I have data in the following format
"stats": {
"team": [],
"outcome": [],
"rank": []
}
I need to determine if there is a combination of 2 or more results present from the above structure then print something.
So the idea is:
(if stats.team.present? && if stats.outcome.present) || (if stats.outcome.present? && if stats.rank.present) || (if stats.team.present? && if stats.rank.present)
A better way is to create a method to add a counter that its incremented if the team, outcome, rank has a count of greater than 0.
And then check if the counter is greater than 2.
Eg:
def my_count
count = 0
count += 1 if stats.team.count > 0
count += 1 if stats.outcome.count > 0
count += 1 if stats.rank.count > 0
if count > 1
return true
end
end
Are these the only 2 options or is there a cleaner way?
Are these the only 2 options or is there a cleaner way?
A ton of cleaner ways, but the best ones will use many?, part of ActiveSupport.
many? is essentially like any?, but instead of asking if "one or more" meet a condition, it asks if two or more. It's by far the most semantically correct implementation of your question:
stats = { team: [], outcome: [], rank: [] }}
stats.many? { |k,v| v.present? } # false
stats = { team: [1], outcome: [1], rank: [] }}
stats.many? { |k,v| v.present? } # true
You can get slightly more clever with stats.values and Symbol#to_proc to shorten this further, but I don't see the need:
stats.values.many?(&:present?)
No need to transform it into an array:
data = {stats: { team: [], outcome: [], rank: [] }}
if data[:stats].reject{|k,v| v.empty?}.size > 1
You can do as
data = {"stats" => { "team" => [], "outcome" => [1], "rank" => [] }}
if data["stats"].values.count([]) > 1
#code
end
First of all, is it hash or object? I will think of hash.
According to your question: some kind of MapReduce may be looking better:
["team", "outcome", "rank"].map{|key| stats[key].present? }.count(true) > 1
You can try map/reduce and can read more here
after map/reduce you can check the output to see if there are any combination
Related
# Example 1
People = ["Terry", "Merry"]
Fruit = ["Apple","Grape","Peach"]
# Possible solutions:
[
{"Terry"=>"Apple","Merry"=>"Grape"},
{"Terry"=>"Apple","Merry"=>"Peach"},
{"Terry"=>"Grape","Merry"=>"Apple"},
{"Terry"=>"Grape","Merry"=>"Peach"},
{"Terry"=>"Peach","Merry"=>"Apple"},
{"Terry"=>"Peach","Merry"=>"Grape"},
]
# Example 2
People = ["Terry", "Merry", "Perry"]
Fruit = ["Apple","Grape"]
# Possible solutions:
[
{"Terry"=>"Apple","Merry"=>"Grape","Perry"=>nil},
{"Terry"=>"Apple","Merry"=>nil,"Perry"=>"Grape"},
{"Terry"=>"Grape","Merry"=>"Apple","Perry"=>nil},
{"Terry"=>"Grape","Merry"=>nil,"Perry"=>"Apple"},
{"Terry"=>nil,"Merry"=>"Apple","Perry"=>"Grape"},
{"Terry"=>nil,"Merry"=>"Grape","Perry"=>"Apple"},
]
Stuck trying to solve this recursively (necessary for this exercise, though let me know if you don't think recursion is possible).
I feel like basically I start by assigning a random person a fruit, and then add that to all possible solutions that arise from the smaller subset of assigning remaining people remaining fruit.
E.g., for Example 1, I assign Terry an Apple, and then aggregate that with the remaining possible options of what Merry can get (either Grape or Peach).
Then just repeat changing up the fruit assigned to the first random person (e.g., with Terry getting Grape then Peach, in Example 1).
I feel like this sounds so straightforward but I'm struggling.
