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nakanohito_piyo
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4610/ 4695
Rydia @NekoMofu http://twitter.com/nakanohito_piyo
Last seen 10 months ago
Member for 10 years, 3 months, 9 days
Difficulty Normal
Ph.D. student of computer science/Japanese translation committer in CheckIO. Feel free to comment!

Best reviews / Newest reviews
string.capwords-nakanohito_piyo 1
I would like to thank antauren for the idea of *string.capwords* : https://py.checkio.org/mission/conversion-into-camelcase/publications/antauren/python-3/first/ More
First-Nagyunagi 1 1
Good for simple/straightforward implementation. "Like" for last line (make dict from lists). 01. def popular_words(text: str, words: list) -> dict: 02. # your code here 03. 04. text_lower = text.lower() 05. text_words = text_lower.split() 06. # create an empty list 07. popular More
First-brownie57 1
Good, but you should not override build-in type "dict". More
First-antauren 1
From random review: I did not come up with *string.capwords*, thanks! One tip, string.capwords() have key *sep*, which makes your code clearer. See my code: https://py.checkio.org/mission/conversion-into-camelcase/publications/nakanohito_piyo/python-3/stringcapwords/ More
First-Vasily__Chibilyaev 1 1
Nice! Also, if you use math.gcd, you can also use functools.reduce to instantly get the answer. https://py.checkio.org/mission/gcd/publications/nakanohito_piyo/python-3/reduce-gcd-args/ More
First-Julio_Malhadas_Neto 1 1
Let's try list comprehension such as t2 = [i for i in text.lower() if i.isalpha()] More
First-kodix09 1 2
Your code wrongly returns True for some case: isometric_strings('', 'a') -> True You should check the length of two strings first. More
Second-Nagyunagi 1 1
Good challenge. See my example: https://py.checkio.org/mission/say-history/publications/nakanohito_piyo/python-3/format/ More
Second-Vasily__Chibilyaev 1 1
Good, clear and interesting substring slicing. Comments: * If w1 == w2, always len(w1)<=len(w2). So w1 == w2 is redundant. * w1[len(w1)-len(w2):]==w2 can be written as w1[-len(w2):]==w2 (using negative indexes) or simply w1.endswith(w2) . Also see my solution: https://py.checkio.org/mission/end-o More
Cut zero-lines then numpy.rot90()-quarkov 1
Great, it is really easy to deal arrays with numpy. More
First-Sawl_Stone
*From Random Review* Good for "transpose" the game board. But you can return function where you find the winner first, rather than append and continue checking. More
First-Nagyunagi 1
Very good for using slicing and upper() function. If-else patterns are saved. More
First-shichou
**key** parameter of list.sort() can be more clear or direct, such as function **abs**. More
First-Nagyunagi 1
Good for C-like answer. For next, try str.**split** function. More
dict slices-veky 1
Interesting solution, but how about this to be categorized as "creative"? More
First-bosse.de.poulet 1
*from random review* Good straightforward solution, but you can simplify your code logic more. line 15--16 and "len(word) > 1 and" in line 31 can be merged such as "if len(word) <= 1: continue" . More
3 shots-DiZ 1
Although I don't understand why this works, this is obviously very creative. More
First-Svetlana_Barvinchenko
Yes this works, but you should not overwrite built-in function *max* to avoid potential bugs. More
Can anyone find wrong case?-nakanohito_piyo 1
This treats round(sqrt(number)) as the sqrt of nearest square number. I didn't prove whether this solution always true, but it passed tests. More
First-michalpiszczek1993
*From random review* Good for straightforward implementation. Comment: You can use n//100 instead of math.floor(n/100), and n = n%100 instead of n -= s*100. More
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