I thought this would be pretty simple, but I seem to be getting things mixed up, and I haven't found anything on stackoverflow that quite matches my question.
I'm trying to write a function that can compare two arrays of file names to make sure their values match. They need to actually match in their position as well, so the order is crucial. In other words:
array1 = ["file1.html", "file2.html", "file3.html", "file4.html"]
array2 = ["file1.html", "file2.html", "file4.html", "file3.html"]
I would want a comparison between these two arrays to return as false, because of the difference in order (even though both arrays actually include the same values). I tried something like this:
matching = true
names1 = array1.map { |x| File.basename(x)}
names2 = array2.map { |x| File.basename(x)}
names1.each_with_index { |file,index|
if file != names2[index]
matching = false
end
}
return matching
This works, but I'm wondering if there's a cleaner, more foolproof way of comparing arrays in this way? Thanks!
You don't need to do anything of the sort. Default array equality operator compares arrays with order
array1 = %w[file1.html file2.html file3.html file4.html]
array2 = %w[file1.html file2.html file4.html file3.html]
array3 = %w[file1.html file2.html file3.html file4.html]
array1 == array2 # => false
array1 == array3 # => true
Example with comparing mapped values (if you have full paths in your arrays)
array1.map{|a| File.basename(a)} == array2.map{|a| File.basename(a) }
# or, as #mudasobwa would suggest
[array1, array2].map{|a| a.map(&File.method(:basename)) }.reduce(:==)
Use the last one if and only if a) you understand it completely and b)
you think it's a good idea.
Related
I know how to iterate through an array and I know how to create hash keys and values. I do not know how to create an array for the value and pass it multiple elements. My desired value for hash below is:
{'abc' => [1, 2, 3] , 'def' => [4,5,6,7]}
How would I achieve this hash, while iterating through array a and b below using each?
a = [1,2,3,4,5,6,7]
c = [1,2,3]
b = ['abc', 'def']
hash = {}
From your guidelines given in the comment:
While iterating through array a, if the element of iteration is included in array c, it is passed to the array value within key 'abc'. Otherwise, it is passed to other array value in key 'def'
You can do this:
hash = {}
hash['abc'] = a.select { |x| c.include?(x) }
hash['def'] = a.reject{ |x| c.include?(x) }
See Enumerable#select and Enumerable#reject. Also can take a look at Enumerable#partition which would be another good choice here, where you want to split an array into two arrays based on some condition:
in_a, not_in_a = a.partition { |x| c.include?(x) }
hash = { 'abc' => in_a, 'def' => not_in_a }
You can also do it with regular each if these fancy enumerable methods are bit too much for you:
hash = { 'abc' => [], 'def' => [] }
a.each do |x|
if c.include?(x)
hash['abc'].push(x)
else
hash['def'].push(x)
end
end
Unfortunately this question turned out not to be as interesting as I was hoping. I was hoping that the problem was this:
Knowing the hash key and a value, how can I make sure the key's value is an array and that the given value is appended to that?
For instance, start with h being {}. I have a key name :k and a value 1. I want h[:k], if it doesn't already exist, to be [1]. But if it does already exist, then it's an array and I want to append 1 to that array; for instance, if h[:k] is already [3,2], now it should be [3,2,1].
I can think of various ways to ensure that, but one possibility is this:
(hash[key] ||= []) << value
To see that this works, let's make it a method:
def add_to_hash_array_value(hash:, key:, value:)
(hash[key] ||= []) << value
end
Now I'll just call that a bunch of times:
h = {}
add_to_hash_array_value(hash:h, key:"abc", value:1)
add_to_hash_array_value(hash:h, key:"abc", value:2)
add_to_hash_array_value(hash:h, key:"def", value:4)
add_to_hash_array_value(hash:h, key:"def", value:5)
add_to_hash_array_value(hash:h, key:"abc", value:3)
puts h #=> {"abc"=>[1, 2, 3], "def"=>[4, 5]}
We got the right answer.
This is nice because suppose I have some way of knowing, given a value, what key it should be appended to. I can just repeatedly apply that decision-making process.
