I've got a simple method that counts total lesson hours in the university schedule for additional modules in the department (students can attend many departments)
def hours_total
#hours_total = user.open_departments.each_with_object({}) do |department, h|
h[department] = (sports_hours[department] || 0) +
(science_hours[department] || 0) +
(intership_sum[department] || 0) +
(art[department] || 0) -
((obligatory_topics[department] || 0) +
(base[department] || 0))
end
end
How can I fix here Cyclomatic complexity for hours_total is too high.? I have no idea how to not repeat || 0 cause in some departments sports_hours[department] can be nil value
The first step I'd take
def hours_total
#hours_total = user.open_departments.each_with_object({}) do |department, h|
positive = [sport_hours, science_hours, internship_sum, art].sum do |pos_h|
pos_h[department].to_i
end
negative = [obligatory_topics, base].sum do |neg_h|
neg_h[department].to_i
end
h[department] = positive - negative
end
end
Note: if your hours can be float values, substitute to_i with to_f.
Now if you and your Rubocop are ok with that, I'd probably leave it. If any of you is unhappy, the positive and negative should be extracted to a method.
Related
Looking to work with a dataset of strings that store money amounts in these formats. For example:
$217.3M
$1.6B
$34M
€1M
€2.8B
I looked at the money gem but it doesn't look like it handles the "M, B, k"'s back to numbers. Looking for a gem that does do that so I can convert exchange rates and compare quantities. I need the opposite of the number_to_human method.
I would start with something like this:
MULTIPLIERS = { 'k' => 10**3, 'm' => 10**6, 'b' => 10**9 }
def human_to_number(human)
number = human[/(\d+\.?)+/].to_f
factor = human[/\w$/].try(:downcase)
number * MULTIPLIERS.fetch(factor, 1)
end
human_to_number('$217.3M') #=> 217300000.0
human_to_number('$1.6B') #=> 1600000000.0
human_to_number('$34M') #=> 34000000.0
human_to_number('€1M') #=> 1000000.0
human_to_number('€2.8B') #=> 2800000000.0
human_to_number('1000') #=> 1000.0
human_to_number('10.88') #=> 10.88
I decided to not be lazy and actually write my own function if anyone else wants this:
def text_to_money(text)
returnarray = []
if (text.count('k') >= 1 || text.count('K') >= 1)
multiplier = 1000
elsif (text.count('M') >= 1 || text.count('m') >= 1)
multiplier = 1000000
elsif (text.count('B') >= 1 || text.count('b') >= 1)
multiplier = 1000000000
else
multiplier = 1
end
num = text.to_s.gsub(/[$,]/,'').to_f
total = num * multiplier
returnarray << [text[0], total]
return returnarray
end
Thanks for the help!
I am building a reputation system where users get points if milestones (10, 100, 1000, ...) are archieved. I have this if statement line:
if (before_points < ((10 || 100 || 1000 || 10000 || 100000 || 1000000))) && (after_points >= ((10 || 100 || 1000 || 10000 || 100000 || 1000000)))
It should return true if the points where either less than 10 or 100 or 1000 ...before, and if the points were more or equal to either 10 or 100 or 1000 ... afterwards.
It works if it was below 10 before, and more than 10 afterwards, and I am not quite sure if it works with 100, but it doesnt work if the points were below 1000 before and more than 1000 afterwards.
Is this the correct way to do this? Is it better to do this with a switch/case?
A more compact way you could do it...
[10, 100, 1000, 10000, 100000, 1000000].any?{|n| before_points < n && after_points >= n}
That expression will return true if a boundary is crossed, or false otherwise
That's not really how logic operation work. The statement:
(10 || 100 || 1000 || 10000 || 100000 || 1000000)
will evaluate to 10. The || operator between 2 or more numbers will return first non-nil value, in this case that's 10, the first value. Related question.
And even if that weren't the case, if the before_points < 10 is true, the before_points < 1000000 would also be true and if only before_points < 1000000 was true, the if statement would still execute just the same as with before_points < 10, so the logic would be wrong.
Depending on what you want to solve, you could either use case or define your milestones in array and iterate values 10,100,...,1000000, setting new milestone each time the condition is still true.
Your assumption is wrong.
if (before_points < ((10 || 100 || ...
will first evaluate the part
10 || 100
which will always return 10 because 10 evaluates to truthy, hence this line
if (before_points < ((10 || 100 || 1000 || 10000 || 100000 || 1000000))) && (after_points >= ((10 || 100 || 1000 || 10000 || 100000 || 1000000)))
is effectively the same of
if (before_points < 10) && (after_points >= 10)
I'm not sure what you want to achieve, but it's probably better to use a case (this is just an example)
case
when before_points < 10 && after_points >= 10
# ...
when before_points < 100 && after_points >= 100
# ...
else
# ...
end
I'm trying to perform a check on my User model using a series of nested if statements. I have two models, User, which has_many :taxes, and tax which belongs_to :user. Within a column of the user table, each user has a specific "state id" which they select when they sign up (for example, state_id: 53 could mean California). Anyway, in my Tax model, I have several if statements which evaluate to see if a user is from a specific region, and if true, the conditions inside the if are performed.
