Parse string values into integer in a nested array

Question

I have a nested array with string values:

a = [["2000", "2004"], ["2005", "2007"]]

I am converting the value to integer like this:

int_val = []

a.each do |a|
  int_val << a.map(&:to_i)
end

int_val #=> [[2000, 2004], [2005, 2007]]

Is there any short solution or one line code for this? [["2000", "2004"], ["2005", "2007"]].map(&:to_i) will not work.

[["2000", "2004"], ["2005", "2007"]].map{|row| row.map(&:to_i)} works... — dawg
– dawg, Commented Apr 10, 2018 at 4:55
@dawg, you may have missed that halfelf gave that as a solution earlier. — Cary Swoveland
– Cary Swoveland, Commented Apr 10, 2018 at 6:36
a.each do |a| – re-using an outer variable name (variable shadowing) can lead to confusion. Don't do that. — Stefan
– Stefan, Commented Apr 10, 2018 at 7:45

halfelf · Accepted Answer · 2018-04-10 04:41:35Z

8

How about "two dimensional" map?

a = [["2000", "2004"], ["2005", "2007"]]
a.map {|i| i.map(&:to_i) }  # => [[2000, 2004], [2005, 2007]]

answered Apr 10, 2018 at 4:41

halfelf

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Comments

Sagar Pandya · Accepted Answer · 2018-04-10 05:42:07Z

4

A single map.

a.map { |i, j| [i.to_i, j.to_i] }

Of course this is not a general solution, but is good for your example.

answered Apr 10, 2018 at 5:42

Sagar Pandya

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Comments

Aleksei Matiushkin · Accepted Answer · 2018-04-10 05:49:01Z

4

Use Enumerable#each_with_object:

a = [["2000", "2004"], ["2005", "2007"]]
a.each_with_object([]) do |a, int_val|
  int_val << a.map(&:to_i)
end
#⇒ [[2000, 2004], [2005, 2007]]

~~It should be less memory-consuming than map.map.~~ (credits @pascalbetz for pointing out that’s not true.)

edited Apr 10, 2018 at 5:49

answered Apr 10, 2018 at 5:02

Aleksei Matiushkin

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4 Comments

Pascal Over a year ago

why would that use less memory? can you expand on that @mudasobwa?

Aleksei Matiushkin Over a year ago

@pascalbetz After a careful re-think, I am to admit I was wrong. Updated the answer.

Pascal Over a year ago

thanks for updating. Actually I was not "pointing out", just wondering why it could use less memory. Still wondering.... might write a benchmark at some point.

Stefan Over a year ago

@pascalbetz it's even slightly less efficient because – unlike map – the array has to be resized as new elements are being added.

Aleksei Matiushkin · Accepted Answer · 2018-04-10 05:50:54Z

1

Out of curiosity:

a = [["2000", "2004"], ["2005", "2007"]]
a.flatten.map(&:to_i).each_slice(2).to_a
#⇒ [[2000, 2004], [2005, 2007]]

answered Apr 10, 2018 at 5:50

Aleksei Matiushkin

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2 Comments

Cary Swoveland Over a year ago

It's curious that this was awarded the greenie considering that it only works when a.size == a[0].size == a[1].size == 2.

Aleksei Matiushkin Over a year ago

Indeed, I was also surprised.

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