However, the trouble with trying to apply that to the original question is that the original question seems not to know exactly how to decide, given a value, what key it should be appended to.
I have been trying to randomize the values in an ordered array (ex:[0,1,2,3]) in Godot. There is supposed to be a shuffle() method for arrays, but it seems to be broken and always returns "null". I have found a workaround that uses a Fisher-Yates shuffle, but the resulting array is considered "unsorted" by the engine, and therefore when I try to use methods such as bsearch() to find a value by it's position, the results are unreliable at best.
My solution was to create a dictionary, comprised of an array containing the random values I have obtained, merged with a second array of equal length with (sorted) numbers (in numerical order) which I can then use as keys to access specific array positions when needed.
Question made simple...
In GDScript, how would you take 2 arrays..
ex: ARRAY1 = [0,1,2,3]
ARRAY2 = [a,b,c,d]
..and merge them to form a dictionary that looks like this:
MergedDictionary = {0:a, 1:b, 2:c, 3:d}
Any help would be greatly appreciated.
Godot does not support "zip" methodology for merging arrays such as Python does, so I am stuck merging them manually. However... there is little to no documentation about how to do this in GDScript, despite my many hours of searching.
Try this:
var a = [1, 2, 3]
var b = ["a", "b", "c"]
var c = {}
if a.size() == b.size():
var i = 0
for element in a:
c[element] = b[i]
i += 1
print("Dictionary c: ", c)
If you want to add elements to a dictionary, you can assign values to the keys like existing keys.
Background:
Hey all, I am experimenting with external APIs and am trying to pull in all of the followers of a User from a site and apply some sorting.
I have refactored a lot of the code, HOWEVER, there is one part that is giving me a really tough time. I am convinced there is an easier way to implement this than what I have included and would be really grateful on any tips to do this in a much more eloquent way.
My goal is simple. I want to collapse an array of arrays of hashes (I hope that is the correct way to explain it) into one array of hashes.
Problem Description:
I have an array named f_collectionswhich has 5 elements. Each element is an array of size 200. Each sub-element of these arrays is a hash of about 10 key-value pairs. My best representation of this is as follows:
f_collections = [ collection1, collection2, ..., collection5 ]
collection1 = [ hash1, hash2, ..., hash200]
hash1 = { user_id: 1, user_name: "bob", ...}
I am trying to collapse this multi-dimensional array into one array of hashes. Since there are five collection arrays, this means the results array would have 1000 elements - all of which would be hashes.
followers = [hash1, hash2, ..., hash1000]
Code (i.e. my attempt which I do not want to keep):
I have gotten this to work with a very ugly piece of code (see below), with nested if statements, blocks, for loops, etc... This thing is a nightmare to read and I have tried my hardest to research ways to do this in a simpler way, I just cannot figure out how. I have tried flatten but it doesn't seem to work.
I am mostly just including this code to show I have tried very hard to solve this problem, and while yes I solved it, there must be a better way!
Note: I have simplified some variables to integers in the code below to make it more readable.
for n in 1..5 do
if n < 5
(0..199).each do |j|
if n == 1
nj = j
else
nj = (n - 1) * 200 + j
end
#followers[nj] = #f_collections[n-1].collection[j]
end
else
(0..199).each do |jj|
njj = (4) * 200 + jj
#followers[njj] = #f_collections[n-1].collection[jj]
end
end
end
Oh... so It is not an array objects that hold collections of hashes. Kind of. Lets give it another try:
flat = f_collection.map do |col|
col.collection
end.flatten
which can be shortened (and is more performant) to:
flat = f_collection.flat_map do |col|
col.collection
end
This works because the items in the f_collection array are objects that have a collection attribute, which in turn is an array.
So it is "array of things that have an array that contains hashes"
Old Answer follows below. I leave it here for documentation purpose. It was based on the assumption that the data structure is an array of array of hashes.
