr.all? { |e| e == ?B }
#=> [5, 27, 45, 48, 70].all? { |e| e == ?B }
#=> false
Notice that false is returned to each right after:
5 == ?B
#=> false
is evaluated. Since false is returned, we move on to the second element of the enumerator:
[13, 19, 41, 51, 69].all? { |e| e == ?B }
#=> false
so we continue. But
["B", "B", "B", "B", "B"].all? { |e| e == ?B }
#=> true
so when true is returned to each, the latter returns true and we are finished.
I have an array in Ruby and I would like to delete the first 10 digits in the array.
array = [1, "a", 3, "b", 2, "c", 4, "d", 5, "a", 1, "z", 7, "e", 21, "q", 30, "a", 4, "t", 7, "m", 5, 1, 2, "q", "s", "l", 13, 46, 31]
It would ideally return
['a', 'b', 'c', 'd', 'a', 'z', 'e', 'q', 0, 'a', 4, t, 7, m, 5 , 1, 2, q, s, 1, 13, 46, 31]
By removing the first 10 digits (1,3,2,4,5,1,7,2,1,3).
Note that 21(2 and 1) and 30(3 and 0) both have 2 digits
Here's what I've tried
digits = array.join().scan(/\d/).first(10).map{|s|s.to_i}
=> [1,3,2,4,5,1,7,2,1,3]
elements = array - digits
This is what I got
["a", "b", "c", "d", "a", "z", "e", 21, "q", 30, "a", "t", "m", "q", "s", "l", 13, 46, 31]
Now it looks like it took the difference instead of subtracting.
I have no idea where to go from here. and now I'm lost. Any help is appreciated.
To delete 10 numbers:
10.times.each {array.delete_at(array.index(array.select{|i| i.is_a?(Integer)}.first))}
array
To delete 10 digits:
array = [1, "a", 3, "b", 2, "c", 4, "d", 5, "a", 1, "z", 7, "e", 21, "q", 30, "a", 4, "t", 7, "m", 5, 1, 2, "q", "s", "l", 13, 46, 31]
i = 10
while (i > 0) do
x = array.select{|item| item.is_a?(Integer)}.first
if x.to_s.length > i
y = array.index(x)
array[y] = x.to_s[0, (i-1)].to_i
else
array.delete_at(array.index(x))
end
i -= x.to_s.length
end
array
Unfortunately not a one-liner:
count = 10
array.each_with_object([]) { |e, a|
if e.is_a?(Integer) && count > 0
str = e.to_s # convert integer to string
del = str.slice!(0, count) # delete up to 'count' characters
count -= del.length # subtract number of actually deleted characters
a << str.to_i unless str.empty? # append remaining characters as integer if any
else
a << e
end
}
#=> ["a", "b", "c", "d", "a", "z", "e", "q", 0, "a", 4, "t", 7, "m", 5, 1, 2, "q", "s", "l", 13, 46, 31]
I would be inclined to do it like this.
Code
def doit(array, max_nbr_to_delete)
cnt = 0
array.map do |e|
if (e.is_a? Integer) && cnt < max_nbr_to_delete
cnt += e.to_s.size
if cnt <= max_nbr_to_delete
nil
else
e.to_s[cnt-max_nbr_to_delete..-1].to_i
end
else
e
end
end.compact
end
Examples
array = [ 1, "a", 3, "b", 2, "c", 4, "d", 5, "a", 1, "z", 7, "e", 21, "q",
30, "a", 4, "t", 7, "m", 5, 1, 2, "q", "s", "l", 13, 46, 31]
doit(array, 10)
#=> ["a", "b", "c", "d", "a", "z", "e", "q", 0, "a", 4,
# "t", 7, "m", 5, 1, 2, "q", "s", "l", 13, 46, 31]
doit(array, 100)
#=> ["a", "b", "c", "d", "a", "z", "e", "q", "a", "t", "m", "q", "s", "l"]
Explanation
Each element e of the array that is not an integer is mapped to e.
For each non-negative integer n having d digits, suppose cnt is the number of digits that map has already been removed from the string. There are three possibilities:
if cnt >= max_nbr_to_delete, no more digits are to be removed, so e (itself) is returned
if cnt + d <= max_nbr_to_delete all d digits of e are to be removed, which is done by mapping e to nil and subsequently removing nil elements
if cnt < max_nbr_to_delete and cnt + d > max_nbr_to_delete, e.to_s[cnt+d-max_nbr_to_delete..-1].to_i is returned (i.e. the first cnt+d-max_nbr_to_delete digits of e are removed).