Combining multiple array/hash selects

Question

I have the following code:

sum = array_of_hashes.select{ |key| (date_range).include? Date.parse(key[:created])}.map { |h| h[:amount] }.sum 
size = array_of_hashes.select{ |key| (date_range).include? Date.parse(key[:created])}.size
total = sum / size

sum selects all hashes with a date that is inside a date range and the adds up all the values of the :amount key.

size counts the number of hashes that are in the date range.

total divides the sum by the size.

How can I combine those so it's not 3 separate items?

toro2k · Accepted Answer · 2014-02-06 16:33:02Z

3

I think it's as simple as:

selected = array_of_hashes.select { ... }
avarage = selected.map { ... }.sum / selected.size

Note: using include? with ranges of dates is pretty inefficient since it needs to traverse the whole dates range, I suggest to use cover? instead.

edited Feb 6, 2014 at 16:33

answered Feb 6, 2014 at 16:10

toro2k

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7 Comments

Shpigford Over a year ago

Do you know of a way to do that in one line (even if it sacrifices a little bit of readability)?

toro2k Over a year ago

Why do you want it on one line?

Shpigford Over a year ago

Asking if it's possible. I you can't do it, no worries. Curious if it can be done, though.

toro2k Over a year ago

@sawa Nothing to be sorry about! :-) cover? and include? behaves almost the same when endpoints are Numerics or Strings of size 1, in any other case include? calls Enumerable#include?. In this answer you can find more information.

toro2k Over a year ago

@sawa Yes, rb_call_super in C is, roughly, super in Ruby.

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Nikos · Accepted Answer · 2014-02-06 16:30:09Z

0

There is really no nice way of doing this more compact. One alternative could be the following:

average = (selected = array_of_hashes.select { ... }.map { ... }).sum/selected.size.to_f

answered Feb 6, 2014 at 16:30

Nikos

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