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First-tofol 1
Hi, I agree with previous comment, but you can also significantly shorten you existing code e. g.: # assign game_result to a shorter variable gr = game_result # remove all the _row, _column, _diag variables as you don't need them (see below), # replace game_result with gr and crea More
1 line-texom512 1
Hi, "crystal clear" but creative :) 1. `map(lambda x:...)` is not good practice. 2. Alternative way to compute signum: `def signum(x): return (x > 0) - (x < 0)` More
Slow and Heavily Described.-BirrellWalsh
Hi, look at [__collections.Counter__](https://docs.python.org/3.5/library/collections.html?highlight=counter#collections.Counter). More
First-telestrial
Hi, I have a few comments: 1. Look at str.islower, str.isupper, str.isdigit. 2. Remove _print_ before publishing. 3. You could omit the last if, as your condition is by itself a boolean: # Instead of if boolean_condition: return True else: return False # you More
Regex solution-kkkkk
What a Perl... oops ... pearl :) More
First-GabrielHahmann
Hi, as _three >= 3_ is already a bool you don't need the last if: return three >= 3 More
First-GabrielHahmann
Hi, I have two comments: 1. Look at slices. Using slices is shorter and more Pythonic than while this way. 2. Why not sum += array[i]? More
First-GabrielHahmann
Hi, two things: 1. m is redundant, you could directly str(number) in for loop: for n in str(number): ... 2. You can write t *= int(n). More
Password-ira.trachuk.1
Hi, I have two comments: 1. letter is a __str__ so you could write: if letter.isdigit(): d = True # and so on 2. Move the _if d and u and l: return True_ inside the for loop. Once all of d, u and l are True you don't need to continue. More
Quicker with RegEx-ira.trachuk.1
Hi, you don't the if. You can directly return the re.match. More
First-GabrielHahmann 1
Hi, in python you can join methods therefore you could write it like this: return ','.join(phrases).replace('right', 'left') More
First-pruvosim
Hi, I have a few comments: 1. Don't use _min_ as variable name. _min_ as a built-in function. 2. taille is redundant and you could immediately return on line 6: if len(data) < 10: return 3. Move line 18 and 19 inside the for loop. Once you meet all three conditions you can return. More
First-krawlof 1
Hi, look at slices in python: # Lines 7-9 you could replace with: suma = sum(array[::2]) You could also write: if array: ... # instead of: if len(array) > 0: ... Look at [this](https://docs.python.org/3/library/stdtypes.html#truth-value-testing). More
First-a691662 1
Hi, this is a bit "selfobfuscating" :) 1. Line 4: True if ... else False is redundant. 2. Look at __set__ type in Python: # You could replace this: reduce(lambda x, y: x+y if x[-1] != y else x, sorted_lst) # with: set(sorted_list) 3. You don't need the alphabet val More
First-Screw
Hi, interesting approach :) I have two comments: 1. Look at negative indexes: # You could write: array[-1] # instead of: array[len(array) - 1] 2. Look at __in__ in Python. E. g.: # You could write: div5 = s[-1] in "50" # instead of: div5 = s[len More
First-ZeLib0ba
Hi, you could write _fibo_check_ like this: def fibo_check(numb): return numb in fibonacii(numb + 1) 1. The _if_ is redundant as numb in ch is already a boolean. 2. ch don't have to be a list as you could iterate over generator. 3. As you use _ch_ only once you could easily eliminate More
mass-Lumiera
Hi, I have two comments: 1. I hope, you've just forgot to remove the semicolon on line 3 :) 2. Look at _sum_ built-in function e. g.: # Lines 5-8 you could replace with: k = sum(array[::2]) # or the whole else block: return sum(array[::2]) * array[-1] # That's it :) More
Проверка паролей-Lumiera
Hi, I have two comments: 1. In fact you don't need the else branch on line 11: if len(data) < 10: return False # unindent the rest by one level. 2. On lines 4-9 you have three time the same pattern. You can extract the pattern and eliminate repetition. You have: if re. More
First-Screw 1
Hi, on thing: _lambda_ and _map_ are redundant: # You can join the phrases first which will result in one big string. s = ','.join(phrases) # And as s is a string you can directly call its replace method. return s.replace('right', 'left') # Finally: As you can chain More
First-jonmosco 1
Hi, line 5 is useless as on the next line you assign a different value to _new_. In fact you can eliminate _new_ completely i. e. replace _new_ with _phrases_ on line 7. More