I want to make a loop on a variable that can be altered inside of the loop.
first_var.sort.each do |first_id, first_value|
second_var.sort.each do |second_id, second_value_value|
difference = first_value - second_value
if difference >= 0
second_var.delete(second_id)
else
second_var[second_id] += first_value
if second_var[second_id] == 0
second_var.delete(second_id)
end
first_var.delete(first_id)
end
end
end
The idea behind this code is that I want to use it for calculating how much money a certain user is going to give some other user. Both of the variables contain hashes. The first_var is containing the users that will get money, and the second_var is containing the users that are going to pay. The loop is supposed to "fill up" a user that should get money, and when a user gets full, or a user is out of money, to just take it out of the loop, and continue filling up the rest of the users.
How do I do this, because this doesn't work?
Okay. What it looks like you have is two hashes, hence the "id, value" split.
If you are looping through arrays and you want to use the index of the array, you would want to use Array.each_index.
If you are looping through an Array of objects, and 'id' and 'value' are attributes, you only need to call some arbitrary block variable, not two.
Lets assume these are two hashes, H1 and H2, of equal length, with common keys. You want to do the following: if H1[key]value is > than H2[key]:value, remove key from H2, else, sum H1:value to H2:value and put the result in H2[key].
H1.each_key do |k|
if H1[k] > H2[k] then
H2.delete(k)
else
H2[k] = H2[k]+H1[k]
end
end
Assume you are looping through two arrays, and you want to sort them by value, and then if the value in A1[x] is greater than the value in A2[x], remove A2[x]. Else, sum A1[x] with A2[x].
b = a2.sort
a1.sort.each_index do |k|
if a1[k] > b[k]
b[k] = nil
else
b[k] = a1[k] + b[k]
end
end
a2 = b.compact
Based on the new info: you have a hash for payees and a hash for payers. Lets call them ees and ers just for convenience. The difficult part of this is that as you modify the ers hash, you might confuse the loop. One way to do this--poorly--is as follows.
e_keys = ees.keys
r_keys = ers.keys
e = 0
r = 0
until e == e_keys.length or r == r_keys.length
ees[e_keys[e]] = ees[e_keys[e]] + ers[r_keys[r]]
x = max_value - ees[e_keys[e]]
ers[r_keys[r]] = x >= 0 ? 0 : x.abs
ees[e_keys[e]] = [ees[e_keys[e]], max_value].min
if ers[r_keys[r]] == 0 then r+= 1 end
if ees[e_keys[e]] == max_value then e+=1 end
end
The reason I say that this is not a great solution is that I think there is a more "ruby" way to do this, but I'm not sure what it is. This does avoid any problems that modifying the hash you are iterating through might cause, however.
Do you mean?
some_value = 5
arrarr = [[],[1,2,5],[5,3],[2,5,7],[5,6,2,5]]
arrarr.each do |a|
a.delete(some_value)
end
arrarr now has the value [[], [1, 2], [3], [2, 7], [6, 2]]
I think you can sort of alter a variable inside such a loop but I would highly recommend against it. I'm guessing it's undefined behaviour.
here is what happened when I tried it
a.each do |x|
p x
a = []
end
prints
1
2
3
4
5
and a is [] at the end
while
a.each do |x|
p x
a = []
end
prints nothing
and a is [] at the end
If you can I'd try using
each/map/filter/select.ect. otherwise make a new array and looping through list a normally.
Or loop over numbers from x to y
1.upto(5).each do |n|
do_stuff_with(arr[n])
end
Assuming:
some_var = [1,2,3,4]
delete_if sounds like a viable candidate for this:
some_var.delete_if { |a| a == 1 }
p some_var
=> [2,3,4]
Related
I have 2 strings:
a = "qwer"
b = "asd"
Result = "qawsedr"
Same is the length of b is greater than a. show alternate the characters.
What is the best way to do this? Should I use loop?
You can get the chars from your a and b string to work with them as arrays and then "merge" them using zip, then join them.
