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:
$(document).ready(function() { | |
p = location.pathname.replace('/', '').split('/')[0]; | |
$("#navbar a").each(function() { | |
if ( $(this).attr('href').replace('/', '') == p ) { | |
$(this).parent().addClass('active'); | |
return; | |
} | |
}); | |
}); |
/* | |
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I hereby claim:
To claim this, I am signing this object:
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: |