I'm working with the Stanford Parser in ruby and want to search for all nodes of a tree with a particular label name.
This is the recursive method i have coded so far
def searchTreeWithLabel(tree,lablename,listOfNodes)
if tree.instance_of?(StanfordParser::Tree)
if tree.lable.toString == lablename then
listOfNodes << tree
else
tree.children.each { |c| searchTreeWithLabel(c, lablename, listOfNodes)}
end
end
listOfNodes
end
i want the method to return a list of Tree nodes that have the label as labelname
I'm not familiar StanfordParser but I imagine you need to take the descent part of the traversal out of the inner conditional and always do it.
Also, did they really implement a toString method? Seriously? It's not .to_s? I mean, I enjoyed Java, before I found Ruby... :-)
my original code was correct but ruby was having some problem with the
if tree.lable.toString == lablename then
statement, turns out tree.value works as well, so now i'm checking
if tree.value == lablename then
and it works.
Related
I've been studying a few searching algorithms and my last problem comes down to binary searching. I watched a few youtube videos to understand the concept and then tried to solve the problem, but keep getting an endless loop error. I've looked through stack overflow, and reddit, and wherever Google would lead me, but I can't quite find a solution that fits my method of coding. Also, please excuse the term 'monkey patching', it's been brought to my attention that the technical term is called 'extending' so the fault lies on my instructors for teaching it to us as 'monkey patching'.
Here's my code:
class Array
def my_bsearch(target)
return nil if self.empty?
middle_idx = self.length/2
left = self.take(middle_idx)
right = self.drop(middle_idx + 1)
return middle_idx if self[middle_idx] == target
until self[middle_idx] == target || self.nil? == nil
if self[middle_idx] < target
right.my_bsearch(target)
elsif self[middle_idx] > target
left.my_bsearch(target)
end
end
end
end
I have a solution, but I don't want to just memorize it-- and I'm having trouble understanding it; as I'm trying to translate it, learn from it, and implement what I'm missing into my own code.
class Array
def my_bsearch(target)
return nil if size == 0
mid = size/2
case self[mid] <=> target
when 0
return mid
when 1
return self.take(mid).my_bsearch(target)
else
search_res = self.drop(mid+1).my_bsearch(target)
search_res.nil? ? nil : mid + 1 + search_res
end
end
end
I guess I understand case/when despite not use to using it. I've tried following it with debugger, but I think I'm hung up on what's going on in the ELSE section. The syntactic sugar, while making this obviously more concise than my logic, isn't straight-forward/clean to someone of my ruby literacy level. So, yeah, my ignorance is most of the problem I guess.
Is there someone who is a little more literate, and patient, able to help me break this down into something I can understand a bit better so I can learn from this?
First, take and drop have sufficiently similar interfaces that you don't actually want your + 1 for drop. It will disregard one element in the array if you do.
Next, self.nil? will always be false (and never nil) for instances of this class. In fact, .nil? is a method exactly to avoid having to ever compare against nil with ==.
You want self.empty?. Furthermore, with the exception of setters, in Ruby messages are sent to self by default. In other words, the only time self. is a useful prefix is when the message ends in = and operates as an lvalue, as in self.instance_var = 'a constant', since without the self., the tokens instance_var = would be interpreted as a local variable rather than an instance variable setting. That's not the case here, so empty? will suffice just as well as self.empty?
So I figured it out, and I decided to answer my own post in hopes to help someone else out if they run into this issue.
So, if I have an Array and the target is the middle_element, then it will report middle_element_idx. That's fine. What if the target is less than middle_element? It recursively searches the left-side of the original Array. When it finds it, it reports the left_side_idx. There's no problem with that because elements in an array are sequentially counted left to right. So, it starts at 0 and goes up.
But what if the target is on the right side of the middle element?
Well, searching for the right side is easy. Relatively the same logic as searching left. Done recursively. And it will return a target_idx if it's found on that right side --however that's the target's idx as it was found in the right-side array! So, you need to take that returned target_idx and add 1 to it and the original middle_element_idx. See below:
def my_bsearch(target)
return nil if self.empty?
middle_idx = self.length/2
left = self.take(middle_idx)
right = self.drop(middle_idx + 1)
if self[middle_idx] == target
return middle_idx
elsif self[middle_idx] > target
return left.my_bsearch(target)
else
searched_right_side = 1 + right.my_bsearch(target)
return nil if searched_right_side.nil? == true
return searched_right_side + middle_idx
end
end
end
Notice how many more lines this solution is? The spaceship operator used in conjunction with case/when and a ternary method will reduce the number of lines significantly.
Based on suggestions/feedback from Tim, I updated it to:
def my_bsearch(target)
return nil if empty?
middle_idx = self.length/2
left = self.take(middle_idx)
right = self.drop(middle_idx)
if self[middle_idx] == target
return middle_idx
elsif self[middle_idx] > target
return left.my_bsearch(target)
else
searched_right_side = right.my_bsearch(target)
return nil if searched_right_side.nil?
return searched_right_side + middle_idx
end
end
end
I'm working on a scrabble-like-app. Given up to 15 letters it finds all possible valid words made from that letters. I decided to store dictionary in a Trie structure. My previous code was making possible combinations and then looking in a Trie if they are valid words. To make it faster, i want to implement John Pirie's suggestion posted here.
I'm using fast_trie gem. Comes with couple of useful methods.
After many hours of trying to implement John Pirie's solution i can't wrap my head around backtracking part of the code. Recursive algorithms seem to be rarely used in ruby - i'm struggling to find any help.
How to make program to go back to previous node, check if there are possible paths not taken before, search them through and go back again?
require 'trie'
trie = Trie.read("local_copy_of_my_trie")
input = "helloworld"
rack = input.chars
$words = [] # after recursive search it conains all possible words made from input
start_node = trie.root
def search_from(node, rack)
current_node = node
current_rack = rack
$words.push(node.full_state) if node.terminal?
search_from(choose_next_node(current_node, current_rack), update_rack(current_rack)) # choose next node backtracks if possible
end
def choose_next_node(node, rack)
possible_paths = []
rack.each { |l| possible_paths.push(l) if $trie.has_children?(node.full_state+l) }
# must backtrack if there are no possibilities
# how to mark path as already taken?
# can i return both node and rack to search_from ?
end
def update_rack(rack)
# how to update rack not knowing the result of choose_next_node here?
end
search_from(start_node, rack)
After working on my code for a while, optimizing the most obvious things, I've resulted in this:
function FindPath(start, finish, path)
--Define a table to hold the paths
local paths = {}
--Make a default argument
path = path or {start}
--Loop through connected nodes
for i,v in ipairs(start:GetConnectedParts()) do
--Determine if backtracking
local loop = false
for i,vv in ipairs(path) do
if v == vv then
loop = true
end
end
if not loop then
--Make a path clone
local npath = {unpack(path)}
npath[#npath+1] = v
if v == finish then
--If we reach the end add the path
return npath
else
--Otherwise add the shortest part extending from this node
paths[#paths+1] = FindPath(v, finish, npath) --NOTED HERE
end
end
end
--Find and return the shortest path
if #paths > 0 then
local lengths = {}
for i,v in ipairs(paths) do
lengths[#lengths+1] = #v
end
local least = math.min(unpack(lengths))
for i,v in ipairs(paths) do
if #v == least then
return v
end
end
end
end
The problem being, the line noted gets some sort of game script timeout error (which I believe is a because of mass recursion with no yielding). I also feel like once that problem is fixed, it'll probably be rather slow even on the scale of a pacman board. Is there a way I can further optimize it, or perhaps a better method I can look into similar to this?
UPDATE: I finally decided to trash my algorithm due to inefficiency, and implemented a Dijkstra algorithm for pathfinding. For anybody interested in the source code it can be found here: http://pastebin.com/Xivf9mwv
You know that Roblox provides you with the PathfindingService? It uses C-side A* pathing to calculate quite quickly. I'd recommend using it
http://wiki.roblox.com/index.php?title=API:Class/PathfindingService
Try to remodel your algorithm to make use of tail calls. This is a great mechanism available in Lua.
A tail call is a type of recursion where your function returns a function call as the last thing it does. Lua has proper tail calls implementation and it will dress this recursion as a 'goto' under the scenes, so your stack will never blow.
Passing 'paths' as one of the arguments of FindPath might help with that.
I saw your edit about ditching the code, but just to help others stumbling on this question:
ipairs is slower than pairs, which is slower than a numeric for-loop.
If performance matters, never use ipairs, but use a for i=1,#tab loop
If you want to clone a table, use a for-loop. Sometimes, you have to use unpack (returning dynamic amount of trailing nils), but this is not such a case. Unpack is also a slow function.
Replacing ipairs with pairs or numeric for-loops and using loops instead of unpack will increase the performance a lot.