For example:
if user_state_id == 53
# Do this
end
The problem is, if the user isn't a match to the first if statement, it doesn't continue to check the remaining statements to see if it matches any of the others. (See Below). If this user had a state_id of 53, it would work normally and run the remaining conditions inside. However, if the user had a state_id of 52 (next in the list) it would not evaluate.
def provincial
if user.state_id == 53 #BC
if calculation = self.income <= 10276
return 0
elsif calculation = self.income > 10276 && self.income <= 37568
return self.income * 0.0506
end
end
if user.state_id == 52 #AB
if calculation = self.income <= 17593
return 0
elsif calculation = self.income < 17593
return self.income * 0.10
end
end
if user.state_id == 60 #ON
if calculation = self.income <= 9574
return 0
elsif calculation = self.income > 9574 && self.income <= 39723
return self.income * 0.0505
end
end
end
I also tried using elsif statements for the primary nested condition checks (elsif user.state_id == 52...) but that did not work either.
I just posting a review of your code by using nested case statements.Sorry for the syntax errors(if any)
case user.state_id
when 53 #BC
case calculation
when <=10276
return 0
when 10277..37568
return self.income * 0.0506
when 52 #AB
case calculation
when 17593
return 0
when < 17593
return self.income * 0.10
when 60 #ON
case calculation
when <= 9754
return 0
when 9575..39793
return self.income * 0.0505
Hope it helps!
In your nested if statement for state_id == 52 you are checking twice whether income is smaller then 17593 (in both if and elseif).
if user.state_id == 52 #AB
if calculation = self.income <= 17593
return 0
elsif calculation = self.income < 17593 # shouldn't it be > 17593?
return self.income * 0.10
end
end
Also you are not covering all the cases! If state_id is equal to 53, and income is bigger than 37568, this will return nil.
Another thing, if you checked in you if statement that self.income <= 10276 (self is not needed btw), you don't need to recheck it again in elsesif (elseif self.income > 10276). There is no other option.
Your calculation variable is completely obsolete - your function returns if it is true and continues when its false.
I'm in the process of learning Ruby, taking a Berkeley's MOOC, and, in some of these MOOC's homework we have an exercise that says:
Define a method sum_to_n? which takes an array of integers and an
additional integer, n, as arguments and returns true if any two
elements in the array of integers sum to n. An empty array should sum
to zero by definition.
I already created two methods that can do the job, but I'm not comfortable with any of them because I think they are not written in the Ruby Way. I hope some of you can help me to learn which would be the right way!
The first method I made uses the each method for both iterations, but what I don't like about this method is that every number is summed with every other number, even with the same number, doing something like this:
arr[1, 2, 3, 4] => 1+1, 1+2, 1+3, 1+4, 2+1, 2+2, 2+3, 2+4, 3+1, 3+2... 4+3, 4+4
As you can see, there's a lot of repeated sums, and I don't want that.
This is the code:
def sum_to_n?(arr, n)
arr.each {|x| arr.each {|y| return true if x + y == n && x != y}}
return true if n == 0 && arr.length == 0
return false
end
With the other method I got what I wanted, just a few sums without repeating any of them or even summing the same numbers, but it looks HORRIBLE, and I'm pretty sure someone would love to kill me for doing it this way, but the method does a great job as you can see:
arr[1, 2, 3, 4] => 1+2, 1+3, 1+4, 2+3, 2+4, 3+4
This is the code:
def sum_to_n?(arr, n)
for i in 0..arr.length - 1
k = i + 1
for k in k..arr.length - 1
sum = arr[i] + arr[k]
if sum == n
return true
end
end
end
return true if n == 0 && arr.length == 0
return false
end
Well, I hope you guys have fun doing a better and prettier method as I did trying.
Thank you for your help.
I'd write it like this:
def sum_to_n?(arr, n)
return true if arr.empty? && n.zero?
arr.combination(2).any? {|a, b| a + b == n }
end
That seems to be a pretty Rubyish solution.