It can be done recursively as follows.
def hmmm(people, fruit)
adj_fruit = fruit + [nil]*([people.size-fruit.size, 0].max)
recurse(adj_fruit).map { |a| people.zip(a).to_h }
end
def recurse(fruit_left, fruit_selected = [])
return [fruit_selected + fruit_left] if fruit_left.size == 1
fruit_left.each_with_object([]) do |f,a|
recurse(fruit_left - [f], fruit_selected + [f]).each { |e| a << e }
end
end
hmmm(["Terry", "Merry"], ["Apple", "Grape", "Peach"])
#=> [{"Terry"=>"Apple", "Merry"=>"Grape"}, {"Terry"=>"Apple", "Merry"=>"Peach"},
# {"Terry"=>"Grape", "Merry"=>"Apple"}, {"Terry"=>"Grape", "Merry"=>"Peach"},
# {"Terry"=>"Peach", "Merry"=>"Apple"}, {"Terry"=>"Peach", "Merry"=>"Grape"}]
Here adj_fruit #=> ["Apple", "Grape", "Peach"]
hmmm(["Terry", "Merry", "Perry"], ["Apple", "Grape"])
#=> [{"Terry"=>"Apple", "Merry"=>"Grape", "Perry"=>nil},
# {"Terry"=>"Apple", "Merry"=>nil, "Perry"=>"Grape"},
# {"Terry"=>"Grape", "Merry"=>"Apple", "Perry"=>nil},
# {"Terry"=>"Grape", "Merry"=>nil, "Perry"=>"Apple"},
# {"Terry"=>nil, "Merry"=>"Apple", "Perry"=>"Grape"},
# {"Terry"=>nil, "Merry"=>"Grape", "Perry"=>"Apple"}]
Here adj_fruit #=> ["Apple", "Grape", nil].
We can see map's receiver in hmmm by removing .map { |a| people.zip(a).to_h } from its last line.
def hmmm(people, fruit)
adj_fruit = fruit + [nil]*([people.size-fruit.size, 0].max)
recurse(adj_fruit)
end
hmmm(["Terry", "Merry"], ["Apple","Grape","Peach"])
#=> [["Apple", "Grape", "Peach"], ["Apple", "Peach", "Grape"],
# ["Grape", "Apple", "Peach"], ["Grape", "Peach", "Apple"],
# ["Peach", "Apple", "Grape"], ["Peach", "Grape", "Apple"]]
A more conventional solution, such as the one following, would not employ recursion.
def hmmm(people, fruit)
(fruit + [nil]*[people.size - fruit.size, 0].max).
permutation(people.size).
map { |a| people.zip(a).to_h }
end
This produces the same return values as those shown above for the recursive solution.
See Array#permutation and Enumerable#zip.
If len(people) <= len(fruit), then you can use
for pieces in itertools.permutations(fruit, len(people)):
assign the pieces of fruit to the people in order
If len(people) > len(fruit), then use
for eaters in itertools.permutations(people, len(fruit))
assign the eaters to the fruit in order, and the others get nothing
I don't know how to combine the two separate cases into a single case
I now see that this was supposed to be solve recursively. Misread the original.
Let's look the possibilities for
assignment(people, fruit):
If len(people) == 0, then you're done, with the empty solution. (Not to be confused with no solution.)
If len(fruit) == 0, then no one gets any fruit. Again, this is an actual solution.
If len(people) <= len(fruit), then the first person gets some piece of fruit, appended onto all possible results of the remainder of the people getting the remainder of the fruit.
If len(people) > len(fruit), then either the first person does or doesn't get a piece of fruit, and recursively the rest of the people get whatever's left.
It's left as an exercise to you how to code this.
For anyone's future reference, this was my answer using recursion.
NOTE that "nil" overcounts; since "nil" is treated as a unique entry, the code reads {"Terry"=>"apple","Merry"=>"nil","Perry"=>"nil"} and {"Terry"=>"apple","Perry"=>"nil","Merry"=>"nil"} as 2 distinct solutions. I did not investigate further because this isn't super realistic for the exercise that this is a part of.
I also didn't investigate further for same reason, but using string "nil" versus nil yielded different results
def pure5(people, fruit, solution = [])
people_count = people.size
fruit_count = fruit.size
diff = people_count - fruit_count
diff.times { fruit << "nil" } if diff > 0
people.each do |p|
fruit.each do |f|
if people.size == 1
obj = {}
obj[p] = f
solution << obj
else
partial_solution = pure5(people - [p], fruit - [f])
partial_solution.each do |s|
s[p] = f
end
solution = solution + partial_solution
end
end
return solution
end
end
I have a code that gets the list of checkins given a span of time. See code below.
from = Time.zone.now.beginning_of_month
to = Time.zone.now.end_of_month
customer_checkins = CustomerCheckin.where(account_id: seld.id, created_at: from..to)
The code would then give me all checkin objects that satisfies the given condition. The next thing that I need to do is to group the list of checkins per customer. So I have this code to do that.
group_customer_id = customer_checkins.group(:customer_id).count
Grouping it by customer id will then result to a hash. See example below.