In the case of strings with different length, the array values must be reversed, so:
def merge_alternately(a, b)
a = a.chars
b = b.chars
if a.length >= b.length
a.zip(b)
else
array = b.zip(a)
array.map{|e| e != array[-1] ? e.reverse : e}
end
end
p merge_alternately('abc', 'def').join
# => "adbecf"
p merge_alternately('ab', 'zsd').join
# => "azbsd"
p merge_alternately('qwer', 'asd').join
# => "qawsedr"
Sebastián's answer gets the job done, but it's needlessly complex. Here's an alternative:
def merge_alternately(a, b)
len = [a.size, b.size].max
Array.new(len) {|n| [ a[n], b[n] ] }.join
end
merge_alternately("ab", "zsd")
# => "azbsd"
The first line gets the size of the longer string. The second line uses the block form of the Array constructor; it yields the indexes from 0 to len-1 to the block, resulting in an array like [["a", "z"], ["b", "s"], [nil, "d"]]. join turns it into a string, conveniently calling to_s on each item, which turns nil into "".
Here's another version that does basically the same thing, but skips the intermediate arrays:
def merge_alternately(a, b)
len = [a.size, b.size].max
len.times.reduce("") {|s, i| s + a[i].to_s + b[i].to_s }
end
len.times yields an Enumerator that yields the indexes from 0 to len-1. reduce starts with an empty string s and in each iteration appends the next characters from a and b (or ""—nil.to_s—if a string runs out of characters).
You can see both on repl.it: https://repl.it/I6c8/1
Just for fun, here's a couple more solutions. This one works a lot like Sebastián's solution, but pads the first array of characters with nils if it's shorter than the second:
def merge_alternately(a, b)
a, b = a.chars, b.chars
a[b.size - 1] = nil if a.size < b.size
a.zip(b).join
end
And it wouldn't be a Ruby answer without a little gsub:
def merge_alternately2(a, b)
if a.size < b.size
b.gsub(/./) { a[$`.size].to_s + $& }
else
a.gsub(/./) { $& + b[$`.size].to_s }
end
end
See these two on repl.it: https://repl.it/I6c8/2
Let's say I have a loop repeating 1000 times.
Previous to the loop, I set 1000 variables, like so:
#n1 = "blabla"
#n2 = "blabla"
#n3 = "blabla"
#n4 = "blabla"
...
I also have a variable #count that counts which iteration the loop is on, ie. it starts at 1 and increases by 1 with each loop. What I want to do is print #n1 if #count = 1, print #n2 if #count = 2, and so forth. In other words, I want ruby to use the value of #count to decide which #n_ variable to use. I don't want to use conditional statements, because I'd need 1000 of them.
Something like this:
#count = 1
if #count < 1001
puts #("n + #count")
#count = #count + 1
end
Is there a way to do this?
Let's say you have an instance called foo, with a thousand instance variables, to loop through them all you could do:
foo.instance_variables.each do |v|
p foo.instance_variable_get(v)
end
That said, you can also use the string name to fetch them:
1000.times do |count|
p foo.instance_variable_get("#n#{count}")
end
Yes, you can do do this:
if #count < 1001
instance_variable_set("##{#count}", #count + 1)
end
It would be more idiomatic to store in a hash, e.g.
h = {}
if #count < 1001
h[#count] = #count + 1
end
While you can use something like instance_variable_get, you would usually use an array or an hash to store your strings:
n = ["blabla", "blabla", "blabla", ... ]
count = 0
if count < 1000
puts n[count]
count += 1
end
Let's say i have two relation arrays of a user's daily buy and sell.
how do i iterate through both of them using .each and still let the the longer array run independently once the shorter one is exhaused. Below i want to find the ratio of someone's daily buys and sells. But can't get the ratio because it's always 1 as i'm iterating through the longer array once for each item of the shorter array.
users = User.all
ratios = Hash.new
users.each do |user|
if user.buys.count > 0 && user.sells.count > 0
ratios[user.name] = Hash.new
buy_array = []
sell_array = []
date = ""
daily_buy = user.buys.group_by(&:created_at)
daily_sell = user.sells.group_by(&:created_at)
daily_buy.each do |buy|
daily_sell.each do |sell|
if buy[0].to_date == sell[0].to_date
date = buy[0].to_date
buy_array << buy[1]
sell_array << sell[1]
end
end
end
ratio_hash[user.name][date] = (buy_array.length.round(2)/sell_array.length)
end
end
Thanks!