I came across this on CodeWars. The accepted answer sure does look very Rubyish, but that is at the cost of performance. Calling arr.combination(2) results in a lot of combinations, it'd be simpler to go over the array element by element and search whether the 'complement' sum - element exists. Here's how that'd look like -
def sum_to_n?(arr, n)
(arr.empty? and n.zero?) or arr.any? { |x| arr.include?(n - x) }
end
Beside #jorg-w-mittag's answer. I found another solution using 'permutation'.
https://stackoverflow.com/a/19351660/66493
def sum_to_n?(arr, n)
(arr.empty? && n.zero?) || arr.permutation(2).any? { |a, b| a + b == n }
end
I didn't know about permutation before.
Still like #jorg-w-mittag answer because its more readable.
This one will do it in O(n.log(n)) rather than O(n²):
a = 1, 2, 3, 4
class Array
def sum_to? n
unless empty?
false.tap {
i, j, sorted = 0, size - 1, sort
loop do
break if i == j
a, b = sorted[i], sorted[j]
sum = a + b
return a, b if sum == n
sum < n ? i += 1 : j -= 1
end
}
end
end
end
a.sum_to? 7 #=> [3, 4]
I had a thought that the beginning of any answer to this question should probably start with pruning the array for superfluous data:
Can't use this:
arr.select! { |e| e <= n } # may be negative values
But this might help:
arr.sort!
while arr[0] + arr[-1] > n # while smallest and largest value > n
arr.delete_at(-1) # delete largest vaue
end
i wonder why no answers here using hash ?
def sum_to_n?(arr, n)
return true if arr.empty? && n.zero?
h = {}
arr.any? { |x| complement = h[n-x]; h[x] = true; complement }
end
puts sum_to_n?([1,2,3,4,5,7], 6) # true
puts sum_to_n?([6,2,3,5,7,9], 6) # false
puts sum_to_n?([3,4,5,3], 6) # true
puts sum_to_n?([3,4,5,7], 6) # false
puts sum_to_n?([], 6) # false
puts sum_to_n?([], 0) # true
I like rohitpaulk's answer but it fails when n doubles x. We should remove x from the array before sending include? n - x.
def sum_to_n?(arr, n)
return true if arr.empty? && n.zero?
arr.any? { |x| arr.tap { arr.delete_at arr.index x }.include? n - x }
end
Lam Phan's answer using a hash is the best
User should insert all the values either positive or negative.
How may i set same sign validation ?
Right i have written this on before_save ..
unless (self.alt_1 >= 0 && self.alt_2 >=0 && self.alt_3 >= 0 &&
self.alt_4 >= 0 && self.alt_5 >= 0 && self.alt_6 >= 0) ||
(self.alt_1 <= 0 && self.alt_2 <=0 && self.alt_3 <= 0 &&
self.alt_4 <= 0 && self.alt_5 <= 0 && self.alt_6 <= 0)
self.errors.add_to_base(_("All values sign should be same."))
end
first_sign = self.alt_1 <=> 0
(2..6).each do |n|
unless (self.send("alt_#{n}") <=> 0) == first_sign
errors.add_to_base(_("All values' signs should be same."))
break
end
end
With this method we first get the sign of alt_1, and then see if the signs of the rest of the elements (alt_2 through alt_6) match. As soon as we find one that doesn't match we add the validation error and stop. It will run a maximum of 6 iterations and a minimum of 2.
Another more clever, but less efficient method, is to use the handy method Enumerable#all?, which returns true if the block passed to it returns true for all elements:
range = 1..6
errors.add_to_base(_("All values' signs should be same.")) unless
range.all? {|n| self.send("alt_#{n}") >= 0 } ||
range.all? {|n| self.send("alt_#{n}") <= 0 }
Here we first check if all of the elements are greater than 0 and then if all of the elements are less than 0. This method iterates a maximum of 12 times and a minimum of 6.
Here's a slightly different approach for you:
irb(main):020:0> def all_same_sign?(ary)
irb(main):021:1> ary.map { |x| x <=> 0 }.each_cons(2).all? { |x| x[0] == x[1] }
irb(main):022:1> end
=> nil
irb(main):023:0> all_same_sign? [1,2,3]
=> true
irb(main):024:0> all_same_sign? [1,2,0]
=> false
irb(main):025:0> all_same_sign? [-1, -5]
=> true
We use the spaceship operator to obtain the sign of each number, and we make sure that each element has the same sign as the element following it. You could also rewrite it to be more lazy by doing
ary.each_cons(2).all? { |x| (x[0] <=> 0) == (x[1] <=> 0) }
but that's less readable in my opinion.
unless
[:<=, :>=].any? do |check|
# Check either <= or >= for all values
[self.alt1, self.alt2, self.alt3, self.alt4, self.alt5, self.alt6].all? do |v|
v.send(check, 0)
end
end
self.errors.add_to_base(_("All values sign should be same."))
end