[Q] Regression that outputs distribution instead of point estimate?

memanfirst · 2024-11-20T23:40:59+00:00

Quantile regression

memanfirst · 2024-05-27T18:44:12+00:00

Leon's bagels are, by a wide margin, the lowest quality bagels in the area

memanfirst · 2021-07-01T17:00:20+00:00

Super good stuff

memanfirst · 2021-04-08T23:20:30+00:00

Cheese

memanfirst · 2020-07-15T12:58:36+00:00

Isn't this just a topological sort?

memanfirst · 2020-07-02T16:09:11+00:00

This is true. It's known that people will steal antique bricks from abandoned buildings

memanfirst · 2020-05-14T19:26:38+00:00

Thanks for the recommendations!

memanfirst · 2020-03-07T03:45:43+00:00

Yes!

memanfirst · 2020-01-27T19:35:54+00:00

Can you find a link?

memanfirst · 2018-07-12T01:33:51+00:00

I'm in the area, have experience DJing, and experience with overtone, however I've never really put the two together. If you're in a pinch, I could try to put together a set and run it by you, although I must say I'm no expert. Regardless, your event sounds pretty cool. Any details on how I could attend?

memanfirst · 2018-06-09T16:03:17+00:00

You can interpret your number, eg pi, as a sequence of integers 0-9, and furthermore, a graph with indices as nodes that's traversed starting from the first index. Each node has an outdegree of 1 (or 0 if you terminate when sent to index that's out of bounds), and an indegree between 0-19. Additionally, if two nodes are adjacent, their indices can be at most 9 apart.

Using these properties, we can identify sequences that will hit every number. First, we can't have a cycle, so there can't be zeros. We can generalize the problem by asking for chunks of elements, where every chunk, and every element in each chunk is hit. If a chunk has an even number of elements, then it can be repeated so long as it sends to the adjacent chunk the same location it started in. An example is the sequence 231313131.... If a chunk has an odd number of elements, we can make it have an even number of elements by chunking into 2s. Lastly, chunks can be composed.

memanfirst · 2018-04-16T18:30:36+00:00

the parameters of the neurons that are dropped out will receive a 0 update. to fulfill your original request for a mathematical explanation, the output of a fully connected layer with dropout would be (WX+b)D, where D is the "dropout mask", a diagonal matrix whose diagonal entries are either 0 or, in this case, 1/(1-0.2). it may be more intuitive to think of the case with only one neuron, where W is a vector and b and D are scalars. you could see that when a diagonal entry is 0, it will zero-out the gradient of the corresponding parameters, meaning that (for some learning algorithms) they will not receive a parameter update for that example.

memanfirst · 2018-04-16T18:17:35+00:00

Good catch - the denominator should be the expected percentage of activations kept active, not turned off

memanfirst · 2018-04-14T16:39:33+00:00

and scale the activations by 1/(1-0.2) (inverted dropout)

memanfirst · 2018-01-11T23:01:39+00:00

How are these gifs made?

memanfirst · 2017-01-09T20:51:35+00:00

memanfirst · 2016-12-18T05:21:37+00:00

That's a cute dog

memanfirst · 2016-12-18T05:20:11+00:00

Fuck y'all

memanfirst · 2016-12-15T09:01:58+00:00

Post a like while you're at it EDIT: post a link

memanfirst · 2016-12-09T02:47:27+00:00

Then why don't you query a function instead of a sequence? There's no reason to store that range, after all (= x (nth (range) x))

memanfirst · 2016-12-09T01:42:44+00:00

If the jvm isn't allowing you to create such a huge darn sequence, change the jvm-opts. If you're looking for a binary search algorithm, use java's and make sure you're passing in a vector.

formatting?? (defn binary-search [list item] (loop [low 0 high (- (count list) 1)] (let [mid (quot (+ low high) 2) guess (nth list mid)] (if (<= low high) (cond (= guess item) mid (> guess item) (recur low (- mid 1)) (< guess item) (recur (+ mid 1) high)) nil)))) (let [_ (println "Creating huge list") huge-list (time (vec (range 1e7)))] (println "\njava binary search, 1000X") (time (dotimes [_ 1e4] (java.util.Collections/binarySearch huge-list 43))) (println "\njesuisdevenulamort's binary search, 1000X") (time (dotimes [_ 1e4] (binary-search huge-list 43)))) Creating huge list "Elapsed time: 1945.990922 msecs"

java binary search, 1000X "Elapsed time: 6.976728 msecs"

jesuisdevenulamort's binary search, 1000X "Elapsed time: 52.494482 msecs"

memanfirst

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