You could concat both arrays and get rid of duplicated elements by doing:
(a_array + b_array).uniq.each do |num|
# code goes here
end
Uniq method API
daily_buy = user.buys.group_by(&:created_at)
daily_sell = user.sells.group_by(&:created_at
buys_and_sells = daily_buy + daily_sell
totals = buys_and_sells.inject({}) do |hsh, transaction|
hsh['buys'] ||= 0;
hsh['sells'] ||= 0;
hsh['buys'] += 1 if transaction.is_a?(Buy)
hsh['sells'] += 1 if transaction.is_a?(Sell)
hsh
end
hsh['buys']/hsh['sells']
I think the above might do it...rather than collecting each thing in to separate arrays, concat them together, then run through each item in the combined array, increasing the count in the appropriate key of the hash returned by the inject.
In this case you can't loop them with each use for loop
this code will give you a hint
ar = [1,2,3,4,5]
br = [1,2,3]
array_l = (ar.length > br.length) ? ar.length : br.length
for i in 0..array_l
if ar[i] and br[i]
puts ar[i].to_s + " " + br[i].to_s
elsif ar[i]
puts ar[i].to_s
elsif br[i]
puts br[i].to_s
end
end
I need to count the number of values that both arrays have.
def process_2arrays(arr1, arr2)
length1 = arr1.count
length2 = arr2.count
arr3 = []
i = 0
while length1 >= i do
ci = arr1[i]
if arr2.include?(ci)
arr3 << ci
damn = arr3.count
i = i + 1
end
return [(damn), (2), (3), (4)]
end
end
When I pass the values to the function it returns [nil, 2, 3, 4]
Whats the problem here?
To find elements that exist in both arrays, use the set intersection method &.
http://ruby-doc.org/core-2.2.0/Array.html#method-i-26
def count_same_elements(a1,a2)
(array1 & array2).length
end
Example
count_same_elements([1,2,3,4],[2,3,4,5])
=> 3
damn is initialized within a do .. end block, specifically the while block. Therefore, its value will live within that scope, and when you call the variable outside the block its value is nil.
If you want to preserve the value, you must initialize the variable to nil outside the block.
i = 0
damn = nil
...
As a side note, your code is lacking the most basic Ruby standards. In Ruby you generally use an iterator, not the while. Moreover, you don't use the return at the end of a method.
This is how you would write your method in Ruby using the iterators and taking advantage of some methods from the core library.
def process_2arrays(arr1, arr2)
arr3 = arr1.select { |e| arr2.include?(e) }
[arr3.size, 2, 3, 4]
end
Changing completely approach, you can use
def process_2arrays(arr1, arr2)
(arr1 & arr2).size
end
What is the best way to incrementally iterate through a pair of hashes in Ruby? Should I convert them to arrays? Should I go an entirely different direction? I am working on a problem where the code is supposed to determine what to bake, and in what quantities, for a bakery given 2 inputs. The number of people to be fed, and their favorite food. They bake 3 things (keys in my_list) and each baked item feeds a set number of people (value in my_list).
def bakery_num(num_of_people, fav_food)
my_list = {"pie" => 8, "cake" => 6, "cookie" => 1}
bake_qty = {"pie_qty" => 0, "cake_qty" => 0, "cookie_qty" => 0}
if my_list.has_key?(fav_food) == false
raise ArgumentError.new("You can't make that food")
end
index = my_list.key_at(fav_food)
until num_of_people == 0
bake_qty[index] = (num_of_people / my_list[index])
num_of_people = num_of_people - bake_qty[index]
index += 1
end
return "You need to make #{pie_qty} pie(s), #{cake_qty} cake(s), and #{cookie_qty} cookie(s)."
end
The goal is to output a list for the bakery that will result in no uneaten food. When doing the math, the modulo would then be divided into the next food item.
Thanks for the help.
What is the best way to incrementally iterate through a pair of hashes in Ruby?
Since the keys of bake_qty conveniently have a '_qty' appended to them from their corresponding keys in my_list, you can use this to your advantage:
max_value = my_list[fav_food]
my_list.each do |key,value|
next if max_value < value
qty = bake_qty[key+'_qty']
...
end
You could use 'inject' method.
until num_of_people == 0
num_of_people = my_list.inject(num_of_people) do |t,(k,v)|
if num_of_people > 0
bake_qty["#{key}_qty"] += num_of_people/v
t - v
end
end
You can sort your hash at the beginning to ensure that your first food is the fav food