If you want to get the lowest value in a table, use this code snippet:
local lowestValue = values[1]
for k,v in pairs(values) do
if v < lowestValue then
lowestValue = k,v
end
end
This could be rewritten for your path example as so:
local least = #path[1]
for k,v in pairs(path) do
if #v < least then
least = v
end
end
I have to admit, you're very inventive. Not a lot of people would use math.min(unpack(tab)) (not counting the fact it's bad)
I have to search an item in an array and return the value of the next item. Example:
a = ['abc.df','-f','test.h']
i = a.find_index{|x| x=~/-f/}
puts a[i+1]
Is there any better way other than working with index?
A classical functional approach uses no indexes (xs.each_cons(2) -> pairwise combinations of xs):
xs = ['abc.df', '-f', 'test.h']
(xs.each_cons(2).detect { |x, y| x =~ /-f/ } || []).last
#=> "test.h"
Using Enumerable#map_detect simplifies it a litte bit more:
xs.each_cons(2).map_detect { |x, y| y if x =~ /-f/ }
#=> "test.h"
The reason something like array.find{something}.next doesn't exist is that it's an array rather than a linked list. Each item is just it's own value; it doesn't have a concept of "the item after me".
#tokland gives a good solution by iterating over the array with each pair of consecutive items, so that when the first item matches, you have your second item handy. There are strong arguments to be made for the functional style, to be sure. Your version is shorter, though, and I'd argue that yours is also more quickly and easily understood at a glance.
If the issue is that you're using it a lot and want something cleaner and more to the point, then of course you could just add it as a singleton method to a:
def a.find_after(&test)
self[find_index(&test).next]
end
Then
a.find_after{|x| x=~/-f/}
is a clear way to find the next item after the first match.
All of that said, I think #BenjaminCox makes the best point about what appears to be your actual goal. If you're parsing command line options, there are libraries that do that well.
I don't know of a cleaner way to do that specific operation. However, it sure looks like you're trying to parse command-line arguments. If so, I'd recommend using the built-in OptionParser module - it'll save a ton of time and hair-pulling trying to parse them yourself.
This article explains how it works.
Your solution working with indexes is fine, as others have commented. You could use Enumerable#drop_while to get an array from your match on and take the second element of that:
a = ['abc.df','-f','test.h']
f_arg = a.drop_while { |e| e !~ /-f/ }[1]
I'm trying to use the two following methods to recursively traverse arrays of arrays until the bottom and then come back up with the match results.
You know how in a tennis tournament they start with 32 matches and pair by pair the winner moves ahead, and at the end there's only one winner? That's what I want to replicate in Ruby.
I created a match_winner that always returns the first array for the sake of simplicity. Then, I send the whole tournament array into winner that calls itself recursively until it finds a simple array corresponding to a single match.
def match_winner(array_of_arrays)
return array_of_arrays[0]
end
def winner(tournament)
if tournament[0][0].is_a?(String)
return match_winner(tournament)
else
tournament.each{|e|
winner(e)
}
end
end
tournament = [
[["one", "two"],["three", "four"]],
[["five", "six"],["seven", "eight"]]
]
puts winner(tournament).inspect
Which outputs:
[[["one", "two"], ["three", "four"]], [["five", "six"], ["seven", "eight"]]]
I tried different permutations and variations on this algorithm but I couldn't make it work correctly and return only the final winner.
Does anyone see anything obviously wrong here?
Now I'm calling winner.
I know that the question looks like it's answered, but I just did the same problem and I have to say that simply changing each to map didn't work for me, because, as the code posted, the result is an array of the first-round winners. What worked for me is:
def winner(tournament)
if tournament[0][0].is_a?(String)
return match_winner(tournament)
else
tournament.map!{|e| #use map!, because we need to apply winner() to new values
e=winner(e) #assign new value somewhere, so recursion can climb back
}
end
end
Maybe more experienced developers can explain why that is. Without these two tips it won't work.
And yes, I know "bang" is a bad coding style, danger danger high voltage, but it's my second day with Ruby and I wanted to get this to work.
And, to understand recursion, you have to understand recursion.
Looks like you want map, not each, and, as a commenter above notes, you didn't call winner in the above code.
When you call:
tournament.each {...}
that method actually returns the tournament, which is thus what winner returns.
What you want is to replace it with
tournament.map {...}
which returns a new array consisting of calling "winner" on each element of tournament.
Assuming you have 2^n number of games always and match_winner works ok:
def winner(game)
if game[0][0][0] == game[0][0][0][0]
match_winner( [ game[0], game[1] ] )
else
match_winner( [winner(game[0]), winner(game[1])] )
end
end