Faibbus profile - python coding challenges

30

First-bravebug 1 1

Hello, naming things explicitely would help make your code clearer. Also line 5 could be simplified as `subj = (x.lower() for x in subj if x.isalpha())` More

57

veky

Closure-veky 1 1

You are only a few comments away from perfection... But anyway, that's impressive. More

22

Sioul

Slices-Sioul 1

Why not extract the upperHalf = (l+1)//2 instead of computing it twice ? More

25

dannedved

achilles_and_the_tortoise-dannedved 1 1

The demonstration of the solution is a nice addition :) More

8

ForcePush

Bruteforce-ForcePush 1

Thank you, I laughed so hard :') More

10

Grzegorz_Wieczerzak

First: regular expression-Grzegorz_Wieczerzak 1

Hello, since you are using `re.match`, you don't need to specify `^` in your regexes. You could also have removed the _(debug)_ print statements. But since you made an effort on the docstrings and variable names, here are some upvotes 😉 More

19

DmytroKaminskiy

Achilles and the Tortoise-DmytroKaminskiy 1

Your formula is correct, but it could be simplified: ``` t2 × adv adv × (a1 - t2) + t2 × adv adv × a1 adv + ------------ = ---------------------------- = ---------- a1 - t2 a1 - t2 a1 - t2 ``` More

5

Lackerlax

First-Lackerlax

Nice :) _(using `if args:` directly might imrpove it a tad)_ More

36

nickie

Straightforward-nickie 1

I find it not-so-great that all these lines of code are duplicated to just change a "<" to ">". On the other hand, I like the idea of caching the keys. More

49

Phil15

I solved the equation, that's all.-Phil15

Hello, your formula is correct, but it could be further simplified: ``` t2 × adv adv × (a1 - t2) + t2 × adv adv × a1 adv + ------------ = ---------------------------- = ---------- a1 - t2 a1 - t2 a1 - t2 ``` More

11

jiizzyy

straight forward-jiizzyy

Nice finite state machine (I got lazy and used `re`). More

11

five_east_west

not the best solution-five_east_west 1

Hello, - nice use of `isupper` 😀 - `pat` could be simplified by using `search` instead of `match` - `pat1` could be simplified as `'[^A-Za-z0-9 ]'` - line 8 could be abbriged to `if pat.match(sub)` - line 12 could be improved with `enumerate`: `for i, word in enumerate(c):` and then use `word` inst More

6

JaneDoe17

First-JaneDoe17

I don't think this solution qualifies as Speedy. You iterate: - once on `array` (let's say its size is `n`); - once on results (size `n//2`) to build `result1`; - once again on `results1` (for the `sum`, size `n//2`). So you could actually save `n` iterations (and some memory) with something al More

26

bsquare

Clear one-line solution-bsquare

Your formula is correct, but it could be simplified: ``` t2 × adv adv × (a1 - t2) + t2 × adv adv × a1 adv + ------------ = ---------------------------- = ---------- a1 - t2 a1 - t2 a1 - t2 ``` More

15

Wojciech__Lazuka

Lamp dictionary-Wojciech__Lazuka 1

Why recreate `light_dict` in every call ? More

17

naexio

First-naexio 1

I bet you can find a more unusual solution ;) More

16

Moonra

First-Moonra 1

I'm not sure about caching in python, but it might be best to sort your data only once. More

13

macobo

threeliner-macobo

Sometimes, 3 lines are better than one ;) I came up with almost the same solution ^^ but yours' even shorter. More

10

ssdev

First-ssdev

Hmm, using my crystal ball, I'd say that your solution might only work in 1/4th of the times... More

22

Faibbus

First, one-liner-Faibbus

checkio = lambda data:(lambda f:(data[0][0] if len(data)==1 else sum([data[0][k]*(-1)**k*checkio(f(data,0,k)) for k in range(len(data))])))(lambda d,i,j:([[d[l][c] for c in range(len(d)) if c != j] for l in range(len(d)) if l != i])) Ok, I hadn't thought about that. Using that, the code would More