Just use #flatten (or #flatten! if you want this to be "inline")
flat = f_collections.flatten
Example
sub1 = [{a: 1}, {a: 2}]
sub2 = [{a: 3}, {a: 4}]
collection = [sub1, sub2]
flat = collection.flatten # returns a new collection
puts flat #> [{:a=>1}, {:a=>2}, {:a=>3}, {:a=>4}]
# or use the "inplace"/"destructive" version
collection.flatten! # modifies existing collection
puts collection #> [{:a=>1}, {:a=>2}, {:a=>3}, {:a=>4}]
Some recommendations for your existing code:
Do not use for n in 1..5, use Ruby-Style enumeration:
["some", "values"].each do |value|
puts value
end
Like this you do not need to hardcode the length (5) of the array (did not realize you removed the variables that specify these magic numbers). If you you want to detect the last iteration you can use each_with_index:
a = ["some", "home", "rome"]
a.each_with_index do |value, index|
if index == a.length - 1
puts "Last value is #{value}"
else
puts "Values before last: #{value}"
end
end
While #flatten will solve your problem you might want to see how DIY-solution could look like:
def flatten_recursive(collection, target = [])
collection.each do |item|
if item.is_a?(Array)
flatten_recursive(item, target)
else
target << item
end
end
target
end
Or an iterative solution (that is limited to two levels):
def flatten_iterative(collection)
target = []
collection.each do |sub|
sub.each do |item|
target << item
end
end
target
end
if I have a two-dimensional array like this: [[1,2,3], [3,2,1], [4,9,3]], I want to be able to find out that there are two identical arrays inside this array, which are [1,2,3] and [3,2,1]. How can I achieve this?
Thank you for all your answers, I was focusing on the leetCode threeSum problem so I didn't leave any comment. But since I am a programming noobie, my answer exceeded the time limit.. so I actually wanted to find the duplicated arrays and remove all the duplicates and leave only one unique array in the multi-dimensional array. I have added some extra code based on #Oleg's answer, and thought I would put my function here :
func removeDuplicates(_ nums: inout [[Int]] ) -> [[Int]]{
let sorted = nums.map{$0.sorted()}
var indexs = [Int]()
for (pos,item) in sorted.enumerated() {
for i in pos+1..<sorted.count {
if item == sorted[i] {
if nums.indices.contains(i){
indexs.append(i)
}
}
}
}
indexs = Array(Set<Int>(indexs))
indexs = indexs.sorted(by: {$0 > $1})
for index in indexs{
nums.remove(at: index)
}
return nums
}
My solution is quite simple and easy to understand.
let input = [[1,2,3], [3,2,1], [4,9,3]]
First let sort all elements of the nested arrays. (It gives us a bit more efficiency.)
let sorted = input.map{$0.sorted()}
Than we should compare each elements.
for (pos,item) in sorted.enumerated() {
for i in pos+1..<sorted.count {
if item == sorted[i] {
print(input[pos])
print(input[i])
}
}
}
Output:
[1, 2, 3]
[3, 2, 1]
One simple and easy brute force approach that comes to my mind is:
Iterate over each row and sort its values. So 1,2,3 will become 123 and 3,2,1 will also become 1,2,3.
Now store it in a key value pair i.e maps. So your key will be 123 and it will map to array 1,2,3 or 3,2,1.
Note:- Your key is all the sorted elements combined together as string without commas.
This way you will know that how may pairs of arrays are there inside a 2d array are identical.
There is a very efficient algorithm using permutation hashing method.
1) preprocess the 2-dim array so that all elements are non-negative. (by subtracting the smallest element from all elements)
2) with each sub-array A:
compute hash[A] = sum(base^A[i] | with all indexes i of sub-array A). Choose base to be a very large prime (1e9+7 for example). You can just ignore the integer-overflow problem when computing, because we use only additions and multiplications here.
3) now you have array "hash" of each sub-array. If the array has 2 identical sub-arrays, then they must have the same hash codes. Find all pairs of sub-arrays having equal hash codes(using hash again, or sorting, ... whatever).
4) For each pair, check again if these sub-arrays actually match (sort and compare, ... whatever). Return true if you can find 2 sub-arrays that actually match, false otherwise.
