odwl profile - python coding challenges

33

PositronicLlama

Functional-PositronicLlama 2 1

I think that list(zip(*matr)) is enough. More

57

veky

for else-veky 2 2

I was sure you would have found something in l.4:) I do agree with the comment that treating boolean as int is less good than an equivalent without. I did the following by using bool as bool count = count + 1 if w[0].isalpha() else 0 More

9

aomoriringo

First-aomoriringo 2

This actually incorrect for large input. Use itertools.count() instead of range(0, 1000) More

24

Juge_Ti

First-Juge_Ti 2

It is unreadable but wow, not even a loop. Did you applied a known math formula? More

26

htamas

First-htamas 2 1

sum(array[-1:] is an interesting way to avoid the empty array. This is probably the shortest readable solution. More

31

Cjkjvfnby

Speedy-Cjkjvfnby 1

Interesting way to get 0. Still very readable. More

14

Kasalehlia

First-Kasalehlia 1

You can avoid the second set creation by using intersection More

15

Gennady

First-Gennady 1

You can use endswith Sorting before by len allow you to avoid the len comparison. Note that by sorting lexical, you could even avoid a loop More

15

drill3r92

regexp-drill3r92 1

1. You don't need the [] around the " 2. you can use () to select the words without the quotes 3. then you don't need to make a comprehension list: return [elem[1:-1] for elem in re.findall(r'["][^"]*["]' , a)] --> return re.findall(r'"([^"]*)"' , a) More