{174621=>9,180262=>1,180263=>1,180272=>1,180273=>3,180274=>3,180275=>4,180276=>3,180277=>2,180278=>4,180279=>4,180280=>3,180281=>5,180282=>8}
I would like now to get the count of customers with the same checkin count - how many customers have 9 checkins, 5 checkins, etc. So given the hash above. I am expecting an output like this:
{9 => 1, 8=> 1, 5 => 1, 4=> 3, 3 => 4, 2=> 1, 1 => 3}
h.each_with_object({}) {|(k,v), h| h[v] = h[v].to_i + 1}
# => {9=>1, 1=>3, 3=>4, 4=>3, 2=>1, 5=>1, 8=>1}
Get the values from hash like:
customer_array = {174621=>9,180262=>1,180263=>1,180272=>1,180273=>3,180274=>3,180275=>4,180276=>3,180277=>2,180278=>4,180279=>4,180280=>3,180281=>5,180282=>8}.values
customer_count = Hash.new(0)
customer_array.each do |v|
customer_count[v] += 1
end
puts customer_count
a = {174621=>9,180262=>1,180263=>1,180272=>1,180273=>3,180274=>3,180275=>4,180276=>3,180277=>2,180278=>4,180279=>4,180280=>3,180281=>5,180282=>8}
result = a.map{|k,v| v}.each_with_object(Hash.new(0)) { |word,counts| counts[word] += 1 }
# => {9=>1, 1=>3, 3=>4, 4=>3, 2=>1, 5=>1, 8=>1}
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
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]
I would like to analyse data in my database to find out how many times certain words appear.
Ideally I would like a list of the top 20 words used in a particular column.
What would be the easiest way of going about this.
Create an autovivified hash and then loop through the rows populating the hash and incrementing the value each time you get the same key (word). Then sort the hash by value.
A word counter...
I wasn't sure if you were asking how to get rails to work on this or how to count words, but I went ahead and did a column-oriented ruby wordcounter anyway.
(BTW, at first I did try the autovivified hash, what a cool trick.)
# col: a column name or number
# strings: a String, Array of Strings, Array of Array of Strings, etc.
def count(col, *strings)
(#h ||= {})[col = col.to_s] ||= {}
[*strings].flatten.each { |s|
s.split.each { |s|
#h[col][s] ||= 0
#h[col][s] += 1
}
}
end
def formatOneCol a
limit = 2
a.sort { |e1,e2| e2[1]<=>e1[1] }.each { |results|
printf("%9d %s\n", results[1], results[0])
return unless (limit -= 1) > 0
}
end
def formatAllCols
#h.sort.each { |a|
printf("\n%9s\n", "Col " + a[0])
formatOneCol a[1]
}
end
count(1,"how now")
count(1,["how", "now", "brown"])
count(1,[["how", "now"], ["brown", "cow"]])
count(2,["you see", "see you",["how", "now"], ["brown", "cow"]])
count(2,["see", ["see", ["see"]]])
count("A_Name Instead","how now alpha alpha alpha")
formatAllCols
$ ruby count.rb
Col 1
3 how
3 now
Col 2
5 see
2 you
Col A_Name Instead
3 alpha
1 how
$
digitalross answer looks too verbose to me, also, as you tag ruby-on-rails and said you use DB.. i'm assuming you need an activerecord model so i'm giving you a full solution
in your model:
def self.top_strs(column_symbol, top_num)
h = Hash.new(0)
find(:all, :select => column_symbol).each do |obj|
obj.send(column_symbol).split.each do |word|
h[word] += 1
end
end
h.map.sort_by(&:second).reverse[0..top_num]
end
for example, model Comment, column body:
Comment.top_strs(:body, 20)