Practically, this method runs extremely fast even though it is very slow theoretically. This is because of the hashing step will prune most of search space, and this hash function is super strong. I am sure 99.99% that if exist, the pair of corresponding sub-arrays having the same hash codes will actually match.
I have two arrays like this:
a = [{'one'=>1, 'two'=>2},{'uno'=>1, 'dos'=>2}]
b = ['english', 'spanish']
I need to add a key-value pair to each hash in a to get this:
a = [{'one'=>1, 'two'=>2, 'language'=>'english'},{'uno'=>1, 'dos'=>2, 'language'=>'spanish'}]
I attempted this:
(0..a.length).each {|c| a[c]['language']=b[c]}
and it does not work. With this:
a[1]['language']=b[1]
(0..a.length).each {|c| puts c}
an error is shown:
NoMethodError (undefined method '[]=' for nil:NilClass)
How can I fix this?
a.zip(b){|h, v| h["language"] = v}
a # => [
# {"one"=>1, "two"=>2, "language"=>"english"},
# {"uno"=>1, "dos"=>2, "language"=>"spanish"}
# ]
When the each iterator over your Range reaches the last element (i.e. a.length), you will attempt to access a nonexisting element of a.
In your example, a.length is 2, so on the last iteration of your each, you will attempt to access a[2], which doesn't exist. (a only contains 2 elements wich indices 0 and 1.) a[2] evaluates to nil, so you will now attempt to call nil['language']=b[2], which is syntactic sugar for nil.[]=('language', b[2]), and since nil doesn't have a []= method, you get a NoMethodError.
The immediate fix is to not iterate off the end of a, by using an exclusive Range:
(0...a.length).each {|c| a[c]['language'] = b[c] }
By the way, the code you posted:
(0..a.length).each {|c| puts c }
should clearly have shown you that you iterate till 2 instead of 1.
That's only the immediate fix, however. The real fix is to simply never iterate over a datastructure manually. That's what iterators are for.
Something like this, where Ruby will keep track of the index for you:
a.each_with_index do |hsh, i| hsh['language'] = b[i] end
Or, without fiddling with indices at all:
a.zip(b.zip(['language'].cycle).map(&:reverse).map(&Array.method(:[])).map(&:to_h)).map {|x, y| x.merge!(y) }
[Note: this last one doesn't mutate the original Arrays and Hashes unlike the other ones.]
The problem you're having is that your (0..a.length) is inclusive. a.length = 2 so you want to modify it to be 0...a.length which is exclusive.
On a side note, you could use Array#each_with_index like this so you don't have to worry about the length and so on.
a.each_with_index do |hash, index|
hash['language'] = b[index]
end
Here is another method you could use
b.each_with_index.with_object(a) do |(lang,i),obj|
obj[i]["language"] = lang
obj
end
#=>[
{"one"=>1, "two"=>2, "language"=>"english"},
{"uno"=>1, "dos"=>2, "language"=>"spanish"}
]
What this does is creates an Enumerator for b with [element,index] then it calls with_object using a as the object. It then iterates over the Enumerator passing in each language and its index along with the a object. It then uses the index from b to find the proper index in a and adds a language key to the hash that is equal to the language.
Please know this is a destructive method where the objects in a will mutate during the process. You could make it non destructive using with_object(a.map(&:dup)) this will dup the hashes in a and the originals will remain untouched.
All that being said I think YAML would be better suited for a task like this but I am not sure what your constraints are. As an example:
yml = <<YML
-
one: 1
two: 2
language: "english"
-
uno: 1
dos: 2
language: "spanish"
YML
require 'yaml'
YAML.load(yml)
#=>[
{"one"=>1, "two"=>2, "language"=>"english"},
{"uno"=>1, "dos"=>2, "language"=>"spanish"}
]
Although using YAML I would change the structure for numbers to be more like language => Array of numbers by index e.g. {"english" => ["zero","one","two"]}. That way you can can access them like ["english"][0] #=> "zero"