Linell Bonnette Linell

## jarascrit.js
$(document).ready(function() {
  p = location.pathname.replace('/', '').split('/')[0];
    $("#navbar a").each(function() {
      if ( $(this).attr('href').replace('/', '') == p ) {
          $(this).parent().addClass('active');
          return;
      }
  });
});

## .phoenix.js
/*

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## keybase.md

      
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                Linell
                / keybase.md
            
            
              Created
              August 11, 2014 19:51
            
          
    Keybase proof

I hereby claim:

I am Linell on github.
I am linell (https://keybase.io/linell) on keybase.
I have a public key whose fingerprint is E285 1A23 F740 2AAF 6EE5  73C4 ACB8 24FC 7396 12F9

To claim this, I am signing this object:

  
## Project Euler Problem One.swift
func isGood(number: Int) -> Bool {
    return (number % 3 == 0 || number % 5 == 0)
}

Array(3...999)
    .filter { (number) in isGood(number) }
    .reduce(0) { (total, number) in total + number }

## random_date.rb
  def random_date
    date1 = DateTime.new(2014,8,1)
    date2 = DateTime.now()
    Time.at( (date2.to_f - date1.to_f)*rand + date1.to_f).to_date
  end

## .git-commit-template.txt
# Type(<scope>): <subject>

# <body>

# <footer>

# Type should be one of the following:
# * feat (new feature)
# * fix (bug fix)
# * docs (changes to documentation)

## pe_problem_one.ex
# The fancy anonymous function syntax. I'm kind of not a fan of it in this case.
divisible? = &(rem(&1, 5) == 0 || rem(&1, 3) == 0)
1..999 |> Enum.filter(divisible?) |> Enum.sum

## jose.ex
"Jose"
|> String.codepoints()
|> Enum.chunk(2)
|> Enum.map(&Enum.join(&1, ""))
 # ["Jo", "se"]

## euler_one.ex
Enum.filter(1..999, fn(x) -> rem(x, 5) == 0 || rem(x, 3) == 0 end)
  |> Enum.reduce(fn(x, acc) -> acc + x end)
# 233168

## thing.sql
-- Works perfectly!
SELECT COUNT(*),
  ST_Union(geom) as location_union,
  ST_SnapToGrid(ST_Transform(ST_SnapToGrid(ST_Transform(geom, _ST_BestSRID(geom)), 10,10),4326),0.0001, 0.0001) AS snap_pt
FROM locations
GROUP BY snap_pt
HAVING COUNT(*) > 1;

-- But what's a decent way to figure out which groups contain a given user?
-- I can use the following query to get the list of users in an example ST_Union from above:
	$(document).ready(function() {
	p = location.pathname.replace('/', '').split('/')[0];
	$("#navbar a").each(function() {
	if ( $(this).attr('href').replace('/', '') == p ) {
	$(this).parent().addClass('active');
	return;
	}
	});
	});
	/*

	▄███████▄ ▄█ █▄ ▄██████▄ ▄████████ ███▄▄▄▄ ▄█ ▀████ ▐████▀
	███ ███ ███ ███ ███ ███ ███ ███ ███▀▀▀██▄ ███ ███▌ ████▀
	███ ███ ███ ███ ███ ███ ███ █▀ ███ ███ ███▌ ███ ▐███
	███ ███ ▄███▄▄▄▄███▄▄ ███ ███ ▄███▄▄▄ ███ ███ ███▌ ▀███▄███▀
	▀█████████▀ ▀▀███▀▀▀▀███▀ ███ ███ ▀▀███▀▀▀ ███ ███ ███▌ ████▀██▄
	███ ███ ███ ███ ███ ███ █▄ ███ ███ ███ ▐███ ▀███
	███ ███ ███ ███ ███ ███ ███ ███ ███ ███ ▄███ ███▄
	▄████▀ ███ █▀ ▀██████▀ ██████████ ▀█ █▀ █▀ ████ ███▄
	func isGood(number: Int) -> Bool {
	return (number % 3 == 0 \|\| number % 5 == 0)
	}

	Array(3...999)
	.filter { (number) in isGood(number) }
	.reduce(0) { (total, number) in total + number }
	def random_date
	date1 = DateTime.new(2014,8,1)
	date2 = DateTime.now()
	Time.at( (date2.to_f - date1.to_f)*rand + date1.to_f).to_date
	end
	# Type(<scope>): <subject>

	# <body>

	# <footer>

	# Type should be one of the following:
	# * feat (new feature)
	# * fix (bug fix)
	# * docs (changes to documentation)
	# The fancy anonymous function syntax. I'm kind of not a fan of it in this case.
	divisible? = &(rem(&1, 5) == 0 \|\| rem(&1, 3) == 0)
	1..999 \|> Enum.filter(divisible?) \|> Enum.sum
	"Jose"
	\|> String.codepoints()
	\|> Enum.chunk(2)
	\|> Enum.map(&Enum.join(&1, ""))
	# ["Jo", "se"]
	Enum.filter(1..999, fn(x) -> rem(x, 5) == 0 \|\| rem(x, 3) == 0 end)
	\|> Enum.reduce(fn(x, acc) -> acc + x end)
	# 233168
	-- Works perfectly!
	SELECT COUNT(*),
	ST_Union(geom) as location_union,
	ST_SnapToGrid(ST_Transform(ST_SnapToGrid(ST_Transform(geom, _ST_BestSRID(geom)), 10,10),4326),0.0001, 0.0001) AS snap_pt
	FROM locations
	GROUP BY snap_pt
	HAVING COUNT(*) > 1;

	-- But what's a decent way to figure out which groups contain a given user?
	-- I can use the following query to get the list of users in an example ST_Union from above: