Friday, February 22, 2019
Hi Sapna,
This is a fun exercise to do and also to do some refactoring after I got to understand what you are aiming to do.
First of all, your code is not optimal yet. Let's look at the output of the example you gave:
{"1"=>"( true ) || ( false ) || ( true )", "2"=>"( true && false ) || ( true && true ) || ( false && true ) || ( false && true ) || ( true && true ) || ( true && false )", "3"=>"( true && false && true ) || ( true && true && false ) || ( false && true && true ) || ( false && true && true ) || ( true && true && false ) || ( true && false && true )"}
The combination of 2 and 3 have duplicated information. For example: (true && false && true) are the same as (true && true && false). I have provided a solution for what I understood from your problem below with some comments.
I am considering that you need to solve this in one method, otherwise, I would split in many different methods to make the code more clear. I am also being very cheeky here to introduce many features of the language that would make your life easier and your code much more understandable.
I hope this helps.
# Evaluate the combinations of conditions.
#
# step_conditions_labels - The Array of conditions.
#
# Examples
#
# conditions_combination(%w[true false true])
# # => {1=>true, 2=>true, 3=>false}
#
# conditions_combination(%w[true false true false true])
# # => {1=>true, 2=>true, 3=>true, 4=>false, 5=>false}
#
# Returns a hash of the evaluation of respective combinations.
def conditions_combination(step_conditions_labels)
# Reduce the Array to the result Hash.
step_conditions_labels.each_with_index.reduce({}) do |hash, (_, index)|
# Update each of the Hash keys with the evalutation of the combination.
hash.update(
index + 1 =>
# Eval is the way to go here as Tales mentioned, but it's a very
# dangerous method. Be careful!
eval(
# Array#combination is what does the trick for you. Check it out:
# The other lines just help on building the condition String.
step_conditions_labels
.combination(index + 1)
.map { |combination| "(#{combination.join(' && ')})" }
.join(' || ')
)
)
end
end
conditions_combination(%w[true false true])
Best regards,
/ Marco
On Fri, Feb 22, 2019 at 6:56 PM Hassan Schroeder <hassan.schroeder@gmail.com> wrote:
On Fri, Feb 22, 2019 at 4:36 AM Sapna Mishra <sapna.spaceo@gmail.com> wrote:
> Issue is now I am not able use this combination to check whether it satisfies the condition as it is consider as string class.
If I saw something like that in a PR I would reject it in ~2 seconds.
What actual problem are you trying to solve?
--
Hassan Schroeder ------------------------ hassan.schroeder@gmail.com
twitter: @hassan
Consulting Availability : Silicon Valley or remote
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